Five weekends
The month of October in 2010 has five Fridays, five Saturdays, and five Sundays.
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
- Write a program to show all months that have this same characteristic of five full weekends from the year 1900 through 2100 (Gregorian calendar).
- Show the number of months with this property (there should be 201).
- Show at least the first and last five dates, in order.
Algorithm suggestions
- Count the number of Fridays, Saturdays, and Sundays in every month.
- Find all of the 31-day months that begin on Friday.
Extra credit
Count and/or show all of the years which do not have at least one five-weekend month (there should be 29).
- Related tasks
360 Assembly
For maximum compatibility, this program uses only the basic instruction set (S/360) and two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible.
* Five weekends 31/05/2016
FIVEWEEK CSECT
USING FIVEWEEK,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) prolog
ST R13,4(R15) "
ST R15,8(R13) "
LR R13,R15 "
LM R10,R11,=AL8(0) nko=0; nok=0
LH R6,Y1 y=y1
LOOPY CH R6,Y2 do y=y1 to y2
BH ELOOPY
MVI YF,X'00' yf=0
LA R7,1 im=1
LOOPIM C R7,=F'7' do im=1 to 7
BH ELOOPIM
LR R1,R7 im
SLA R1,1 *2 (H)
LH R2,ML-2(R1) ml(im)
ST R2,M m=ml(im)
MVC D,=F'1' d=1
L R4,M m
C R4,=F'2' if m<=2
BH MSUP2
L R8,M m
LA R8,12(R8) mw=m+12
LR R9,R6 y
BCTR R9,0 yw=y-1
B EMSUP2
MSUP2 L R8,M mw=m
LR R9,R6 yw=y
EMSUP2 LR R4,R9 ym
SRDA R4,32 .
D R4,=F'100' yw/100
ST R5,J j=yw/100
ST R4,K k=yw//100
LR R4,R8 mw
LA R4,1(R4) mw+1
MH R4,=H'26' (mw+1)*26
SRDA R4,32 .
D R4,=F'10' (mw+1)*26/10
LR R2,R5 "
A R2,D d
A R2,K d+k
L R3,K k
SRA R3,2 k/4
AR R2,R3 (mw+1)*26/10+k/4
L R3,J j
SRA R3,2 j/4
AR R2,R3 (mw+1)*26/10+k/4+j/4
LA R5,5 5
M R4,J 5*j
AR R2,R5 (mw+1)*26/10+k/4+j/4
SRDA R2,32 .
D R2,=F'7' (d+(mw+1)*26/10+k+k/4+j/4+5*j)/7
C R2,=F'6' if dow=friday
BNE NOFRIDAY
XDECO R6,XDEC y
MVC PG+0(4),XDEC+8 output y
LR R1,R7 im
MH R1,=H'3' *3
LA R14,MN-3(R1) @mn(im)
MVC PG+5(3),0(R14) output mn(im)
XPRNT PG,8 print buffer
LA R11,1(R11) nok=nok+1
MVI YF,X'01' yf=1
NOFRIDAY LA R7,1(R7) im=im+1
B LOOPIM
ELOOPIM L R4,YF yf
CLI YF,X'00' if yf=0
BNE EYFNE0
LA R10,1(R10) nko=nko+1
EYFNE0 LA R6,1(R6) y=y+1
B LOOPY
ELOOPY XDECO R11,XDEC nok
MVC PG+0(4),XDEC+8 output nok
MVC PG+4(12),=C' occurrences'
XPRNT PG,80 print buffer
XDECO R10,XDEC nko
MVC PG+0(4),XDEC+8 output nko
MVC PG+4(33),=C' years with no five weekend month'
XPRNT PG,80 print buffer
L R13,4(0,R13) epilog
LM R14,R12,12(R13) "
XR R15,R15 "
BR R14 exit
Y1 DC H'1900' year start
Y2 DC H'2100' year stop
ML DC H'1',H'3',H'5',H'7',H'8',H'10',H'12'
MN DC C'jan',C'mar',C'may',C'jul',C'aug',C'oct',C'dec'
YF DS X year flag
M DS F month
D DS F day
J DS F j=yw/100
K DS F j=mod(yw,100)
PG DC CL80'....-' buffer
XDEC DS CL12 temp for XDECO
YREGS
END FIVEWEEK
- Output:
1901-mar 1902-aug 1903-may 1904-jan 1904-jul 1905-dec 1907-mar 1908-may ... 2094-jan 2094-oct 2095-jul 2097-mar 2098-aug 2099-may 2100-jan 2100-oct 201 occurrences 29 years with no five weekend month
Action!
Day of the week is determined using Sakamoto method.
;https://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week#Sakamoto.27s_methods
BYTE FUNC DayOfWeek(INT y BYTE m,d) ;1<=m<=12, y>1752
BYTE ARRAY t=[0 3 2 5 0 3 5 1 4 6 2 4]
BYTE res
IF m<3 THEN
y==-1
FI
res=(y+y/4-y/100+y/400+t(m-1)+d) MOD 7
RETURN (res)
PROC Main()
BYTE ARRAY m31=[1 3 5 7 8 10 12]
INT ARRAY years(250)
BYTE ARRAY months(250)
INT y
BYTE i,m,mCount,yCount,found,c
mCount=0 yCount=0 c=0
FOR y=1900 TO 2100
DO
found=0
FOR i=0 TO 6
DO
m=m31(i)
IF DayOfWeek(y,m,1)=5 THEN
years(mCount)=y
months(mCount)=m
found=1
mCount==+1
FI
OD
IF found=0 THEN
yCount==+1
FI
OD
Print("5-weekend months in 1900-2100: ") PrintBE(mCount)
Print("non 5-weekend years in 1900-2100: ") PrintBE(yCount)
PutE()
FOR i=0 TO 4
DO
PrintI(years(i)) Put('/) PrintBE(months(i))
OD
PrintE("...")
FOR i=mCount-5 TO mCount-1
DO
PrintI(years(i)) Put('/) PrintBE(months(i))
OD
RETURN
- Output:
Screenshot from Atari 8-bit computer
5-weekend months in 1900-2100: 201 non 5-weekend years in 1900-2100: 29 1901/3 1902/8 1903/5 1904/1 1904/7 ... 2097/3 2098/8 2099/5 2100/1 2100/10
Ada
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Calendar.Formatting; use Ada.Calendar;
use Ada.Calendar.Formatting;
procedure Five_Weekends is
Months : Natural := 0;
begin
for Year in Year_Number range 1901..2100 loop
for Month in Month_Number range 1..12 loop
begin
if Day_Of_Week (Formatting.Time_Of (Year, Month, 31)) = Sunday then
Put_Line (Year_Number'Image (Year) & Month_Number'Image (Month));
Months := Months + 1;
end if;
exception
when Time_Error =>
null;
end;
end loop;
end loop;
Put_Line ("Number of months:" & Integer'Image (Months));
end Five_Weekends;
- Sample output:
1901 3 1902 8 1903 5 1904 1 1904 7 1905 12 1907 3 1908 5 1909 1 1909 10 1910 7 1911 12 1912 3 1913 8 1914 5 1915 1 1915 10 1916 12 1918 3 1919 8 1920 10 1921 7 1922 12 1924 8 1925 5 1926 1 1926 10 1927 7 1929 3 1930 8 1931 5 1932 1 1932 7 1933 12 1935 3 1936 5 1937 1 1937 10 1938 7 1939 12 1940 3 1941 8 1942 5 1943 1 1943 10 1944 12 1946 3 1947 8 1948 10 1949 7 1950 12 1952 8 1953 5 1954 1 1954 10 1955 7 1957 3 1958 8 1959 5 1960 1 1960 7 1961 12 1963 3 1964 5 1965 1 1965 10 1966 7 1967 12 1968 3 1969 8 1970 5 1971 1 1971 10 1972 12 1974 3 1975 8 1976 10 1977 7 1978 12 1980 8 1981 5 1982 1 1982 10 1983 7 1985 3 1986 8 1987 5 1988 1 1988 7 1989 12 1991 3 1992 5 1993 1 1993 10 1994 7 1995 12 1996 3 1997 8 1998 5 1999 1 1999 10 2000 12 2002 3 2003 8 2004 10 2005 7 2006 12 2008 8 2009 5 2010 1 2010 10 2011 7 2013 3 2014 8 2015 5 2016 1 2016 7 2017 12 2019 3 2020 5 2021 1 2021 10 2022 7 2023 12 2024 3 2025 8 2026 5 2027 1 2027 10 2028 12 2030 3 2031 8 2032 10 2033 7 2034 12 2036 8 2037 5 2038 1 2038 10 2039 7 2041 3 2042 8 2043 5 2044 1 2044 7 2045 12 2047 3 2048 5 2049 1 2049 10 2050 7 2051 12 2052 3 2053 8 2054 5 2055 1 2055 10 2056 12 2058 3 2059 8 2060 10 2061 7 2062 12 2064 8 2065 5 2066 1 2066 10 2067 7 2069 3 2070 8 2071 5 2072 1 2072 7 2073 12 2075 3 2076 5 2077 1 2077 10 2078 7 2079 12 2080 3 2081 8 2082 5 2083 1 2083 10 2084 12 2086 3 2087 8 2088 10 2089 7 2090 12 2092 8 2093 5 2094 1 2094 10 2095 7 2097 3 2098 8 2099 5 2100 1 2100 10 Number of months: 201
ALGOL 68
five_weekends: BEGIN
INT m, year, nfives := 0, not5 := 0;
BOOL no5weekend;
MODE MONTH = STRUCT(
INT n,
[3]CHAR name
) # MODE MONTH #;
[]MONTH month = (
MONTH(13, "Jan"),
MONTH(3, "Mar"),
MONTH(5, "May"),
MONTH(7, "Jul"),
MONTH(8, "Aug"),
MONTH(10, "Oct"),
MONTH(12, "Dec")
);
FOR year FROM 1900 TO 2100 DO
IF year = 1905 THEN printf($"..."l$) FI;
no5weekend := TRUE;
FOR m TO UPB month DO
IF n OF month(m) = 13 THEN
IF day_of_week(1, n OF month(m), year-1) = 6 THEN
IF year<1905 OR year > 2096 THEN printf(($g, 5zl$, name OF month(m), year)) FI;
nfives +:= 1;
no5weekend := FALSE
FI
ELSE
IF day_of_week(1, n OF month(m), year) = 6 THEN
IF year<1905 OR year > 2096 THEN printf(($g, 5zl$, name OF month(m), year)) FI;
nfives +:= 1;
no5weekend := FALSE
FI
FI
OD;
IF no5weekend THEN not5 +:= 1 FI
OD;
printf(($g, g(0)l$, "Number of months with five weekends between 1900 and 2100 = ", nfives));
printf(($g, g(0)l$, "Number of years between 1900 and 2100 with no five weekend months = ", not5));
# contains #
PROC day_of_week = (INT d, m, y)INT: BEGIN
INT j, k;
j := y OVER 100;
k := y MOD 100;
(d + (m+1)*26 OVER 10 + k + k OVER 4 + j OVER 4 + 5*j) MOD 7
END # function day_of_week #;
SKIP
END # program five_weekends #
- Output:
Mar 1901 Aug 1902 May 1903 Jan 1904 Jul 1904 ... Mar 2097 Aug 2098 May 2099 Jan 2100 Oct 2100 Number of months with five weekends between 1900 and 2100 = 201 Number of years between 1900 and 2100 with no five weekend months = 29
Amazing Hopper
VERSION 1
#include <jambo.h>
#context Obtiene suma de días de weekend
Args ( Domingo, Viernes, Sábado )
Not zero( Val( [3:end,Domingo] Of 'calendario' )), Get summatory
Not zero( Val( [3:end,Viernes] Of 'calendario' )), Get summatory, Add it
Not zero( Val( [3:end,Sábado ] Of 'calendario' )), Get summatory, Add it
Return\\
#define __PRNNL__ {"\n"}print
#synon __PRNNL__ *Print it
#defn Paralosdías(*) #GENCODE $$$*$$$ #ATCMLIST; #ENDGEN;
#enum 1,DOMINGO,6,VIERNES,7,SABADO
Main
Set stack 15
Init zero (calendario, candidato, total, columna)
/* Configura meses */
Meses={}, mes largo = {}
Let list ( Meses := "Enero ","Febrero ","Marzo ","Abril ","Mayo ",\
"Junio ","Julio ","Agosto ","Septiembre","Octubre ",\
"Noviembre ","Diciembre " )
Let list ( mes largo := 1, 3, 5, 7, 8, 10, 12 )
/* Busca los meses con weekend larguísimo */
Loop for (año = 1900, #( año <= 2100), ++año)
Loop for( i=1, #(i<=7), ++i)
Let ( calendario := Calendar( [i] Of 'mes largo' ---Backup to 'candidato'---,año,1) )
Para los días 'DOMINGO, VIERNES, SABADO' Obtiene suma de días de weekend
When ( Is equal to '15' ){
++total, Print (año," : ", [candidato] Of 'Meses'," | ")
When ( columna++ Is equal to '3' ) { Prnl, columna=0 }
}
Back
Back
Set ( Utf8("\nTotal de años con weekend de 5 días = "), total ), and Print it
End
Subrutines
- Output:
1901 : Marzo | 1902 : Agosto | 1903 : Mayo | 1904 : Enero | 1904 : Julio | 1905 : Diciembre | 1907 : Marzo | 1908 : Mayo | 1909 : Enero | 1909 : Octubre | 1910 : Julio | 1911 : Diciembre | 1912 : Marzo | 1913 : Agosto | 1914 : Mayo | 1915 : Enero | 1915 : Octubre | 1916 : Diciembre | 1918 : Marzo | 1919 : Agosto | 1920 : Octubre | 1921 : Julio | 1922 : Diciembre | 1924 : Agosto | 1925 : Mayo | 1926 : Enero | 1926 : Octubre | 1927 : Julio | 1929 : Marzo | 1930 : Agosto | 1931 : Mayo | 1932 : Enero | 1932 : Julio | 1933 : Diciembre | 1935 : Marzo | 1936 : Mayo | 1937 : Enero | 1937 : Octubre | 1938 : Julio | 1939 : Diciembre | 1940 : Marzo | 1941 : Agosto | 1942 : Mayo | 1943 : Enero | 1943 : Octubre | 1944 : Diciembre | 1946 : Marzo | 1947 : Agosto | 1948 : Octubre | 1949 : Julio | 1950 : Diciembre | 1952 : Agosto | 1953 : Mayo | 1954 : Enero | 1954 : Octubre | 1955 : Julio | 1957 : Marzo | 1958 : Agosto | 1959 : Mayo | 1960 : Enero | 1960 : Julio | 1961 : Diciembre | 1963 : Marzo | 1964 : Mayo | 1965 : Enero | 1965 : Octubre | 1966 : Julio | 1967 : Diciembre | 1968 : Marzo | 1969 : Agosto | 1970 : Mayo | 1971 : Enero | 1971 : Octubre | 1972 : Diciembre | 1974 : Marzo | 1975 : Agosto | 1976 : Octubre | 1977 : Julio | 1978 : Diciembre | 1980 : Agosto | 1981 : Mayo | 1982 : Enero | 1982 : Octubre | 1983 : Julio | 1985 : Marzo | 1986 : Agosto | 1987 : Mayo | 1988 : Enero | 1988 : Julio | 1989 : Diciembre | 1991 : Marzo | 1992 : Mayo | 1993 : Enero | 1993 : Octubre | 1994 : Julio | 1995 : Diciembre | 1996 : Marzo | 1997 : Agosto | 1998 : Mayo | 1999 : Enero | 1999 : Octubre | 2000 : Diciembre | 2002 : Marzo | 2003 : Agosto | 2004 : Octubre | 2005 : Julio | 2006 : Diciembre | 2008 : Agosto | 2009 : Mayo | 2010 : Enero | 2010 : Octubre | 2011 : Julio | 2013 : Marzo | 2014 : Agosto | 2015 : Mayo | 2016 : Enero | 2016 : Julio | 2017 : Diciembre | 2019 : Marzo | 2020 : Mayo | 2021 : Enero | 2021 : Octubre | 2022 : Julio | 2023 : Diciembre | 2024 : Marzo | 2025 : Agosto | 2026 : Mayo | 2027 : Enero | 2027 : Octubre | 2028 : Diciembre | 2030 : Marzo | 2031 : Agosto | 2032 : Octubre | 2033 : Julio | 2034 : Diciembre | 2036 : Agosto | 2037 : Mayo | 2038 : Enero | 2038 : Octubre | 2039 : Julio | 2041 : Marzo | 2042 : Agosto | 2043 : Mayo | 2044 : Enero | 2044 : Julio | 2045 : Diciembre | 2047 : Marzo | 2048 : Mayo | 2049 : Enero | 2049 : Octubre | 2050 : Julio | 2051 : Diciembre | 2052 : Marzo | 2053 : Agosto | 2054 : Mayo | 2055 : Enero | 2055 : Octubre | 2056 : Diciembre | 2058 : Marzo | 2059 : Agosto | 2060 : Octubre | 2061 : Julio | 2062 : Diciembre | 2064 : Agosto | 2065 : Mayo | 2066 : Enero | 2066 : Octubre | 2067 : Julio | 2069 : Marzo | 2070 : Agosto | 2071 : Mayo | 2072 : Enero | 2072 : Julio | 2073 : Diciembre | 2075 : Marzo | 2076 : Mayo | 2077 : Enero | 2077 : Octubre | 2078 : Julio | 2079 : Diciembre | 2080 : Marzo | 2081 : Agosto | 2082 : Mayo | 2083 : Enero | 2083 : Octubre | 2084 : Diciembre | 2086 : Marzo | 2087 : Agosto | 2088 : Octubre | 2089 : Julio | 2090 : Diciembre | 2092 : Agosto | 2093 : Mayo | 2094 : Enero | 2094 : Octubre | 2095 : Julio | 2097 : Marzo | 2098 : Agosto | 2099 : Mayo | 2100 : Enero | 2100 : Octubre | Total de años con weekend de 5 días = 201
VERSION 2
Esta versión genera el mismo resultado, sin hacer uso de la macro "calendar", con la diferencia que ahora le añadí el conteo de los días que no tienen el famoso "five weekend":
#include <jambo.h>
#define __PRNNL__ {"\n"}print
#synon __PRNNL__ *Print it
#synon Set *Set
Main
Set stack 15
Init zero (candidato, total, sin weekend largo, sw, columna, fecha)
/* Configura meses */
mes largo = {}
Let list ( mes largo := 1, 3, 5, 7, 8, 10, 12 )
/* Busca los meses con weekend larguísimo */
Loop for (año = 1900, #( año <= 2100), ++año)
Loop for( i=1, #(i<=7), ++i)
Let ( candidato := [i] Of 'mes largo' )
Let ( fecha := Multicat ("1/",Str(candidato),"/",Str(año)) )
When ( Strday 'fecha' Is equal to '"Viernes"', \
And ( Days of month 'fecha' Compared to '31', Are equals? )) {
++total, sw=1
Print (año," : ", Just left (13, Month name 'candidato')," | ")
When ( columna++ Is equal to '3' ) { Prnl, columna=0 }
}
Back
When ( Not( sw ) ) { ++ sin weekend largo }, sw=0
Back
now Set ( Utf8("\nTotal de años con weekend de 5 días = "), total )
and Set ( Utf8("\nAños sin weekend de 5 días: "), sin weekend largo) then Print it
End
- Output:
1901 : Marzo | 1902 : Agosto | 1903 : Mayo | 1904 : Enero | 1904 : Julio | 1905 : Diciembre | 1907 : Marzo | 1908 : Mayo | ... 2093 : Mayo | 2094 : Enero | 2094 : Octubre | 2095 : Julio | 2097 : Marzo | 2098 : Agosto | 2099 : Mayo | 2100 : Enero | 2100 : Octubre | Total de años con weekend de 5 días = 201 Años sin weekend de 5 días: 29
AppleScript
Imperative
set fiveWeekendMonths to {}
set noFiveWeekendYears to {}
set someDate to current date
set day of someDate to 1
repeat with someYear from 1900 to 2100
set year of someDate to someYear
set foundOne to false
repeat with someMonth in {January, March, May, July, ¬
August, October, December}
set month of someDate to someMonth
if weekday of someDate is Friday then
set foundOne to true
set end of fiveWeekendMonths to ¬
(someYear as text) & "-" & (someMonth as text)
end
end repeat
if not foundOne then
set end of noFiveWeekendYears to someYear
end
end repeat
set text item delimiters to ", "
set monthList to ¬
(items 1 thru 5 of fiveWeekendMonths as text) & ", ..." & linefeed & ¬
" ..., " & (items -5 thru end of fiveWeekendMonths as text)
set monthCount to count fiveWeekendMonths
set yearCount to count noFiveWeekendYears
set resultText to ¬
"Months with five weekends (" & monthCount & "): " & linefeed & ¬
" " & monthList & linefeed & linefeed & ¬
"Years with no such months (" & yearCount & "): "
set y to 1
repeat while y < yearCount
set final to y+11
if final > yearCount then
set final to yearCount
end
set resultText to ¬
resultText & linefeed & ¬
" " & (items y through final of noFiveWeekendYears as text)
set y to y + 12
end
resultText
- Output:
(can always add "display dialog resultText" for GUI output)
Months with five weekends (201): 1901-March, 1902-August, 1903-May, 1904-January, 1904-July, ... ..., 2097-March, 2098-August, 2099-May, 2100-January, 2100-October Years with no such months (29): 1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063 2068, 2074, 2085, 2091, 2096
Functional
-- TEST -----------------------------------------------------------------------
on run
fiveWeekends(1900, 2100)
end run
-- FIVE WEEKENDS --------------------------------------------------------------
-- fiveWeekends :: Int -> Int -> Record
on fiveWeekends(fromYear, toYear)
set lstYears to enumFromTo(fromYear, toYear)
-- yearMonthString :: (Int, Int) -> String
script yearMonthString
on |λ|(lstYearMonth)
((item 1 of lstYearMonth) as string) & " " & ¬
item (item 2 of lstYearMonth) of ¬
{"January", "", "March", "", "May", "", ¬
"July", "August", "", "October", "", "December"}
end |λ|
end script
-- addLongMonthsOfYear :: [(Int, Int)] -> [(Int, Int)]
script addLongMonthsOfYear
on |λ|(lstYearMonth, intYear)
-- yearMonth :: Int -> (Int, Int)
script yearMonth
on |λ|(intMonth)
{intYear, intMonth}
end |λ|
end script
lstYearMonth & ¬
map(yearMonth, my longMonthsStartingFriday(intYear))
end |λ|
end script
-- leanYear :: Int -> Bool
script leanYear
on |λ|(intYear)
0 = length of longMonthsStartingFriday(intYear)
end |λ|
end script
set lstFullMonths to map(yearMonthString, ¬
foldl(addLongMonthsOfYear, {}, lstYears))
set lstLeanYears to filter(leanYear, lstYears)
{{|number|:length of lstFullMonths}, ¬
{firstFive:(items 1 thru 5 of lstFullMonths)}, ¬
{lastFive:(items -5 thru -1 of lstFullMonths)}, ¬
{leanYearCount:length of lstLeanYears}, ¬
{leanYears:lstLeanYears}}
end fiveWeekends
-- longMonthsStartingFriday :: Int -> [Int]
on longMonthsStartingFriday(intYear)
-- startIsFriday :: Int -> Bool
script startIsFriday
on |λ|(iMonth)
weekday of calendarDate(intYear, iMonth, 1) is Friday
end |λ|
end script
filter(startIsFriday, [1, 3, 5, 7, 8, 10, 12])
end longMonthsStartingFriday
-- calendarDate :: Int -> Int -> Int -> Date
on calendarDate(intYear, intMonth, intDay)
tell (current date)
set {its year, its month, its day, its time} to ¬
{intYear, intMonth, intDay, 0}
return it
end tell
end calendarDate
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- enumFromTo :: Enum a => a -> a -> [a]
on enumFromTo(m, n)
set {intM, intN} to {fromEnum(m), fromEnum(n)}
if intM > intN then
set d to -1
else
set d to 1
end if
set lst to {}
if class of m is text then
repeat with i from intM to intN by d
set end of lst to chr(i)
end repeat
else
repeat with i from intM to intN by d
set end of lst to i
end repeat
end if
return lst
end enumFromTo
-- fromEnum :: Enum a => a -> Int
on fromEnum(x)
set c to class of x
if c is boolean then
if x then
1
else
0
end if
else if c is text then
if x ≠ "" then
id of x
else
missing value
end if
else
x as integer
end if
end fromEnum
-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
- Output:
{{|number|:201},
{firstFive:{"1901 March", "1902 August", "1903 May", "1904 January", "1904 July"}},
{lastFive:{"2097 March", "2098 August", "2099 May", "2100 January", "2100 October"}},
{leanYearCount:29},
{leanYears:{1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979,
1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074,
2085, 2091, 2096}}}
Using integer math
An alternative to the date object manipulation methods above.
on monthsWithFiveWeekends(startYear, endYear)
set Dec1 to (current date)
tell Dec1 to set {its day, its month, its year} to {1, December, startYear - 1}
set daysFromBaseFriday to (Dec1's weekday as integer) - Friday
set longMonths to {"January", "March", "May", "July", "August", "October", "December"}
set daysBetween to {31, 59, 61, 61, 31, 61, 61} -- Days since starts of preceding long months.
set hits to {}
set hitlessYears to {}
repeat with y from startYear to endYear
set noHIts to true
-- Find long months that begin on Fridays.
repeat with i from 1 to 7
set daysFromBaseFriday to daysFromBaseFriday + (daysBetween's item i)
if ((i = 2) and (y mod 4 = 0) and ((y mod 100 > 0) or (y mod 400 = 0))) then ¬
set daysFromBaseFriday to daysFromBaseFriday + 1 -- Leap year.
if (daysFromBaseFriday mod 7 = 0) then
set end of hits to (longMonths's item i) & (space & y)
set noHIts to false
end if
end repeat
if (noHIts) then set end of hitlessYears to y
end repeat
return {hits:hits, hitlessYears:hitlessYears}
end monthsWithFiveWeekends
on join(lst, delim)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to delim
set txt to lst as text
set AppleScript's text item delimiters to astid
return txt
end join
on task()
set {startYear, endYear} to {1900, 2100}
set theResults to monthsWithFiveWeekends(startYear, endYear)
set output to {((count theResults's hits) as text) & " of the months from " & startYear & ¬
" to " & endYear & " have five weekends,", ¬
"the first and last five of these months being:"}
set end of output to join(theResults's hits's items 1 thru 5, ", ") & " …"
set end of output to "… " & join(theResults's hits's items -5 thru -1, ", ")
set hitlessCount to (count theResults's hitlessYears)
set end of output to linefeed & hitlessCount & " of the years have no such months:"
set cut to (hitlessCount + 1) div 2
set end of output to join(theResults's hitlessYears's items 1 thru cut, ", ")
set end of output to join(theResults's hitlessYears's items (cut + 1) thru -1, ", ")
return join(output, linefeed)
end task
task()
- Output:
"201 of the months from 1900 to 2100 have five weekends,
the first and last five of these months being:
March 1901, August 1902, May 1903, January 1904, July 1904 …
… March 2097, August 2098, May 2099, January 2100, October 2100
29 of the years have no such months:
1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001
2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096"
Arturo
longMonths: [1 3 5 7 8 10 12]
dates: []
yearsWithout: 0
loop 1900..2100 'year [
found?: false
loop longMonths 'month [
dt: to :date .format:"d-M-YYYY" ~"1-|month|-|year|"
if friday? dt [
'dates ++ @[@[dt\Month year]]
found?: true
]
]
if not? found? ->
inc 'yearsWithout
]
print.lines map first.n:5 dates 'd -> ~"|to :string d\0|, |to :string d\1|"
print "..."
print.lines map last.n:5 dates 'd -> ~"|to :string d\0|, |to :string d\1|"
print ""
print ["Found" size dates "months in total."]
print ["There are" yearsWithout "years without any month that has 5 full weekends."]
- Output:
March, 1901 August, 1902 May, 1903 January, 1904 July, 1904 ... March, 2097 August, 2098 May, 2099 January, 2100 October, 2100 Found 201 months in total. There are 29 years without any month that has 5 full weekends.
AutoHotkey
Year := 1900
End_Year := 2100
31_Day_Months = 01,03,05,07,08,10,12
While Year <= End_Year
{
Loop, Parse, 31_Day_Months, CSV
{
FormatTime, Day, %Year%%A_LoopField%01, dddd
IfEqual, Day, Friday
{
All_Months_With_5_Weekends .= A_LoopField . "/" . Year ", "
5_Weekend_Count++
Year_Has_5_Weekend_Month := 1
}
}
IfEqual, Year_Has_5_Weekend_Month, 0
{
All_Years_Without_5_Weekend .= Year ", "
No_5_Weekend_Count ++
}
Year ++
Year_Has_5_Weekend_Month := 0
}
; Trim the spaces and comma off the last item.
StringTrimRight, All_Months_With_5_Weekends, All_Months_With_5_Weekends, 5
StringTrimRight, All_Years_Without_5_Weekend, All_Years_Without_5_Weekend, 4
MsgBox,
(
Months with 5 day weekends between 1900 and 2100 : %5_Weekend_Count%
%All_Months_With_5_Weekends%
)
MsgBox,
(
Years with no 5 day weekends between 1900 and 2100 : %No_5_Weekend_Count%
%All_Years_Without_5_Weekend%
)
AutoIt
Call the Funktion with $ret > 1 And $ret < 3 to determine Returned Array 1 - All found Weekend Dates 2 - Years Without 5 Weekend Months 3 - Year and Month with 5 Weekends
#include <Date.au3>
#include <Array.au3>
$array = Five_weekends(1)
_ArrayDisplay($array)
$array = Five_weekends(2)
_ArrayDisplay($array)
$array = Five_weekends(3)
_ArrayDisplay($array)
Func Five_weekends($ret = 1)
If $ret < 1 Or $ret > 3 Then Return SetError(1, 0, 0)
Local $avDateArray[1]
Local $avYearArray[1]
Local $avMonthArray[1]
For $iYear = 1900 To 2100
Local $checkyear = False
For $iMonth = 1 To 12
If _DateDaysInMonth($iYear, $iMonth) <> 31 Then ContinueLoop ; Month has less then 31 Days
If _DateToDayOfWeek($iYear, $iMonth, "01") <> 6 Then ContinueLoop ;First Day is not a Friday
_ArrayAdd($avMonthArray, $iYear & "-" & _DateToMonth($iMonth))
$checkyear = True
For $s = 1 To 31
Local $Date = _DateToDayOfWeek($iYear, $iMonth, $s)
If $Date = 6 Or $Date = 7 Or $Date = 1 Then ; if Date is Friday, Saturday or Sunday
_ArrayAdd($avDateArray, $iYear & "\" & StringFormat("%02d", $iMonth) & "\" & StringFormat("%02d", $s))
EndIf
Next
Next
If Not $checkyear Then _ArrayAdd($avYearArray, $iYear)
Next
$avDateArray[0] = UBound($avDateArray) - 1
$avYearArray[0] = UBound($avYearArray) - 1
$avMonthArray[0] = UBound($avMonthArray) - 1
If $ret = 1 Then
Return $avDateArray
ElseIf $ret = 2 Then
Return $avYearArray
ElseIf $ret = 3 Then
Return $avMonthArray
EndIf
EndFunc ;==>Five_weekends
AWK
# usage: awk -f 5weekends.awk cal.txt
# Filter a file of month-calendars, such as
# ...
## January 1901
## Mo Tu We Th Fr Sa Su
## 1 2 3 4 5 6
## 7 8 9 10 11 12 13
## 14 15 16 17 18 19 20
## 21 22 23 24 25 26 27
## 28 29 30 31
# ...
## March 1901
## Mo Tu We Th Fr Sa Su
## 1 2 3
## 4 5 6 7 8 9 10
## 11 12 13 14 15 16 17
## 18 19 20 21 22 23 24
## 25 26 27 28 29 30 31
# ...
# This file is generated by a script for the unix-shell,
# see http://rosettacode.org/wiki/Five_weekends#UNIX_Shell
BEGIN { print("# Month with 5 weekends:")
badYears = numW5 = 0;
lastW5 = -1
}
0+$2>33 { if( $2>currYear ) { # calendar-header: month, year
if( lastW5==numW5 ) {
badYears++; sep="\n"
if( badYears % 10 ) { sep=" " }
bY=bY currYear sep; # collect years in string
##print badYears,":", currYear
}
lastW5=numW5
}
WE=0; currYear=$2; currMY = $1 " " $2;
##print currMY;
next
}
/^Mo/ { next } # skip lines with weekday-names
{ $0 = substr($0,13) } # cut inputline, leave Fr,Sa,Su only
NF>2 { WE++; # 3 fields left => complete weekend found
if( WE>4 ) {
numW5++; printf("%4d : %s\n", numW5, currMY)
}
}
END { print("# Found", numW5, "month with 5 weekends.")
print("# Found", badYears, "years with no month having 5 weekends:")
print(bY)
}
See also: unix-shell and Calendar.
- Output:
# Month with 5 weekends: 1 : March 1901 2 : August 1902 3 : May 1903 4 : January 1904 5 : July 1904 6 : December 1905 ... 196 : July 2095 197 : March 2097 198 : August 2098 199 : May 2099 200 : January 2100 201 : October 2100 # Found 201 month with 5 weekends. # Found 29 years with no month having 5 weekends: 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
BBC BASIC
INSTALL @lib$+"DATELIB"
DIM Month$(12)
Month$() = "","January","February","March","April","May","June", \
\ "July","August","September","October","November","December"
num% = 0
FOR year% = 1900 TO 2100
PRINT ; year% ": " ;
oldnum% = num%
FOR month% = 1 TO 12
IF FN_dim(month%,year%) = 31 IF FN_dow(FN_mjd(1,month%,year%)) = 5 THEN
num% += 1
PRINT Month$(month%), ;
ENDIF
NEXT
IF num% = oldnum% PRINT "(none)" ELSE PRINT
NEXT year%
PRINT "Total = " ; num%
- Output:
1900: (none) 1901: March 1902: August 1903: May 1904: January July 1905: December 1906: (none) ... 2093: May 2094: January October 2095: July 2096: (none) 2097: March 2098: August 2099: May 2100: January October Total = 201
C
#include <stdio.h>
#include <time.h>
static const char *months[] = {"January", "February", "March", "April", "May",
"June", "July", "August", "September", "October", "November", "December"};
static int long_months[] = {0, 2, 4, 6, 7, 9, 11};
int main() {
int n = 0, y, i, m;
struct tm t = {0};
printf("Months with five weekends:\n");
for (y = 1900; y <= 2100; y++) {
for (i = 0; i < 7; i++) {
m = long_months[i];
t.tm_year = y-1900;
t.tm_mon = m;
t.tm_mday = 1;
if (mktime(&t) == -1) { /* date not supported */
printf("Error: %d %s\n", y, months[m]);
continue;
}
if (t.tm_wday == 5) { /* Friday */
printf(" %d %s\n", y, months[m]);
n++;
}
}
}
printf("%d total\n", n);
return 0;
}
- Output:
(note that the C library may not be able to handle all dates; the output may vary across systems)
Error: 1900 January Error: 1900 March Error: 1900 May Error: 1900 July Error: 1900 August Error: 1900 October Error: 1900 December Error: 1901 January Error: 1901 March Error: 1901 May Error: 1901 July Error: 1901 August Error: 1901 October Error: 1901 December 1902 August 1903 May 1904 January 1904 July 1905 December ... 2097 March 2098 August 2099 May 2100 January 2100 October 200 total
Not your typical method. Requires ncal
.
#include <stdio.h>
#include <string.h>
int check_month(int y, int m)
{
char buf[1024], *ptr;
int bytes, *a = &m;
sprintf(buf, "ncal -m %d -M %d", m, y);
FILE *fp = popen(buf, "r");
if (!fp) return -1;
bytes = fread(buf, 1, 1024, fp);
fclose(fp);
buf[bytes] = 0;
#define check_day(x) \
ptr = strstr(buf, x);\
if (5 != sscanf(ptr, x" %d %d %d %d %d", a, a, a, a, a)) return 0
check_day("Fr");
check_day("Sa");
check_day("Su");
return 1;
}
int main()
{
int y, m, cnt = 0;
for (y = 1900; y <= 2100; y++) {
for (m = 1; m <= 12; m++) {
if (check_month(y, m) <= 0) continue;
printf("%d-%02d ", y, m);
if (++cnt % 16 == 0) printf("\n");
}
}
printf("\nTotal: %d\n", cnt);
return 0;
}
- Output:
1901-03 1902-08 1903-05 1904-01 1904-07 1905-12 1907-03 1908-05 1909-01 1909-10 1910-07 1911-12 1912-03 1913-08 1914-05 1915-01 . . . 2093-05 2094-01 2094-10 2095-07 2097-03 2098-08 2099-05 2100-01 2100-10 Total: 201
C#
With iteration
using System;
namespace _5_Weekends
{
class Program
{
const int FIRST_YEAR = 1900;
const int LAST_YEAR = 2100;
static int[] _31_MONTHS = { 1, 3, 5, 7, 8, 10, 12 };
static void Main(string[] args)
{
int totalNum = 0;
int totalNo5Weekends = 0;
for (int year = FIRST_YEAR; year <= LAST_YEAR; year++)
{
bool has5Weekends = false;
foreach (int month in _31_MONTHS)
{
DateTime firstDay = new DateTime(year, month, 1);
if (firstDay.DayOfWeek == DayOfWeek.Friday)
{
totalNum++;
has5Weekends = true;
Console.WriteLine(firstDay.ToString("yyyy - MMMM"));
}
}
if (!has5Weekends) totalNo5Weekends++;
}
Console.WriteLine("Total 5-weekend months between {0} and {1}: {2}", FIRST_YEAR, LAST_YEAR, totalNum);
Console.WriteLine("Total number of years with no 5-weekend months {0}", totalNo5Weekends);
}
}
}
- Output:
1901 - March 1902 - August 1903 - May 1904 - January 1904 - July 1905 - December ... 2095 - July 2097 - March 2098 - August 2099 - May 2100 - January 2100 - October Total 5-weekend months between 1900 and 2100: 201 Total number of years with no 5-weekend months 29
With LINQ
using System;
using System.Linq;
using System.Collections.Generic;
public class Program
{
public static void Main()
{
const int startYear = 1900, endYear = 2100;
var query = (
from year in startYear.To(endYear)
from month in 1.To(12)
where DateTime.DaysInMonth(year, month) == 31
select new DateTime(year, month, 1) into date
where date.DayOfWeek == DayOfWeek.Friday
select date)
.ToList();
Console.WriteLine("Count: " + query.Count);
Console.WriteLine();
Console.WriteLine("First and last 5:");
for (int i = 0; i < 5; i++)
Console.WriteLine(query[i].ToString("MMMM yyyy"));
Console.WriteLine("...");
for (int i = query.Count - 5; i < query.Count; i++)
Console.WriteLine(query[i].ToString("MMMM yyyy"));
Console.WriteLine();
Console.WriteLine("Years without 5 weekends:");
Console.WriteLine(string.Join(" ", startYear.To(endYear).Except(query.Select(dt => dt.Year))));
}
}
public static class IntExtensions
{
public static IEnumerable<int> To(this int start, int end) => Enumerable.Range(start, end - start + 1);
}
- Output:
Count: 201 First and last 5: March 1901 August 1902 May 1903 January 1904 July 1904 ... March 2097 August 2098 May 2099 January 2100 October 2100 Years without 5 weekends: 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
C++
#include <vector>
#include <boost/date_time/gregorian/gregorian.hpp>
#include <algorithm>
#include <iostream>
#include <iterator>
using namespace boost::gregorian ;
void print( const date &d ) {
std::cout << d.year( ) << "-" << d.month( ) << "\n" ;
}
int main( ) {
greg_month longmonths[ ] = {Jan, Mar , May , Jul ,
Aug , Oct , Dec } ;
int monthssize = sizeof ( longmonths ) / sizeof (greg_month ) ;
typedef std::vector<date> DateVector ;
DateVector weekendmonster ;
std::vector<unsigned short> years_without_5we_months ;
for ( unsigned short i = 1900 ; i < 2101 ; i++ ) {
bool months_found = false ; //does a given year have 5 weekend months ?
for ( int j = 0 ; j < monthssize ; j++ ) {
date d ( i , longmonths[ j ] , 1 ) ;
if ( d.day_of_week( ) == Friday ) { //for the month to have 5 weekends
weekendmonster.push_back( d ) ;
if ( months_found == false )
months_found = true ;
}
}
if ( months_found == false ) {
years_without_5we_months.push_back( i ) ;
}
}
std::cout << "Between 1900 and 2100 , there are " << weekendmonster.size( )
<< " months with 5 complete weekends!\n" ;
std::cout << "Months with 5 complete weekends are:\n" ;
std::for_each( weekendmonster.begin( ) , weekendmonster.end( ) , print ) ;
std::cout << years_without_5we_months.size( ) << " years had no months with 5 complete weekends!\n" ;
std::cout << "These are:\n" ;
std::copy( years_without_5we_months.begin( ) , years_without_5we_months.end( ) ,
std::ostream_iterator<unsigned short>( std::cout , "\n" ) ) ;
std::cout << std::endl ;
return 0 ;
}
- Sample output:
Between 1900 and 2100 , there are 201 months with 5 complete weekends! Months with 5 complete weekends are: 1901-Mar 1902-Aug 1903-May 1904-Jan 1904-Jul 1905-Dec 1907-Mar 1908-May 1909-Jan 1909-Oct 1910-Jul 1911-Dec 1912-Mar 1913-Aug 1914-May 1915-Jan 1915-Oct 1916-Dec 1918-Mar 1919-Aug 1920-Oct 1921-Jul 1922-Dec 1924-Aug 1925-May 1926-Jan 1926-Oct 1927-Jul 1929-Mar 1930-Aug 1931-May 1932-Jan 1932-Jul 1933-Dec 1935-Mar 1936-May 1937-Jan 1937-Oct 1938-Jul 1939-Dec 1940-Mar 1941-Aug 1942-May 1943-Jan 1943-Oct 1944-Dec 1946-Mar 1947-Aug 1948-Oct 1949-Jul 1950-Dec 1952-Aug 1953-May 1954-Jan 1954-Oct 1955-Jul 1957-Mar 1958-Aug 1959-May 1960-Jan 1960-Jul 1961-Dec 1963-Mar 1964-May 1965-Jan 1965-Oct 1966-Jul 1967-Dec 1968-Mar 1969-Aug 1970-May 1971-Jan 1971-Oct 1972-Dec 1974-Mar 1975-Aug 1976-Oct 1977-Jul 1978-Dec 1980-Aug 1981-May 1982-Jan 1982-Oct 1983-Jul 1985-Mar 1986-Aug 1987-May 1988-Jan 1988-Jul 1989-Dec 1991-Mar 1992-May 1993-Jan 1993-Oct 1994-Jul 1995-Dec 1996-Mar 1997-Aug 1998-May 1999-Jan 1999-Oct 2000-Dec 2002-Mar 2003-Aug 2004-Oct 2005-Jul 2006-Dec 2008-Aug 2009-May 2010-Jan 2010-Oct 2011-Jul 2013-Mar 2014-Aug 2015-May 2016-Jan 2016-Jul 2017-Dec 2019-Mar 2020-May 2021-Jan 2021-Oct 2022-Jul 2023-Dec 2024-Mar 2025-Aug 2026-May 2027-Jan 2027-Oct 2028-Dec 2030-Mar 2031-Aug 2032-Oct 2033-Jul 2034-Dec 2036-Aug 2037-May 2038-Jan 2038-Oct 2039-Jul 2041-Mar 2042-Aug 2043-May 2044-Jan 2044-Jul 2045-Dec 2047-Mar 2048-May 2049-Jan 2049-Oct 2050-Jul 2051-Dec 2052-Mar 2053-Aug 2054-May 2055-Jan 2055-Oct 2056-Dec 2058-Mar 2059-Aug 2060-Oct 2061-Jul 2062-Dec 2064-Aug 2065-May 2066-Jan 2066-Oct 2067-Jul 2069-Mar 2070-Aug 2071-May 2072-Jan 2072-Jul 2073-Dec 2075-Mar 2076-May 2077-Jan 2077-Oct 2078-Jul 2079-Dec 2080-Mar 2081-Aug 2082-May 2083-Jan 2083-Oct 2084-Dec 2086-Mar 2087-Aug 2088-Oct 2089-Jul 2090-Dec 2092-Aug 2093-May 2094-Jan 2094-Oct 2095-Jul 2097-Mar 2098-Aug 2099-May 2100-Jan 2100-Oct 29 years had no months with 5 complete weekends! These are: 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
Using C++20
#include <chrono>
#include <iostream>
#include <vector>
int main() {
const std::vector<std::chrono::month> long_months = { std::chrono::January, std::chrono::March,
std::chrono::May, std::chrono::July, std::chrono::August, std::chrono::October, std::chrono::December };
int month_count = 0;
int blank_years = 0;
for ( int y = 1900; y <= 2100; ++y ) {
bool blank_year = true;
for ( std::chrono::month m : long_months ) {
std::chrono::year_month_day first_of_month{std::chrono::year{y}, m, std::chrono::day{1}};
if ( std::chrono::weekday{first_of_month} == std::chrono::Friday ) {
std::cout << first_of_month.year() << " " << first_of_month.month() << std::endl;
month_count++;
blank_year = false;
}
}
if ( blank_year ) {
blank_years++;
}
}
std::cout << "Found " << month_count << " months with five Fridays, Saturdays and Sundays." << std::endl;
std::cout << "There were " << blank_years << " years with no such months." << std::endl;
}
- Output:
1901 Mar 1902 Aug 1903 May 1904 Jan 1904 Jul // elided... // 2097 Mar 2098 Aug 2099 May 2100 Jan 2100 Oct Found 201 months with five Fridays, Saturdays and Sundays. There were 29 years with no such months.
Ceylon
module.ceylon
module rosetta.fiveweekends "1.0.0" {
import ceylon.time "1.2.2";
}
run.ceylon
import ceylon.time {
date,
Date
}
import ceylon.time.base {
january,
december,
friday,
Month
}
shared void run() {
[Date[],Integer[]] result = fiveWeekendsRecursive();
value fiveWeekendFirstOfMonths = result[0];
Integer[] yearsWithNoFiveWeekendMonths = result[1];
print("# five weekend months = ``fiveWeekendFirstOfMonths.size``");
print("# years without five weekend months = ``yearsWithNoFiveWeekendMonths.size``");
yearsWithNoFiveWeekendMonths.each(print);
}
[Date[], Integer[]] fiveWeekendsRecursive()
=> fiveWeekendsRecursiveInner{ year = 1900;
month = january;
fiveWeekendFirstOfMonths = [];
yearsWithNoFiveWeekendMonths = []; };
[Date[], Integer[]] fiveWeekendsRecursiveInner(Integer year,
Month month,
Date[] fiveWeekendFirstOfMonths,
Integer[] yearsWithNoFiveWeekendMonths) {
if (year > 2100) {
return [fiveWeekendFirstOfMonths,yearsWithNoFiveWeekendMonths];
}
Date firstOfMonth = date{ year = year; month = month; day = 1; };
Boolean isFiveWeekendMonth =
(month.numberOfDays() == 31 && friday == firstOfMonth.dayOfWeek);
Boolean hasNoFiveWeekends =
month == december &&
! isFiveWeekendMonth &&
fiveWeekendFirstOfMonths.filter((date) => date.year == year).size == 0;
return fiveWeekendsRecursiveInner(if (month == december) then year+1 else year,
if (month == december) then january else month.plusMonths(1),
if (isFiveWeekendMonth)
then fiveWeekendFirstOfMonths.withTrailing(firstOfMonth)
else fiveWeekendFirstOfMonths,
if (hasNoFiveWeekends)
then yearsWithNoFiveWeekendMonths.withTrailing(year)
else yearsWithNoFiveWeekendMonths);
}
- Output:
# five weekend months = 201 # years without five weekend months = 29 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
Clojure
(import java.util.GregorianCalendar
java.text.DateFormatSymbols)
(->> (for [year (range 1900 2101)
month [0 2 4 6 7 9 11] ;; 31 day months
:let [cal (GregorianCalendar. year month 1)
day (.get cal GregorianCalendar/DAY_OF_WEEK)]
:when (= day GregorianCalendar/FRIDAY)]
(println month "-" year))
count
(println "Total Months: " ,))
COBOL
program-id. five-we.
data division.
working-storage section.
1 wk binary.
2 int-date pic 9(8).
2 dow pic 9(4).
2 friday pic 9(4) value 5.
2 mo-sub pic 9(4).
2 months-with-5 pic 9(4) value 0.
2 years-no-5 pic 9(4) value 0.
2 5-we-flag pic 9(4) value 0.
88 5-we-true value 1 when false 0.
1 31-day-mos pic 9(14) value 01030507081012.
1 31-day-table redefines 31-day-mos.
2 mo-no occurs 7 pic 99.
1 cal-date.
2 yr pic 9(4).
2 mo pic 9(2).
2 da pic 9(2) value 1.
procedure division.
perform varying yr from 1900 by 1
until yr > 2100
set 5-we-true to false
perform varying mo-sub from 1 by 1
until mo-sub > 7
move mo-no (mo-sub) to mo
compute int-date = function
integer-of-date (function numval (cal-date))
compute dow = function mod
((int-date - 1) 7) + 1
if dow = friday
perform output-date
add 1 to months-with-5
set 5-we-true to true
end-if
end-perform
if not 5-we-true
add 1 to years-no-5
end-if
end-perform
perform output-counts
stop run
.
output-counts.
display "Months with 5 weekends: " months-with-5
display "Years without 5 weekends: " years-no-5
.
output-date.
display yr "-" mo
.
end program five-we.
- Output:
1901-03 1902-08 1903-05 1904-01 1904-07 . . . 2097-03 2098-08 2099-05 2100-01 2100-10 Months with 5 weekends: 0201 Years without 5 weekends: 0029
CoffeeScript
startsOnFriday = (month, year) ->
# 0 is Sunday, 1 is Monday, ... 5 is Friday, 6 is Saturday
new Date(year, month, 1).getDay() == 5
has31Days = (month, year) ->
new Date(year, month, 31).getDate() == 31
checkMonths = (year) ->
month = undefined
count = 0
month = 0
while month < 12
if startsOnFriday(month, year) and has31Days(month, year)
count += 1
console.log year + ' ' + month + ''
month += 1
count
fiveWeekends = ->
startYear = 1900
endYear = 2100
year = undefined
monthTotal = 0
yearsWithoutFiveWeekends = []
total = 0
year = startYear
while year <= endYear
monthTotal = checkMonths(year)
total += monthTotal
# extra credit
if monthTotal == 0
yearsWithoutFiveWeekends.push year
year += 1
console.log 'Total number of months: ' + total + ''
console.log ''
console.log yearsWithoutFiveWeekends + ''
console.log 'Years with no five-weekend months: ' + yearsWithoutFiveWeekends.length + ''
return
fiveWeekends()
- Output:
1901 2
1902 7
1903 4
1904 0
1904 6
1905 11
1907 2
1908 4
1909 0
1909 9
1910 6
1911 11
1912 2
1913 7
1914 4
1915 0
1915 9
1916 11
1918 2
1919 7
1920 9
1921 6
1922 11
1924 7
..
Total number of months: 201
1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096
Years with no five-weekend months: 29
Common Lisp
;; Given a date, get the day of the week. Adapted from
;; http://lispcookbook.github.io/cl-cookbook/dates_and_times.html
(defun day-of-week (day month year)
(nth-value
6
(decode-universal-time
(encode-universal-time 0 0 0 day month year 0)
0)))
(defparameter *long-months* '(1 3 5 7 8 10 12))
(defun sundayp (day month year)
(= (day-of-week day month year) 6))
(defun ends-on-sunday-p (month year)
(sundayp 31 month year))
;; We use the "long month that ends on Sunday" rule.
(defun has-five-weekends-p (month year)
(and (member month *long-months*)
(ends-on-sunday-p month year)))
;; For the extra credit problem.
(defun has-at-least-one-five-weekend-month-p (year)
(let ((flag nil))
(loop for month in *long-months* do
(if (has-five-weekends-p month year)
(setf flag t)))
flag))
(defun solve-it ()
(let ((good-months '())
(bad-years 0))
(loop for year from 1900 to 2100 do
;; First form in the PROGN is for the extra credit.
(progn (unless (has-at-least-one-five-weekend-month-p year)
(incf bad-years))
(loop for month in *long-months* do
(when (has-five-weekends-p month year)
(push (list month year) good-months)))))
(let ((len (length good-months)))
(format t "~A months have five weekends.~%" len)
(format t "First 5 months: ~A~%" (subseq good-months (- len 5) len))
(format t "Last 5 months: ~A~%" (subseq good-months 0 5))
(format t "Years without a five-weekend month: ~A~%" bad-years))))
- Output:
201 months have five weekends. First 5 months: ((7 1904) (1 1904) (5 1903) (8 1902) (3 1901)) Last 5 months: ((10 2100) (1 2100) (5 2099) (8 2098) (3 2097)) Years without a five-weekend month: 29
D
import std.stdio, std.datetime, std.algorithm, std.range;
Date[] m5w(in Date start, in Date end) pure /*nothrow*/ {
typeof(return) res;
// adjust to 1st day
for (Date when = Date(start.year, start.month, 1);
when < end;
when.add!"months"(1))
// Such month must have 3+4*7 days and start at friday
// for 5 FULL weekends.
if (when.daysInMonth == 31 &&
when.dayOfWeek == DayOfWeek.fri)
res ~= when;
return res;
}
bool noM5wByYear(in int year) pure {
return m5w(Date(year, 1, 1), Date(year, 12, 31)).empty;
}
void main() {
immutable m = m5w(Date(1900, 1, 1), Date(2100, 12, 31));
writeln("There are ", m.length,
" months of which the first and last five are:");
foreach (d; m[0 .. 5] ~ m[$ - 5 .. $])
writeln(d.toSimpleString()[0 .. $ - 3]);
immutable n = iota(1900, 2101).filter!noM5wByYear().walkLength();
writefln("\nThere are %d years in the range that do not have " ~
"months with five weekends.", n);
}
- Output:
There are 201 months of which the first and last five are: 1901-Mar 1902-Aug 1903-May 1904-Jan 1904-Jul 2097-Mar 2098-Aug 2099-May 2100-Jan 2100-Oct There are 29 years in the range that do not have months with five weekends.
Simpler Version
void main() {
import std.stdio, std.datetime, std.traits;
enum first_year = 1900;
enum last_year = 2100;
uint totalNo5Weekends;
immutable(Date)[] fiveWeekendMonths;
foreach (immutable year; first_year .. last_year + 1) {
bool has5Weekends = false;
foreach (immutable month; EnumMembers!Month) {
immutable firstDay = Date(year, month, 1);
if (firstDay.daysInMonth == 31 &&
firstDay.dayOfWeek == DayOfWeek.fri) {
has5Weekends = true;
fiveWeekendMonths ~= firstDay;
}
}
if (!has5Weekends)
totalNo5Weekends++;
}
writefln("Total 5-weekend months between %d and %d: %d",
first_year, last_year, fiveWeekendMonths.length);
foreach (immutable date; fiveWeekendMonths[0 .. 5])
writeln(date.month, ' ', date.year);
"...".writeln;
foreach (immutable date; fiveWeekendMonths[$ - 5 .. $])
writeln(date.month, ' ', date.year);
writeln("\nTotal number of years with no 5-weekend months: ",
totalNo5Weekends);
}
- Output:
Total 5-weekend months between 1900 and 2100: 201 mar 1901 aug 1902 may 1903 jan 1904 jul 1904 ... mar 2097 aug 2098 may 2099 jan 2100 oct 2100 Total number of years with no 5-weekend months: 29
Dart
main() {
var total = 0;
var empty = <int>[];
for (var year = 1900; year < 2101; year++) {
var months =
[1, 3, 5, 7, 8, 10, 12].where((m) => DateTime(year, m, 1).weekday == 5);
print('$year\t$months');
total += months.length;
if (months.isEmpty) empty.add(year);
}
print('Total: $total');
print('Year with none: $empty');
}
- Sample output:
1900 () 1901 (3) 1902 (8) 1903 (5) 1904 (1, 7) ... 2098 (8) 2099 (5) 2100 (1, 10) Total: 201 Year with none: [1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096]
Delphi
program FiveWeekends;
{$APPTYPE CONSOLE}
uses SysUtils, DateUtils;
var
lMonth, lYear: Integer;
lDate: TDateTime;
lFiveWeekendCount: Integer;
lYearsWithout: Integer;
lFiveWeekendFound: Boolean;
begin
for lYear := 1900 to 2100 do
begin
lFiveWeekendFound := False;
for lMonth := 1 to 12 do
begin
lDate := EncodeDate(lYear, lMonth, 1);
if (DaysInMonth(lDate) = 31) and (DayOfTheWeek(lDate) = DayFriday) then
begin
Writeln(FormatDateTime('mmm yyyy', lDate));
Inc(lFiveWeekendCount);
lFiveWeekendFound := True;
end;
end;
if not lFiveWeekendFound then
Inc(lYearsWithout);
end;
Writeln;
Writeln(Format('Months with 5 weekends: %d', [lFiveWeekendCount]));
Writeln(Format('Years with no 5 weekend months: %d', [lYearsWithout]));
end.
DuckDB
create or replace function day_before(day) as (6+day) % 7;
create or replace function weekday_of_last_day_of_month(year, month) as (
if(month=12,
day_before( extract('dayofweek' from make_date(year+1, 1, 1) )),
day_before( extract('dayofweek' from make_date( year, month+1, 1 ) ) ))
);
# The only case where the month has 5 full weekends is when the last day
# of the month falls on a Sunday (0) and the month has 31 days.
#
create or replace function five_weekends(start, stop) as table (
select year, month
from range(start, stop) as t(year),
unnest([1,3,5,7,8,10,12]) as u(month) -- months with 31 days
where weekday_of_last_day_of_month(year, month) = 0
);
from five_weekends(1900, 2101) order by year, month;
- Output:
┌───────┬───────┐ │ year │ month │ │ int64 │ int32 │ ├───────┼───────┤ │ 1901 │ 3 │ │ 1902 │ 8 │ │ 1903 │ 5 │ │ 1904 │ 1 │ │ 1904 │ 7 │ │ 1905 │ 12 │ │ 1907 │ 3 │ │ 1908 │ 5 │ │ 1909 │ 1 │ │ 1909 │ 10 │ │ 1910 │ 7 │ │ 1911 │ 12 │ │ 1912 │ 3 │ │ 1913 │ 8 │ │ 1914 │ 5 │ │ 1915 │ 1 │ │ 1915 │ 10 │ │ 1916 │ 12 │ │ 1918 │ 3 │ │ 1919 │ 8 │ │ · │ · │ │ · │ · │ │ · │ · │ │ 2081 │ 8 │ │ 2082 │ 5 │ │ 2083 │ 1 │ │ 2083 │ 10 │ │ 2084 │ 12 │ │ 2086 │ 3 │ │ 2087 │ 8 │ │ 2088 │ 10 │ │ 2089 │ 7 │ │ 2090 │ 12 │ │ 2092 │ 8 │ │ 2093 │ 5 │ │ 2094 │ 1 │ │ 2094 │ 10 │ │ 2095 │ 7 │ │ 2097 │ 3 │ │ 2098 │ 8 │ │ 2099 │ 5 │ │ 2100 │ 1 │ │ 2100 │ 10 │ ├───────┴───────┤ │ 201 rows │ │ (40 shown) │ └───────────────┘
EasyLang
func leap year .
return if year mod 4 = 0 and (year mod 100 <> 0 or year mod 400 = 0)
.
func weekday year month day .
normdoom[] = [ 3 7 7 4 2 6 4 1 5 3 7 5 ]
c = year div 100
r = year mod 100
s = r div 12
t = r mod 12
c_anchor = (5 * (c mod 4) + 2) mod 7
doom = (s + t + (t div 4) + c_anchor) mod 7
anchor = normdoom[month]
if leap year = 1 and month <= 2
anchor = (anchor + 1) mod1 7
.
return (doom + day - anchor + 7) mod 7 + 1
.
mdays[] = [ 31 28 31 30 31 30 31 31 30 31 30 31 ]
mon$[] = [ "Jan" "Feb" "Mar" "Apr" "May" "Jun" "Jul" "Aug" "Sep" "Oct" "Nov" "Dec" ]
#
for year = 1900 to 2100
for m to 12
if mdays[m] = 31 and weekday year m 1 = 6
print year & "-" & mon$[m]
sum += 1
.
.
.
print "Total : " & sum
- Output:
1901-Mar 1902-Aug 1903-May 1904-Jan 1904-Jul . . 2097-Mar 2098-Aug 2099-May 2100-Jan 2100-Oct Total : 201
Elixir
defmodule Date do
@months { "January", "February", "March", "April", "May", "June",
"July", "August", "September", "October", "November", "December" }
def five_weekends(year) do
for m <-[1,3,5,7,8,10,12], :calendar.day_of_the_week(year, m, 31) == 7, do: elem(@months, m-1)
end
end
months = Enum.map(1900..2100, fn year -> {year, Date.five_weekends(year)} end)
{none, months5} = Enum.partition(months, fn {_,m} -> Enum.empty?(m) end)
count = Enum.reduce(months5, 0, fn {year, months}, acc ->
IO.puts "#{year} : #{Enum.join(months, ", ")}"
acc + length(months)
end)
IO.puts "Found #{count} month with 5 weekends."
IO.puts "\nFound #{length(none)} years with no month having 5 weekends:"
IO.puts "#{inspect Enum.map(none, fn {y,_}-> y end)}"
- Output:
1901 : March 1902 : August 1903 : May 1904 : January, July 1905 : December ... 2095 : July 2097 : March 2098 : August 2099 : May 2100 : January, October Found 201 month with 5 weekends. Found 29 years with no month having 5 weekends: [1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096]
Erlang
Two examples, both based on first building a nested list-of-lists like [Year1, Year2, ..., YearN], where each year is a sublist of year and month tuples, like [{YearN, 1}, {YearN, 2}, ..., {YearN, 12}].
First, a pretty compact example intended for use with escript:
#!/usr/bin/env escript
%%
%% Calculate number of months with five weekends between years 1900-2100
%%
main(_) ->
Years = [ [{Y,M} || M <- lists:seq(1,12)] || Y <- lists:seq(1900,2100) ],
{CountedYears, {Has5W, TotM5W}} = lists:mapfoldl(
fun(Months, {Has5W, Tot}) ->
WithFive = [M || M <- Months, has_five(M)],
CountM5W = length(WithFive),
{{Months,CountM5W}, {Has5W++WithFive, Tot+CountM5W}}
end, {[], 0}, Years),
io:format("There are ~p months with five full weekends.~n"
"Showing top and bottom 5:~n",
[TotM5W]),
lists:map(fun({Y,M}) -> io:format("~p-~p~n", [Y,M]) end,
lists:sublist(Has5W,1,5) ++ lists:nthtail(TotM5W-5, Has5W)),
No5W = [Y || {[{Y,_M}|_], 0} <- CountedYears],
io:format("The following ~p years do NOT have any five-weekend months:~n",
[length(No5W)]),
lists:map(fun(Y) -> io:format("~p~n", [Y]) end, No5W).
has_five({Year, Month}) ->
has_five({Year, Month}, calendar:last_day_of_the_month(Year, Month)).
has_five({Year, Month}, Days) when Days =:= 31 ->
calendar:day_of_the_week({Year, Month, 1}) =:= 5;
has_five({_Year, _Month}, _DaysNot31) ->
false.
Second, a more verbose Erlang module:
-module(five_weekends).
-export([report/0, print_5w_month/1, print_year_with_no_5w_month/1]).
report() ->
Years = make_nested_period_list(1900, 2100),
{CountedYears, {All5WMonths, CountOf5WMonths}} = lists:mapfoldl(
fun(SingleYearSublist, {All5WMonths, CountOf5WMonths}) ->
MonthsWith5W = [Month || Month <- SingleYearSublist, if_has_5w(Month)],
CountOf5WMonthsFor1Year = length(MonthsWith5W),
{ % Result of map for this year sublist:
{SingleYearSublist,CountOf5WMonthsFor1Year},
% Accumulate total result for our fold:
{All5WMonths ++ MonthsWith5W, CountOf5WMonths + CountOf5WMonthsFor1Year}
}
end, {[], 0}, Years),
io:format("There are ~p months with five full weekends.~n"
"Showing top and bottom 5:~n",
[CountOf5WMonths]),
lists:map(fun print_5w_month/1, take_nth_first_and_last(5, All5WMonths)),
YearsWithout5WMonths = find_years_without_5w_months(CountedYears),
io:format("The following ~p years do NOT have any five-weekend months:~n",
[length(YearsWithout5WMonths)]),
lists:map(fun print_year_with_no_5w_month/1, YearsWithout5WMonths).
make_nested_period_list(FromYear, ToYear) ->
[ make_monthtuple_sublist_for_year(Year) || Year <- lists:seq(FromYear, ToYear) ].
make_monthtuple_sublist_for_year(Year) ->
[ {Year, Month} || Month <- lists:seq(1,12) ].
if_has_5w({Year, Month}) ->
if_has_5w({Year, Month}, calendar:last_day_of_the_month(Year, Month)).
if_has_5w({Year, Month}, Days) when Days =:= 31 ->
calendar:day_of_the_week({Year, Month, 1}) =:= 5;
if_has_5w({_Year, _Month}, _DaysNot31) ->
false.
print_5w_month({Year, Month}) ->
io:format("~p-~p~n", [Year, Month]).
print_year_with_no_5w_month(Year) ->
io:format("~p~n", [Year]).
take_nth_first_and_last(N, List) ->
Len = length(List),
lists:sublist(List, 1, N) ++ lists:nthtail(Len - N, List).
find_years_without_5w_months(List) ->
[Y || {[{Y,_M}|_], 0} <- List].
- Output:
There are 201 months with five full weekends. Showing top and bottom 5: 1901-3 1902-8 1903-5 1904-1 1904-7 2097-3 2098-8 2099-5 2100-1 2100-10 The following 29 years do NOT have any five-weekend months: 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
ERRE
PROGRAM FIVE_WEEKENDS
DIM M$[12]
PROCEDURE MODULO(X,Y->MD)
IF Y=0 THEN
MD=X
ELSE
MD=X-Y*INT(X/Y)
END IF
END PROCEDURE
PROCEDURE WD(M,D,Y->RES%)
IF M=1 OR M=2 THEN
M+=12
Y-=1
END IF
MODULO(365*Y+INT(Y/4)-INT(Y/100)+INT(Y/400)+D+INT((153*M+8)/5),7->RES)
RES%=RES+1.0
END PROCEDURE
BEGIN
M$[]=("","JANUARY","FEBRUARY","MARCH","APRIL","MAY","JUNE","JULY","AUGUST","SEPTEMBER","OCTOBER","NOVEMBER","DECEMBER")
PRINT(CHR$(12);) ! CLS
FOR YEAR=1900 TO 2100 DO
FOREACH MONTH IN (1,3,5,7,8,10,12) DO ! months with 31 days
WD(MONTH,1,YEAR->RES%)
IF RES%=6 THEN ! day #6 is Friday
PRINT(YEAR;": ";M$[MONTH])
CNT%=CNT%+1
! IF CNT% MOD 20=0 THEN GET(K$) END IF ! press a key for next page
END IF
END FOR
END FOR
PRINT("Total =";CNT%)
END PROGRAM
- Output:
1901: MARCH 1902: AUGUST 1903: MAY 1904: JANUARY 1904: JULY 1905: DECEMBER ... 2093: MAY 2094: JANUARY 2094: OCTOBER 2095: JULY 2097: MARCH 2098: AUGUST 2099: MAY 2100: JANUARY 2100: OCTOBER Total = 201
Euphoria
--Five Weekend task from Rosetta Code wiki
--User:Lnettnay
include std/datetime.e
atom numbermonths = 0
sequence longmonths = {1, 3, 5, 7, 8, 10, 12}
sequence yearsmonths = {}
atom none = 0
datetime dt
for year = 1900 to 2100 do
atom flag = 0
for month = 1 to length(longmonths) do
dt = new(year, longmonths[month], 1)
if weeks_day(dt) = 6 then --Friday is day 6
flag = 1
numbermonths += 1
yearsmonths = append(yearsmonths, {year, longmonths[month]})
end if
end for
if flag = 0 then
none += 1
end if
end for
puts(1, "Number of months with five full weekends from 1900 to 2100 = ")
? numbermonths
puts(1, "First five and last five years, months\n")
for count = 1 to 5 do
? yearsmonths[count]
end for
for count = length(yearsmonths) - 4 to length(yearsmonths) do
? yearsmonths[count]
end for
puts(1, "Number of years that have no months with five full weekends = ")
? none
- Output:
Number of months with five full weekends from 1900 to 2100 = 201 First five and last five years, months {1901,3} {1902,8} {1903,5} {1904,1} {1904,7} {2097,3} {2098,8} {2099,5} {2100,1} {2100,10} Number of years that have no months with five full weekends = 29
F#
open System
[<EntryPoint>]
let main argv =
let (yearFrom, yearTo) = (1900, 2100)
let monthsWith5We year =
[1; 3; 5; 7; 8; 10; 12] |>
List.filter (fun month -> DateTime(year, month, 1).DayOfWeek = DayOfWeek.Friday)
let ym5we =
[yearFrom .. yearTo]
|> List.map (fun year -> year, (monthsWith5We year))
let countMonthsWith5We =
ym5we
|> List.sumBy (snd >> List.length)
let countYearsWithout5WeMonths =
ym5we
|> List.sumBy (snd >> List.isEmpty >> (function|true->1|_->0))
let allMonthsWith5we =
ym5we
|> List.filter (snd >> List.isEmpty >> not)
printfn "%d months in the range of years from %d to %d have 5 weekends."
countMonthsWith5We yearFrom yearTo
printfn "%d years in the range of years from %d to %d have no month with 5 weekends."
countYearsWithout5WeMonths yearFrom yearTo
printfn "Months with 5 weekends: %A ... %A"
(List.take 5 allMonthsWith5we)
(List.take 5 (List.skip ((List.length allMonthsWith5we) - 5) allMonthsWith5we))
0
- Output:
201 months in the range of years from 1900 to 2100 have 5 weekends. 29 years in the range of years from 1900 to 2100 have no month with 5 weekends. Months with 5 weekends: [(1901, [3]); (1902, [8]); (1903, [5]); (1904, [1; 7]); (1905, [12])] ... [(2095, [7]); (2097, [3]); (2098, [8]); (2099, [5]); (2100, [1; 10])]
Factor
USING: calendar calendar.format formatting io kernel math
sequences ;
: timestamps>my ( months -- )
[ { MONTH bl YYYY nl } formatted 2drop ] each ;
: month-range ( start-year #months -- seq )
[ <year> ] [ <iota> ] bi* [ months time+ ] with map ;
: find-five-weekend-months ( months -- months' )
[ [ friday? ] [ days-in-month ] bi 31 = and ] filter ;
1900 12 201 * month-range find-five-weekend-months
[ length "%d five-weekend months found.\n" printf ]
[ 5 head timestamps>my "..." print ]
[ 5 tail* timestamps>my ] tri
- Output:
201 five-weekend months found. Mar 1901 Aug 1902 May 1903 Jan 1904 Jul 1904 ... Mar 2097 Aug 2098 May 2099 Jan 2100 Oct 2100
Fortran
Using Zeller's congruence
program Five_weekends
implicit none
integer :: m, year, nfives = 0, not5 = 0
logical :: no5weekend
type month
integer :: n
character(3) :: name
end type month
type(month) :: month31(7)
month31(1) = month(13, "Jan")
month31(2) = month(3, "Mar")
month31(3) = month(5, "May")
month31(4) = month(7, "Jul")
month31(5) = month(8, "Aug")
month31(6) = month(10, "Oct")
month31(7) = month(12, "Dec")
do year = 1900, 2100
no5weekend = .true.
do m = 1, size(month31)
if(month31(m)%n == 13) then
if(Day_of_week(1, month31(m)%n, year-1) == 6) then
write(*, "(a3, i5)") month31(m)%name, year
nfives = nfives + 1
no5weekend = .false.
end if
else
if(Day_of_week(1, month31(m)%n, year) == 6) then
write(*,"(a3, i5)") month31(m)%name, year
nfives = nfives + 1
no5weekend = .false.
end if
end if
end do
if(no5weekend) not5 = not5 + 1
end do
write(*, "(a, i0)") "Number of months with five weekends between 1900 and 2100 = ", nfives
write(*, "(a, i0)") "Number of years between 1900 and 2100 with no five weekend months = ", not5
contains
function Day_of_week(d, m, y)
integer :: Day_of_week
integer, intent(in) :: d, m, y
integer :: j, k
j = y / 100
k = mod(y, 100)
Day_of_week = mod(d + (m+1)*26/10 + k + k/4 + j/4 + 5*j, 7)
end function Day_of_week
end program Five_weekends
- Output:
Mar 1901 Aug 1902 May 1903 Jan 1904 Jul 1904 ... Mar 2097 Aug 2098 May 2099 Jan 2100 Oct 2100 Number of months with five weekends between 1900 and 2100 = 201 Number of years between 1900 and 2100 with no five weekend months = 29
FreeBASIC
' version 23-06-2015
' compile with: fbc -s console
Function wd(m As Integer, d As Integer, y As Integer) As Integer
' Zellerish
' 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday
' 4 = Thursday, 5 = Friday, 6 = Saturday
If m < 3 Then ' If m = 1 Or m = 2 Then
m += 12
y -= 1
End If
Return (y + (y \ 4) - (y \ 100) + (y \ 400) + d + ((153 * m + 8) \ 5)) Mod 7
End Function
' ------=< MAIN >=------
' only months with 31 day can have five weekends
' these months are: January, March, May, July, August, October, December
' in nr: 1, 3, 5, 7, 8, 10, 12
' the 1e day needs to be on a friday (= 5)
Dim As String month_names(1 To 12) => {"January","February","March",_
"April","May","June","July","August",_
"September","October","November","December"}
Dim As Integer m, yr, total, i, j, yr_without(200)
Dim As String answer
For yr = 1900 To 2100 ' Gregorian calendar
answer = ""
For m = 1 To 12 Step 2
If m = 9 Then m = 8
If wd(m , 1 , yr) = 5 Then
answer = answer + month_names(m) + ", "
total = total + 1
End If
Next
If answer <> "" Then
Print Using "#### | "; yr;
Print Left(answer, Len(answer) -2) ' get rid of extra " ,"
Else
i = i + 1
yr_without(i) = yr
End If
Next
Print
Print "nr of month for 1900 to 2100 that has five weekends";total
Print
Print i;" years don't have months with five weekends"
For j = 1 To i
Print yr_without(j); " ";
If j Mod 8 = 0 Then Print
Next
Print
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
- Output:
1901 | March 1902 | August 1903 | May 1904 | January, July 1905 | December ... 2095 | July 2097 | March 2098 | August 2099 | May 2100 | January, October nr of month from 1900 to 2100 that has five weekends 201 29 years don't have months with five weekends 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
Gambas
Click this link to run this code
Public Sub Main()
Dim aMonth As Short[] = [1, 3, 5, 7, 8, 10, 12] 'All 31 day months
Dim aMMStore As New String[] 'To store results
Dim siYear, siMonth, siCount As Short 'Various variables
Dim dDay As Date 'To store the day to check
Dim sTemp As String 'Temp string
For siYear = 1900 To 2100 'Loop through each year
For siMonth = 0 To 6 'Loop through each 31 day month
dDay = Date(siYear, aMonth[siMonth], 1) 'Get the date of the 1st of the month
If WeekDay(dDay) = 5 Then aMMStore.Add(Format(dDay, "mmmm yyyy")) 'If the 1st is a Friday then store the result
Next
Next
For Each sTemp In aMMStore 'For each item in the stored array..
Inc siCount 'Increase siCount
If siCount < 6 Then Print aMMStore[siCount] 'If 1 of the 1st 5 dates then print it
If siCount = 6 Then Print String$(14, "-") 'Print a separator
If siCount > aMMStore.Max - 4 Then Print aMMStore[siCount - 1] 'If 1 of the last 5 dates then print it
Next
Print gb.NewLine & "Total months = " & Str(siCount) 'Print the number of months found
siCount = 0 'Reset siCount
sTemp = aMMStore.Join(",") 'Put all the stored dates in one string joined by commas
aMMStore.Clear 'Clear the store for reuse
For siYear = 1900 To 2100 'Loop through each year
If Not InStr(sTemp, Str(siYear)) Then 'If the year is not in the stored string then..
Inc siCount 'Increase siCount (Amount of years that don't have 5 weekend months)
aMMStore.Add(Str(siYear)) 'Add to the store
End If
Next
Print gb.NewLine & "There are " & Str(siCount) &
" years that do not have at least one five-weekend month" 'Print the amount of years with no 5 weekend months
Print aMMStore.Join(",") 'Print the years with no 5 weekend months
End
Output:
August 1902 May 1903 January 1904 July 1904 December 1905 -------------- March 2097 August 2098 May 2099 January 2100 October 2100 Total months = 201 There are 29 years that do not have at least one five-weekend month 1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096
GAP
# return a list of two lists :
# first is the list of months with five weekends between years y1 and y2 (included)
# second is the list of years without such months, in the same interval
FiveWeekends := function(y1, y2)
local L, yL, badL, d, m, y;
L := [ ];
badL := [ ];
for y in [y1 .. y2] do
yL := [ ];
for m in [1, 3, 5, 7, 8, 10, 12] do
if WeekDay([1, m, y]) = "Fri" then
d := StringDate([1, m, y]);
Add(yL, d{[4 .. 11]});
fi;
od;
if Length(yL) = 0 then
Add(badL, y);
else
Append(L, yL);
fi;
od;
return [ L, badL ];
end;
r := FiveWeekends(1900, 2100);;
n := Length(r[1]);
# 201
Length(r[2]);
# 29
r[1]{[1 .. 5]};
# [ "Mar-1901", "Aug-1902", "May-1903", "Jan-1904", "Jul-1904" ]
r[1]{[n-4 .. n]};
# [ "Mar-2097", "Aug-2098", "May-2099", "Jan-2100", "Oct-2100" ]
Go
Using second algorithm suggestion:
package main
import (
"fmt"
"time"
)
func main() {
var n int // for task item 2
var first, last time.Time // for task item 3
haveNone := make([]int, 0, 29) // for extra credit
fmt.Println("Months with five weekends:") // for task item 1
for year := 1900; year <= 2100; year++ {
var hasOne bool // for extra credit
for _, month := range []time.Month{1, 3, 5, 7, 8, 10, 12} {
t := time.Date(year, month, 1, 0, 0, 0, 0, time.UTC)
if t.Weekday() == time.Friday {
// task item 1: show month
fmt.Println(" ", t.Format("2006 January"))
n++
hasOne = true
last = t
if first.IsZero() {
first = t
}
}
}
if !hasOne {
haveNone = append(haveNone, year)
}
}
fmt.Println(n, "total\n") // task item 2: number of months
// task item 3
fmt.Println("First five dates of weekends:")
for i := 0; i < 5; i++ {
fmt.Println(" ", first.Format("Monday, January 2, 2006"))
first = first.Add(7 * 24 * time.Hour)
}
fmt.Println("Last five dates of weekends:")
for i := 0; i < 5; i++ {
fmt.Println(" ", last.Format("Monday, January 2, 2006"))
last = last.Add(7 * 24 * time.Hour)
}
// extra credit
fmt.Println("\nYears with no months with five weekends:")
for _, y := range haveNone {
fmt.Println(" ", y)
}
fmt.Println(len(haveNone), "total")
}
- Output:
Months with five weekends: 1901 March 1902 August 1903 May 1904 January 1904 July ... 2097 March 2098 August 2099 May 2100 January 2100 October 201 total First five dates of weekends: Friday, March 1, 1901 Friday, March 8, 1901 Friday, March 15, 1901 Friday, March 22, 1901 Friday, March 29, 1901 Last five dates of weekends: Friday, October 1, 2100 Friday, October 8, 2100 Friday, October 15, 2100 Friday, October 22, 2100 Friday, October 29, 2100 Years with no months with five weekends: 1900 1906 1917 1923 1928 .... 2068 2074 2085 2091 2096 29 total
Groovy
Solution:
enum Day {
Sun, Mon, Tue, Wed, Thu, Fri, Sat
static Day valueOf(Date d) { Day.valueOf(d.format('EEE')) }
}
def date = Date.&parse.curry('yyyy-M-dd')
def isLongMonth = { firstDay -> (firstDay + 31).format('dd') == '01'}
def fiveWeekends = { years ->
years.collect { year ->
(1..12).collect { month ->
date("${year}-${month}-01")
}.findAll { firstDay ->
isLongMonth(firstDay) && Day.valueOf(firstDay) == Day.Fri
}
}.flatten()
}
Test:
def ym = { it.format('yyyy-MM') }
def years = 1900..2100
def fiveWeekendMonths = fiveWeekends(years)
println "Number of five weekend months: ${fiveWeekendMonths.size()}"
fiveWeekendMonths.each { println (ym(it)) }
- Output:
Number of five weekend months: 201 1901-03 1902-08 1903-05 1904-01 1904-07 1905-12 1907-03 1908-05 1909-01 1909-10 1910-07 1911-12 1912-03 1913-08 1914-05 1915-01 1915-10 1916-12 1918-03 1919-08 1920-10 1921-07 1922-12 1924-08 1925-05 1926-01 1926-10 1927-07 1929-03 1930-08 1931-05 1932-01 1932-07 1933-12 1935-03 1936-05 1937-01 1937-10 1938-07 1939-12 1940-03 1941-08 1942-05 1943-01 1943-10 1944-12 1946-03 1947-08 1948-10 1949-07 1950-12 1952-08 1953-05 1954-01 1954-10 1955-07 1957-03 1958-08 1959-05 1960-01 1960-07 1961-12 1963-03 1964-05 1965-01 1965-10 1966-07 1967-12 1968-03 1969-08 1970-05 1971-01 1971-10 1972-12 1974-03 1975-08 1976-10 1977-07 1978-12 1980-08 1981-05 1982-01 1982-10 1983-07 1985-03 1986-08 1987-05 1988-01 1988-07 1989-12 1991-03 1992-05 1993-01 1993-10 1994-07 1995-12 1996-03 1997-08 1998-05 1999-01 1999-10 2000-12 2002-03 2003-08 2004-10 2005-07 2006-12 2008-08 2009-05 2010-01 2010-10 2011-07 2013-03 2014-08 2015-05 2016-01 2016-07 2017-12 2019-03 2020-05 2021-01 2021-10 2022-07 2023-12 2024-03 2025-08 2026-05 2027-01 2027-10 2028-12 2030-03 2031-08 2032-10 2033-07 2034-12 2036-08 2037-05 2038-01 2038-10 2039-07 2041-03 2042-08 2043-05 2044-01 2044-07 2045-12 2047-03 2048-05 2049-01 2049-10 2050-07 2051-12 2052-03 2053-08 2054-05 2055-01 2055-10 2056-12 2058-03 2059-08 2060-10 2061-07 2062-12 2064-08 2065-05 2066-01 2066-10 2067-07 2069-03 2070-08 2071-05 2072-01 2072-07 2073-12 2075-03 2076-05 2077-01 2077-10 2078-07 2079-12 2080-03 2081-08 2082-05 2083-01 2083-10 2084-12 2086-03 2087-08 2088-10 2089-07 2090-12 2092-08 2093-05 2094-01 2094-10 2095-07 2097-03 2098-08 2099-05 2100-01 2100-10
Extra Credit:
Test:
def yearsWith = fiveWeekendMonths.collect { it.format('yyyy') as int } as Set
def yearsWithout = (years as Set) - yearsWith
println "\nNumber of years without a five weekend month: ${yearsWithout.size()}"
yearsWithout.each { println it }
- Output:
Number of years without a five weekend month: 29 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
Harbour
PROCEDURE Main()
LOCAL y, m, d, nFound, cNames, nTot := 0, nNotFives := 0
LOCAL aFounds := {}
SET DATE ANSI
FOR y := 1900 TO 2100
nFound := 0 ; cNames := ""
FOR m := 1 TO 12
d := CtoD( hb_NtoS( y ) +"/" + hb_NtoS( m ) + "/1" )
IF CDoW( d ) == "Friday"
IF DaysInMonth( m ) == 31
nFound++
cNames += CMonth( d ) + " "
ENDIF
ENDIF
NEXT
IF nFound > 0
AAdd( aFounds, hb_NtoS( y ) + " : " + hb_NtoS( nFound ) + " ( " + Rtrim( cNames ) + " )" )
nTot += nFound
ELSE
nNotFives++
ENDIF
NEXT
? "Total months with five weekends: " + hb_NtoS( nTot )
? "(see bellow the first and last five years/months with five weekends)"
?
AEval( aFounds, { | e, n | Iif( n < 6, Qout( e ), NIL ) } )
Qout("...")
AEval( aFounds, { | e, n | Iif( n > Len(aFounds)-5, Qout( e ), NIL ) } )
?
? "Years with no five weekends months: " + hb_NtoS( nNotFives )
RETURN
- Output:
Total months with five weekends: 201 (see bellow the first and last five years/months with five weekends) 1901 : 1 ( March ) 1902 : 1 ( August ) 1903 : 1 ( May ) 1904 : 2 ( January July ) 1905 : 1 ( December ) ... 2095 : 1 ( July ) 2097 : 1 ( March ) 2098 : 1 ( August ) 2099 : 1 ( May ) 2100 : 2 ( January October ) Years with no five weekends months: 29
Haskell
Without using date libraries
Not using any helper libraries, this code profits from Haskell's lazy evaluation. Knowing that the first day of 1900 was a Monday, we make an infinit list of days and split it in years and months. To do this, we take and drop days from the head of the list of days.
year 1900 = [Monday, Tuesday, Wednesday...] (365 items) year 1904 = [Friday, Saturday, Sunday...] (366 items -- leap year)
and also:
year 1900 = [(January, [Monday, Tuesday..]), (February, [Thursday, Friday..]), ...]
Now it is easy to get all months with 31 days that start on Friday.
import Data.List (intercalate)
data DayOfWeek = Monday | Tuesday | Wednesday | Thursday | Friday |
Saturday | Sunday
deriving (Eq, Show)
-- the whole thing bases upon an infinite list of weeks
daysFrom1_1_1900 :: [DayOfWeek]
daysFrom1_1_1900 = concat $ repeat [Monday, Tuesday, Wednesday,
Thursday, Friday, Saturday, Sunday]
data Month = January | February | March | April | May | June | July |
August | September | October | November | December
deriving (Show)
type Year = Int
type YearCalendar = (Year, [DayOfWeek])
type MonthlyCalendar = (Year, [(Month, [DayOfWeek])])
-- makes groups of 365 or 366 days for each year (infinite list)
yearsFrom :: [DayOfWeek] -> Year -> [YearCalendar]
yearsFrom s i = (i, yeardays) : yearsFrom rest (i + 1)
where
yeardays = take (leapOrNot i) s
yearlen = length yeardays
rest = drop yearlen s
leapOrNot n = if isLeapYear n then 366 else 365
yearsFrom1900 :: [YearCalendar]
yearsFrom1900 = yearsFrom daysFrom1_1_1900 1900
-- makes groups of days for each month of the year
months :: YearCalendar -> MonthlyCalendar
months (y, d) = (y, [(January, january), (February, february),
(March, march), (April, april), (May, may), (June, june),
(July, july), (August, august), (September, september),
(October, october), (November, november), (December, december)])
where
leapOrNot = if isLeapYear y then 29 else 28
january = take 31 d
february = take leapOrNot $ drop 31 d
march = take 31 $ drop (31 + leapOrNot) d
april = take 30 $ drop (62 + leapOrNot) d
may = take 31 $ drop (92 + leapOrNot) d
june = take 30 $ drop (123 + leapOrNot) d
july = take 31 $ drop (153 + leapOrNot) d
august = take 31 $ drop (184 + leapOrNot) d
september = take 30 $ drop (215 + leapOrNot) d
october = take 31 $ drop (245 + leapOrNot) d
november = take 30 $ drop (276 + leapOrNot) d
december = take 31 $ drop (306 + leapOrNot) d
-- see if a year is a leap year
isLeapYear n
| n `mod` 100 == 0 = n `mod` 400 == 0
| otherwise = n `mod` 4 == 0
-- make a list of the months of a year that have 5 weekends
-- (they must have 31 days and the first day must be Friday)
-- if the year doesn't contain any 5-weekended months, then
-- return the year and an empty list
whichFiveWeekends :: MonthlyCalendar -> (Year, [Month])
whichFiveWeekends (y, ms) = (y, map (\(m, _) -> m) found) -- extract the months & leave out their days
where found = filter (\(m, a@(d:ds)) -> and [length a == 31,
d == Friday]) ms
-- take all days from 1900 until 2100, grouping them by years, then by
-- months, and calculating whether they have any 5-weekended months
-- or not
calendar :: [MonthlyCalendar]
calendar = map months $ yearsFrom1900
fiveWeekends1900To2100 :: [(Year, [Month])]
fiveWeekends1900To2100 = takeWhile (\(y, _) -> y <= 2100) $
map whichFiveWeekends calendar
main = do
-- count the number of years with 5 weekends
let answer1 = foldl (\c (_, m) -> c + length m) 0 fiveWeekends1900To2100
-- take only the years with 5-weekended months
answer2 = filter (\(_, m) -> not $ null m) fiveWeekends1900To2100
-- take only the years without 5-weekended months
answer30 = filter (\(_, m) -> null m) fiveWeekends1900To2100
-- count how many years without 5-weekended months there are
answer31 = length answer30
-- show the years without 5-weekended months
answer32 = intercalate ", " $ map (\(y, m) -> show y) answer30
putStrLn $ "There are " ++ show answer1 ++ " months with 5 weekends between 1900 and 2100."
putStrLn "\nThe first ones are:"
mapM_ (putStrLn . formatMonth) $ take 5 $ answer2
putStrLn "\nThe last ones are:"
mapM_ (putStrLn . formatMonth) $ reverse $ take 5 $ reverse answer2
putStrLn $ "\n" ++ show answer31 ++ " years don't have at least one five-weekened month"
putStrLn "\nThose are:"
putStrLn answer32
formatMonth :: (Year, [Month]) -> String
formatMonth (y, m) = show y ++ ": " ++ intercalate ", " [ show x | x <- m ]
- Output:
There are 201 months with 5 weekends between 1900 and 2100. The first ones are: 1901: March 1902: August 1903: May 1904: January, July 1905: December The last ones are: 2095: July 2097: March 2098: August 2099: May 2100: January, October 29 years don't have at least one five-weekended month Those are: 1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096
Using Data.Time
import Data.List (intercalate)
import Data.List.Split (chunksOf)
import Data.Time (Day, fromGregorian, gregorianMonthLength)
import Data.Time.Calendar.WeekDate (toWeekDate)
---------------- MONTHS WITH FIVE WEEKENDS ---------------
fiveFridayMonths :: Integer -> [(Integer, Int)]
fiveFridayMonths y =
[1 .. 12]
>>= \m ->
[ (y, m)
| isFriday (fromGregorian y m 1),
31 == gregorianMonthLength y m
]
isFriday :: Day -> Bool
isFriday d = 5 == day
where
(_, _, day) = toWeekDate d
--------------------------- TEST -------------------------
main :: IO ()
main = do
let years = [1900 .. 2100]
xs = fiveFridayMonths <$> years
lean =
concat $
zipWith
(\months year -> [year | null months])
xs
years
n = (length . concat) xs
(putStrLn . intercalate "\n\n")
[ "How many five-weekend months 1900-2100 ?",
'\t' : show n,
"First five ?",
'\t' : show (concat (take 5 xs)),
"Last five ?",
'\t' : show (concat (drop (n - 5) xs)),
"How many lean years ? (No five-weekend months)",
'\t' : show (length lean),
"Which years are lean ?",
unlines $
('\t' :) . unwords . fmap show
<$> chunksOf 5 lean
]
- Output:
How many five-weekend months 1900-2100 ? 201 First five ? [(1901,3),(1902,8),(1903,5),(1904,1),(1904,7)] Last five ? [(2097,3),(2098,8),(2099,5),(2100,10),(2100,1)] How many lean years ? (No five-weekend months) 29 Which years are lean ? 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
Icon and Unicon
printf.icn provides formatting datetime.icn provides julian
- Output:
There are 201 months from 1900 to 2100 with five full weekends. The first and last five such months are: 1901-3-1 1902-8-1 1903-5-1 1904-1-1 1904-7-1 ... 2098-8-1 2099-5-1 2100-1-1 2100-10-1 There are 29 years without such months as follows: 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
Inform 7
Inform 7 has no built-in date functions, so this solution implements date types and a day-of-week function.
Calendar is a room.
When play begins:
let happy month count be 0;
let sad year count be 0;
repeat with Y running from Y1900 to Y2100:
if Y is a sad year, increment the sad year count;
repeat with M running through months:
if M of Y is a happy month:
say "[M] [year number of Y].";
increment the happy month count;
say "Found [happy month count] month[s] with five weekends and [sad year count] year[s] with no such months.";
end the story.
Section - Years
A year is a kind of value. Y1 specifies a year.
To decide which number is year number of (Y - year):
decide on Y / Y1.
To decide if (N - number) is divisible by (M - number):
decide on whether or not the remainder after dividing N by M is zero.
Definition: a year (called Y) is a leap year:
let YN be the year number of Y;
if YN is divisible by 400, yes;
if YN is divisible by 100, no;
if YN is divisible by 4, yes;
no.
Section - Months
A month is a kind of value. The months are defined by the Table of Months.
Table of Months
month month number
January 1
February 2
March 3
April 4
May 5
June 6
July 7
August 8
September 9
October 10
November 11
December 12
A month has a number called length. The length of a month is usually 31.
September, April, June, and November have length 30. February has length 28.
To decide which number is number of days in (M - month) of (Y - year):
let L be the length of M;
if M is February and Y is a leap year, decide on L + 1;
otherwise decide on L.
Section - Weekdays
A weekday is a kind of value. The weekdays are defined by the Table of Weekdays.
Table of Weekdays
weekday weekday number
Saturday 0
Sunday 1
Monday 2
Tuesday 3
Wednesday 4
Thursday 5
Friday 6
To decide which weekday is weekday of the/-- (N - number) of (M - month) of (Y - year):
let MN be the month number of M;
let YN be the year number of Y;
if MN is less than 3:
increase MN by 12;
decrease YN by 1;
let h be given by Zeller's Congruence;
let WDN be the remainder after dividing h by 7;
decide on the weekday corresponding to a weekday number of WDN in the Table of Weekdays.
Equation - Zeller's Congruence
h = N + ((MN + 1)*26)/10 + YN + YN/4 + 6*(YN/100) + YN/400
where h is a number, N is a number, MN is a number, and YN is a number.
To decide which number is number of (W - weekday) days in (M - month) of (Y - year):
let count be 0;
repeat with N running from 1 to the number of days in M of Y:
if W is the weekday of the N of M of Y, increment count;
decide on count.
Section - Happy Months and Sad Years
To decide if (M - month) of (Y - year) is a happy month:
if the number of days in M of Y is 31 and the weekday of the 1st of M of Y is Friday, decide yes;
decide no.
To decide if (Y - year) is a sad year:
repeat with M running through months:
if M of Y is a happy month, decide no;
decide yes.
- Output:
March 1901. August 1902. May 1903. January 1904. July 1904. ... March 2097. August 2098. May 2099. January 2100. October 2100. Found 201 months with five weekends and 29 years with no such months.
J
require 'types/datetime numeric'
find5wkdMonths=: verb define
years=. range 2{. y
months=. 1 3 5 7 8 10 12
m5w=. (#~ 0 = weekday) >,{years;months;31 NB. 5 full weekends iff 31st is Sunday(0)
>'MMM YYYY' fmtDate toDayNo m5w
)
Usage:
# find5wkdMonths 1900 2100 NB. number of months found
201
(5&{. , '...' , _5&{.) find5wkdMonths 1900 2100 NB. First and last 5 months found
Mar 1901
Aug 1902
May 1903
Jan 1904
Jul 1904
...
Mar 2097
Aug 2098
May 2099
Jan 2100
Oct 2100
# (range -. {:"1@(_ ". find5wkdMonths)) 1900 2100 NB. number of years without 5 weekend months
29
Java
In Java 1.5+ you can add import static java.util.Calendar.*;
to the list of imports and remove all other occurrences of Calendar.
from the rest of the code (e.g. Calendar.FRIDAY
would simply become FRIDAY
). It's more portable (and probably more clear) to leave the Calendar.
's in.
import java.util.Calendar;
import java.util.GregorianCalendar;
public class FiveFSS {
private static boolean[] years = new boolean[201];
private static int[] month31 = {Calendar.JANUARY, Calendar.MARCH, Calendar.MAY,
Calendar.JULY, Calendar.AUGUST, Calendar.OCTOBER, Calendar.DECEMBER};
public static void main(String[] args) {
StringBuilder months = new StringBuilder();
int numMonths = 0;
for (int year = 1900; year <= 2100; year++) {
for (int month : month31) {
Calendar date = new GregorianCalendar(year, month, 1);
if (date.get(Calendar.DAY_OF_WEEK) == Calendar.FRIDAY) {
years[year - 1900] = true;
numMonths++;
//months are 0-indexed in Calendar
months.append((date.get(Calendar.MONTH) + 1) + "-" + year +"\n");
}
}
}
System.out.println("There are "+numMonths+" months with five weekends from 1900 through 2100:");
System.out.println(months);
System.out.println("Years with no five-weekend months:");
for (int year = 1900; year <= 2100; year++) {
if(!years[year - 1900]){
System.out.println(year);
}
}
}
}
- Output:
(middle results cut out)
There are 201 months with five weekends from 1900 through 2100: 3-1901 8-1902 5-1903 1-1904 7-1904 12-1905 3-1907 5-1908 1-1909 10-1909 7-1910 ... 12-2090 8-2092 5-2093 1-2094 10-2094 7-2095 3-2097 8-2098 5-2099 1-2100 10-2100 Years with no five-weekend months: 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
JavaScript
ES5
Imperative
function startsOnFriday(month, year)
{
// 0 is Sunday, 1 is Monday, ... 5 is Friday, 6 is Saturday
return new Date(year, month, 1).getDay() === 5;
}
function has31Days(month, year)
{
return new Date(year, month, 31).getDate() === 31;
}
function checkMonths(year)
{
var month, count = 0;
for (month = 0; month < 12; month += 1)
{
if (startsOnFriday(month, year) && has31Days(month, year))
{
count += 1;
document.write(year + ' ' + month + '<br>');
}
}
return count;
}
function fiveWeekends()
{
var
startYear = 1900,
endYear = 2100,
year,
monthTotal = 0,
yearsWithoutFiveWeekends = [],
total = 0;
for (year = startYear; year <= endYear; year += 1)
{
monthTotal = checkMonths(year);
total += monthTotal;
// extra credit
if (monthTotal === 0)
yearsWithoutFiveWeekends.push(year);
}
document.write('Total number of months: ' + total + '<br>');
document.write('<br>');
document.write(yearsWithoutFiveWeekends + '<br>');
document.write('Years with no five-weekend months: ' + yearsWithoutFiveWeekends.length + '<br>');
}
fiveWeekends();
- Sample output:
1901 2 1902 7 1903 4 1904 0 1904 6 ... 2097 2 2098 7 2099 4 2100 0 2100 9 Total number of months: 201 1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096 Years with no five-weekend months: 29
Here is an alternative solution that uses the offset between the first day of every month, generating the same solution but without relying on the Date object.
var Months = [
'Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun',
'Jul', 'Aug', 'Sept', 'Oct', 'Nov', 'Dec'
];
var leap = 0,
// Relative offsets between first day of each month
offset = [3,0,3,2,3,2,3,3,2,3,2,3],
// Months that contain 31 days
longMonths = [1,3,5,7,8,10,12],
startYear = 1900,
year = startYear,
endYear = 2100,
// Jan 1, 1900 starts on a Monday
day = 1,
totalPerYear = 0,
total = 0,
without = 0;
for (; year < endYear + 1; year++) {
leap = totalPerYear = 0;
if (year % 4 === 0) {
if (year % 100 === 0) {
if (year % 400 === 0) {
leap = 1;
}
} else {
leap = 1;
}
}
for (var i = 0; i < offset.length; i++) {
for (var j = 0; day === 5 && j < longMonths.length; j++) {
if (i + 1 === longMonths[j]) {
console.log(year + '-' + Months[i]);
totalPerYear++;
total++;
break;
}
}
// February -- if leap year, then +1 day
if (i == 1) {
day = (day + leap) % 7;
} else {
day = (day + offset[i]) % 7;
}
}
if (totalPerYear === 0) {
without++;
}
}
console.log('Number of months that have five full weekends from 1900 to 2100: ' + total);
console.log('Number of years without any five full weekend months: ' + without);
- Output:
1901-Mar 1902-Aug 1903-May 1904-Jan 1904-Jul ... 2097-Mar 2098-Aug 2099-May 2100-Jan 2100-Oct Number of months that have five full weekends from 1900 to 2100: 201 Number of years without any five full weekend months: 29
Functional
(function () {
'use strict';
// longMonthsStartingFriday :: Int -> Int
function longMonthsStartingFriday(y) {
return [0, 2, 4, 6, 7, 9, 11]
.filter(function (m) {
return (new Date(Date.UTC(y, m, 1)))
.getDay() === 5;
});
}
// range :: Int -> Int -> [Int]
function range(m, n) {
return Array.apply(null, Array(n - m + 1))
.map(function (x, i) {
return m + i;
});
}
var lstNames = [
'January', '', 'March', '', 'May', '',
'July', 'August', '', 'October', '', 'December'
],
lstYears = range(1900, 2100),
lstFullMonths = lstYears
.reduce(function (a, y) {
var strYear = y.toString();
return a.concat(
longMonthsStartingFriday(y)
.map(function (m) {
return strYear + ' ' + lstNames[m];
})
);
}, []),
lstLeanYears = lstYears
.filter(function (y) {
return longMonthsStartingFriday(y)
.length === 0;
});
return JSON.stringify({
number: lstFullMonths.length,
firstFive: lstFullMonths.slice(0, 5),
lastFive: lstFullMonths.slice(-5),
leanYearCount: lstLeanYears.length
},
null, 2
);
})();
- Output:
{ "number": 201, "firstFive": [ "1901 March", "1902 August", "1903 May", "1904 January", "1904 July" ], "lastFive": [ "2097 March", "2098 August", "2099 May", "2100 January", "2100 October" ], "leanYearCount": 29 }
ES6
(() => {
// longMonthsStartingFriday :: Int -> [Int]
const longMonthsStartingFriday = y =>
filter(m => (new Date(Date.UTC(y, m, 1)))
.getDay() === 5, [0, 2, 4, 6, 7, 9, 11]);
// Years -> YearMonths
// fullMonths :: [Int] -> [String]
const fullMonths = xs =>
foldl((a, y) => a.concat(
map(m => `${y.toString()} ${[
'January', '', 'March', '', 'May', '',
'July', 'August', '', 'October', '', 'December'
][m]}`, longMonthsStartingFriday(y))
), [], xs);
// leanYears :: [Int] -> [Int]
const leanYears = years =>
filter(y => longMonthsStartingFriday(y)
.length === 0, years);
// GENERIC ----------------------------------------------------------------
// A list of functions applied to a list of arguments
// <*> :: [(a -> b)] -> [a] -> [b]
const ap = (fs, xs) => //
[].concat.apply([], fs.map(f => //
[].concat.apply([], xs.map(x => [f(x)]))));
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
// filter :: (a -> Bool) -> [a] -> [a]
const filter = (f, xs) => xs.filter(f);
// foldl :: (b -> a -> b) -> b -> [a] -> b
const foldl = (f, a, xs) => xs.reduce(f, a);
// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map(f);
// show :: a -> String
const show = x => JSON.stringify(x, null, 2);
// TEST -------------------------------------------------------------------
const [lstFullMonths, lstLeanYears] = ap(
[fullMonths, leanYears], [enumFromTo(1900, 2100)]
);
return show({
number: lstFullMonths.length,
firstFive: lstFullMonths.slice(0, 5),
lastFive: lstFullMonths.slice(-5),
leanYearCount: lstLeanYears.length
});
})();
- Output:
{ "number": 201, "firstFive": [ "1901 March", "1902 August", "1903 May", "1904 January", "1904 July" ], "lastFive": [ "2097 March", "2098 August", "2099 May", "2100 January", "2100 October" ], "leanYearCount": 29 }
jq
Foundations: Zeller's Congruence
# Use Zeller's Congruence to determine the day of the week, given
# year, month and day as integers in the conventional way.
# Emit 0 for Saturday, 1 for Sunday, etc.
#
def day_of_week(year; month; day):
if month == 1 or month == 2 then
[month + 12, year - 1]
else
[month, year]
end
| day + (13*(.[0] + 1)/5|floor)
+ (.[1]%100) + ((.[1]%100)/4|floor)
+ (.[1]/400|floor) - 2*(.[1]/100|floor)
| . % 7
;
Nuts and Bolts:
def weekday_of_last_day_of_month(year; month):
def day_before(day): (6+day) % 7;
if month==12 then day_before( day_of_week(year+1; 1; 1) )
else day_before( day_of_week( year; month+1; 1 ) )
end
;
# The only case where the month has 5 weekends is when the last day
# of the month falls on a Sunday and the month has 31 days.
#
def five_weekends(from; to):
reduce range(from; to) as $year
([]; reduce (1,3,5,7,8,10,12) as $month # months with 31 days
(.;
weekday_of_last_day_of_month($year; $month) as $day
| if $day == 1 then . + [[ $year, $month]] else . end ))
;
# Input [year, month] as conventional integers; print e.g. "Jan 2001"
def pp:
def month:
["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"][.-1];
"\(.[1] | month) \(.[0])"
;
The Task:
five_weekends(1900;2101)
| "There are \(length) months with 5 weekends from 1900 to 2100 inclusive;",
"the first and last five are as follows:",
( .[0: 5][] | pp),
"...",
( .[length-5: ][] | pp),
"In this period, there are \( [range(1900;2101)] - map( .[0] ) | length ) years which have no five-weekend months."
- Output:
$ jq -r -n -f Five_Weekends.jq
There are 201 months with 5 weekends from 1900 to 2100 inclusive;
the first and last five are as follows:
Mar 1901
Aug 1902
May 1903
Jan 1904
Jul 1904
...
Mar 2097
Aug 2098
May 2099
Jan 2100
Oct 2100
In this period, there are 29 years which have no five-weekend months.
Julia
isweekend(dt::Date) = Dates.dayofweek(dt) ∈ (Dates.Friday, Dates.Saturday, Dates.Sunday)
function hasfiveweekend(month::Integer, year::Integer)
dmin = Date(year, month, 1)
dmax = dmin + Dates.Day(Dates.daysinmonth(dmin) - 1)
return count(isweekend, dmin:dmax) ≥ 15
end
months = collect((y, m) for y in 1900:2100, m in 1:12 if hasfiveweekend(m, y))
println("Number of months with 5 full-weekends: $(length(months))")
println("First five such months:")
for (y, m) in months[1:5] println(" - $y-$m") end
println("Last five such months:")
for (y, m) in months[end-4:end] println(" - $y-$m") end
# extra credit
yrs = getindex.(months, 1)
nyrs = 2100 - 1899 - length(unique(yrs))
println("Number of year with not one 5-full-weekend month: $nyrs")
- Output:
Number of months with 5 full-weekends: 201 First five such months: - 1904-1 - 1909-1 - 1915-1 - 1926-1 - 1932-1 Last five such months: - 2062-12 - 2073-12 - 2079-12 - 2084-12 - 2090-12 Number of year with not one 5-full-weekend month: 29
k
cal_j:(_jd[19000101]+!(-/_jd 21010101 19000101)) / enumerate the calendar
is_we:(cal_j!7) _lin 4 5 6 / identify friday saturdays and sundays
m:__dj[cal_j]%100 / label the months
mi:&15=+/'is_we[=m] / group by month and sum the weekend days
`0:,"There are ",($#mi)," months with five weekends"
m5:(?m)[mi]
`0:$5#m5
`0:,"..."
`0:$-5#m5
y:1900+!201 / enumerate the years in the range
y5:?_ m5%100 / label the years of the months
yn5:y@&~y _lin y5 / find any years not in the 5 weekend month list
`0:,"There are ",($#yn5)," years without any five-weekend months"
`0:,1_,/",",/:$yn5
- Output:
There are 201 months with five weekends 190103 190208 190305 190401 190407 ... 209703 209808 209905 210001 210010 There are 29 years without any five-weekend months 1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096
Kotlin
// version 1.0.6
import java.util.*
fun main(args: Array<String>) {
val calendar = GregorianCalendar(1900, 0, 1)
val months31 = arrayOf(1, 3, 5, 7, 8, 10, 12)
val monthsWithFive = mutableListOf<String>()
val yearsWithNone = mutableListOf<Int>()
for (year in 1900..2100) {
var countInYear = 0 // counts months in a given year with 5 weekends
for (month in 1..12) {
if ((month in months31) && (Calendar.FRIDAY == calendar[Calendar.DAY_OF_WEEK])) {
countInYear++
monthsWithFive.add("%02d".format(month) + "-" + year)
}
calendar.add(Calendar.MONTH, 1)
}
if (countInYear == 0) yearsWithNone.add(year)
}
println("There are ${monthsWithFive.size} months with 5 weekends")
println("The first 5 are ${monthsWithFive.take(5)}")
println("The final 5 are ${monthsWithFive.takeLast(5)}")
println()
println("There are ${yearsWithNone.size} years with no months which have 5 weekends, namely:")
println(yearsWithNone)
}
- Output:
There are 201 months with 5 weekends The first 5 are [03-1901, 08-1902, 05-1903, 01-1904, 07-1904] The final 5 are [03-2097, 08-2098, 05-2099, 01-2100, 10-2100] There are 29 years with no months which have 5 weekends, namely: [1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096]
Lasso
local(
months = array(1, 3, 5, 7, 8, 10, 12),
fivemonths = array,
emptyears = array,
checkdate = date,
countyear
)
#checkdate -> day = 1
loop(-from = 1900, -to = 2100) => {
#countyear = false
#checkdate -> year = loop_count
with month in #months
do {
#checkdate -> month = #month
if(#checkdate -> dayofweek == 6) => {
#countyear = true
#fivemonths -> insert(#checkdate -> format(`YYYY MMM`))
}
}
if(not #countyear) => {
#emptyears -> insert(loop_count)
}
}
local(
monthcount = #fivemonths -> size,
output = 'Total number of months ' + #monthcount + '<br /> Starting five months '
)
loop(5) => {
#output -> append(#fivemonths -> get(loop_count) + ', ')
}
#output -> append('<br /> Ending five months ')
loop(-from = #monthcount - 5, -to = #monthcount) => {
#output -> append(#fivemonths -> get(loop_count) + ', ')
}
#output -> append('<br /> Years with no five weekend months ' + #emptyears -> size + '<br />')
with year in #emptyears do {
#output -> append(#year + ', ')
}
#output
- Result:
Total number of months 201 Starting five months 1901 Mar, 1902 Aug, 1903 May, 1903 Jan, 1904 Jul, Ending five months 2095 Jul, 2097 Mar, 2098 Aug, 2099 May, 2099 Jan, 2100 Oct, Years with no five weekend months 29 1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096,
Lua
local months={"JAN","MAR","MAY","JUL","AUG","OCT","DEC"}
local daysPerMonth={31+28,31+30,31+30,31,31+30,31+30,0}
function find5weMonths(year)
local list={}
local startday=((year-1)*365+math.floor((year-1)/4)-math.floor((year-1)/100)+math.floor((year-1)/400))%7
for i,v in ipairs(daysPerMonth) do
if startday==4 then list[#list+1]=months[i] end
if i==1 and year%4==0 and year%100~=0 or year%400==0 then
startday=startday+1
end
startday=(startday+v)%7
end
return list
end
local cnt_months=0
local cnt_no5we=0
for y=1900,2100 do
local list=find5weMonths(y)
cnt_months=cnt_months+#list
if #list==0 then
cnt_no5we=cnt_no5we+1
end
print(y.." "..#list..": "..table.concat(list,", "))
end
print("Months with 5 weekends: ",cnt_months)
print("Years without 5 weekends in the same month:",cnt_no5we)
- Output:
1900 0: 1901 1: MAR 1902 1: AUG 1903 1: MAY 1904 2: JAN, JUL 1905 1: DEC 1906 0: 1907 1: MAR 1908 1: MAY 1909 2: JAN, OCT ... ... ... 2090 1: DEC 2091 0: 2092 1: AUG 2093 1: MAY 2094 2: JAN, OCT 2095 1: JUL 2096 0: 2097 1: MAR 2098 1: AUG 2099 1: MAY 2100 2: JAN, OCT Months with 5 weekends: 201 Years without 5 weekends in the same month: 29
Maple
five_weekends:= proc()
local i, month, count;
#Only months with 31 days can possibly satisfy the condition
local long_months := [1,3,5,7,8,10,12];
local months := ["January","February","March","April","May","June","July","August","September","October","November","December"];
count := 0;
for i from 1900 to 2100 by 1 do
for month in long_months do
if Calendar:-DayOfWeek(Date(i, month, 1)) = 6 then
printf("%d-%s\n", i, months[month]);
count++;
end if;
end do;
end do;
printf("%d months have five full weekends.\n", count);
end proc;
five_weekends();
- Output:
1901-March 1902-August 1903-May 1904-January 1904-July 1905-December 1907-March 1908-May 1909-January 1909-October 1910-July 1911-December 1912-March 1913-August 1914-May 1915-January 1915-October 1916-December 1918-March 1919-August 1920-October 1921-July 1922-December 1924-August 1925-May 1926-January 1926-October 1927-July 1929-March 1930-August 1931-May 1932-January 1932-July 1933-December 1935-March 1936-May 1937-January 1937-October 1938-July 1939-December 1940-March 1941-August 1942-May 1943-January 1943-October 1944-December 1946-March 1947-August 1948-October 1949-July 1950-December 1952-August 1953-May 1954-January 1954-October 1955-July 1957-March 1958-August 1959-May 1960-January 1960-July 1961-December 1963-March 1964-May 1965-January 1965-October 1966-July 1967-December 1968-March 1969-August 1970-May 1971-January 1971-October 1972-December 1974-March 1975-August 1976-October 1977-July 1978-December 1980-August 1981-May 1982-January 1982-October 1983-July 1985-March 1986-August 1987-May 1988-January 1988-July 1989-December 1991-March 1992-May 1993-January 1993-October 1994-July 1995-December 1996-March 1997-August 1998-May 1999-January 1999-October 2000-December 2002-March 2003-August 2004-October 2005-July 2006-December 2008-August 2009-May 2010-January 2010-October 2011-July 2013-March 2014-August 2015-May 2016-January 2016-July 2017-December 2019-March 2020-May 2021-January 2021-October 2022-July 2023-December 2024-March 2025-August 2026-May 2027-January 2027-October 2028-December 2030-March 2031-August 2032-October 2033-July 2034-December 2036-August 2037-May 2038-January 2038-October 2039-July 2041-March 2042-August 2043-May 2044-January 2044-July 2045-December 2047-March 2048-May 2049-January 2049-October 2050-July 2051-December 2052-March 2053-August 2054-May 2055-January 2055-October 2056-December 2058-March 2059-August 2060-October 2061-July 2062-December 2064-August 2065-May 2066-January 2066-October 2067-July 2069-March 2070-August 2071-May 2072-January 2072-July 2073-December 2075-March 2076-May 2077-January 2077-October 2078-July 2079-December 2080-March 2081-August 2082-May 2083-January 2083-October 2084-December 2086-March 2087-August 2088-October 2089-July 2090-December 2092-August 2093-May 2094-January 2094-October 2095-July 2097-March 2098-August 2099-May 2100-January 2100-October 201 months have five full weekends.
Mathematica / Wolfram Language
years = {1900, 2100}; months = {1 ,3 ,5 ,7 ,8 ,10 ,12};
result = Select[Tuples[{Range@@years, months}], (DateString[# ~ Join ~ 1, "DayNameShort"] == "Fri")&];
Print[result // Length," months with 5 weekends" ];
Print["First months: ", DateString[#,{"MonthName"," ","Year"}]& /@ result[[1 ;; 5]]];
Print["Last months: " , DateString[#,{"MonthName"," ","Year"}]& /@ result[[-5 ;; All]]];
Print[# // Length, " years without 5 weekend months:\n", #] &@
Complement[Range @@ years, Part[Transpose@result, 1]];
- Output:
201 months with 5 weekends First months: {March 1901, August 1902, May 1903, January 1904, July 1904} Last months: {March 2097, August 2098, May 2099, January 2100, October 2100} 29 years without 5 weekend months {1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096}
MATLAB / Octave
longmonth = [1 3 5 7 8 10 12];
i = 1;
for y = 1900:2100
for m = 1:numel(longmonth)
[num,name] = weekday(datenum(y,longmonth(m),1));
if num == 6
x(i,:) = datestr(datenum(y,longmonth(m),1),'mmm yyyy'); %#ok<SAGROW>
i = i+1;
end
end
end
fprintf('There are %i months with 5 weekends between 1900 and 2100.\n',length(x))
fprintf('\n The first 5 months are:\n')
for j = 1:5
fprintf('\t %s \n',x(j,:))
end
fprintf('\n The final 5 months are:\n')
for j = length(x)-4:length(x)
fprintf('\t %s \n',x(j,:))
end
- Output:
There are 201 months with 5 weekends between 1900 and 2100. The first 5 months are: Mar 1901 Aug 1902 May 1903 Jan 1904 Jul 1904 The final 5 months are: Mar 2097 Aug 2098 May 2099 Jan 2100 Oct 2100
Maxima
left(a, n) := makelist(a[i], i, 1, n)$
right(a, n) := block([m: length(a)], makelist(a[i], i, m - n + 1, m))$
a: [ ]$
for year from 1900 thru 2100 do
for month in [1, 3, 5, 7, 8, 10, 12] do
if weekday(year, month, 1) = 'friday then
a: endcons([year, month], a)$
length(a);
201
left(a, 5);
[[1901,3],[1902,8],[1903,5],[1904,1],[1904,7]]
right(a, 5);
[[2097,3],[2098,8],[2099,5],[2100,1],[2100,10]]
MUMPS
FIVE
;List and count the months between 1/1900 and 12/2100 that have 5 full weekends
;Extra credit - list and count years with no months with five full weekends
;Using the test that the 31st of a month is on a Sunday
;Uses the VA's public domain routine %DTC (Part of the Kernel) named here DIDTC
NEW YEAR,MONTH,X,Y,CNTMON,NOT,NOTLIST
; YEAR is the year we're testing
; MONTH is the month we're testing
; X is the date in "internal" format, as an input to DOW^DIDTC
; Y is the day of the week (0=Sunday, 1=Monday...) output from DOW^DIDTC
; CNTMON is a count of the months that have 5 full weekends
; NOT is a flag if there were no months with 5 full weekends yet that year
; NOTLIST is a list of years that do not have any months with 5 full weekends
SET CNTMON=0,NOTLIST=""
WRITE !!,"The following months have five full weekends:"
FOR YEAR=200:1:400 DO ;years since 12/31/1700 epoch
. SET NOT=0
. FOR MONTH="01","03","05","07","08","10","12" DO
. . SET X=YEAR_MONTH_"31"
. . DO DOW^DIDTC
. . IF (Y=0) DO
. . . SET NOT=NOT+1,CNTMON=CNTMON+1
. . . WRITE !,MONTH_"-"_(YEAR+1700)
. SET:(NOT=0) NOTLIST=NOTLIST_$SELECT($LENGTH(NOTLIST)>1:",",1:"")_(YEAR+1700)
WRITE !,"For a total of "_CNTMON_" months."
WRITE !!,"There are "_$LENGTH(NOTLIST,",")_" years with no five full weekends in any month."
WRITE !,"They are: "_NOTLIST
KILL YEAR,MONTH,X,Y,CNTMON,NOT,NOTLIST
QUIT
F ;Same logic as the main entry point, shortened format
N R,M,X,Y,C,N,L S C=0,L=""
W !!,"The following months have five full weekends:"
F R=200:1:400 D
. S N=0 F M="01","03","05","07","08","10","12" S X=R_M_"31" D DOW^DIDTC I 'Y S N=N+1,C=C+1 W !,M_"-"_(R+1700)
. S:'N L=L_$S($L(L):",",1:"")_(R+1700)
W !,"For a total of "_C_" months.",!!,"There are "_$L(L,",")_" years with no five full weekends in any month.",!,"They are: "_L
Q
Usage:
USER>d ^FIVE The following months have five full weekends: 03-1901 08-1902 05-1903 01-1904 07-1904 12-1905 . . . . . . 01-2094 10-2094 07-2095 03-2097 08-2098 05-2099 01-2100 10-2100 For a total of 201 months. There are 29 years with no five full weekends in any month. They are: 1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096
NetRexx
/* NetRexx ************************************************************
* 30.08.2012 Walter Pachl derived from Rexx version 3
* omitting dead code left there
**********************************************************************/
options replace format comments java crossref savelog symbols
Numeric digits 20
nr5fwe=0
years_without_5fwe=0
mnl='Jan Mar May Jul Aug Oct Dec'
ml='1 3 5 7 8 10 12'
Loop j=1900 To 2100
year_has_5fwe=0
Loop mi=1 To ml.words()
m=ml.word(mi)
jd=greg2jul(j,m,1)
IF jd//7=4 Then Do /* 1st m j is a Friday */
nr5fwe=nr5fwe+1
year_has_5fwe=1
If j<=1905 | 2095<=j Then
Say mnl.word(mi) j 'has 5 full weekends'
End
End
If j=1905 Then Say '...'
if year_has_5fwe=0 Then years_without_5fwe=years_without_5fwe+1
End
Say ' '
Say nr5fwe 'occurrences of 5 full weekends in a month'
Say years_without_5fwe 'years without 5 full weekends'
exit
method greg2jul(yy,mm,d) public static returns Rexx
/***********************************************************************
* Converts a Gregorian date to the corresponding Julian day number
* 19891101 Walter Pachl REXXified algorithm published in CACM
* (Fliegel & vanFlandern, CACM Vol.11 No.10 October 1968)
***********************************************************************/
numeric digits 12
/***********************************************************************
* The published formula:
* res=d-32075+1461*(yy+4800+(mm-14)%12)%4+,
* 367*(mm-2-((mm-14)%12)*12)%12-3*((yy+4900+(mm-14)%12)%100)%4
***********************************************************************/
mma=(mm-14)%12
yya=yy+4800+mma
result=d-32075+1461*yya%4+367*(mm-2-mma*12)%12-3*((yya+100)%100)%4
Return result /* return the result */
Output: see Rexx version 3
NewLISP
#!/usr/local/bin/newlisp
(context 'KR)
(define (Kraitchik year month day)
; See https://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week#Kraitchik.27s_variation
; Function adapted for specific task (not for general usage).
(if (or (= 1 month) (= 2 month))
(dec year)
)
;- - - -
(setf m-table '(_ 1 4 3 6 1 4 6 2 5 0 3 5)) ; - - - First element of list is dummy!
(setf m (m-table month))
;- - - -
(setf c-table '(0 5 3 1))
(setf century%4 (mod (int (slice (string year) 0 2)) 4))
(setf c (c-table century%4))
;- - - -
(setf yy* (slice (string year) -2))
(if (= "0" (yy* 0))
(setf yy* (yy* 1))
)
(setf yy (int yy*))
(setf y (mod (+ (/ yy 4) yy) 7))
;- - - -
(setf dow-table '(6 0 1 2 3 4 5))
(dow-table (mod (+ day m c y) 7))
)
(context 'MAIN)
(setf Fives 0)
(setf NotFives 0)
(setf Report '())
(setf months-table '((1 "Jan") (3 "Mar") (5 "May") (7 "Jul") (8 "Aug") (10 "Oct") (12 "Dec")))
(for (y 1900 2100)
(setf FivesFound 0)
(setf Names "")
(dolist (m '(1 3 5 7 8 10 12))
(setf Dow (KR:Kraitchik y m 1))
(if (= 5 Dow)
(begin
(++ FivesFound)
(setf Names (string Names " " (lookup m months-table)))
)
)
)
(if (zero? FivesFound)
(++ NotFives)
(begin
(setf Report (append Report (list (list y FivesFound (string "(" Names " )")))))
(setf Fives (+ Fives FivesFound))
)
)
)
;- - - - Display all report data
;(dolist (x Report)
; (println (x 0) ": " (x 1) " " (x 2))
;)
;- - - - Display only first five and last five records
(dolist (x (slice Report 0 5))
(println (x 0) ": " (x 1) " " (x 2))
)
(println "...")
(dolist (x (slice Report -5))
(println (x 0) ": " (x 1) " " (x 2))
)
(println "\nTotal months with five weekends: " Fives)
(println "Years with no five weekends months: " NotFives)
(exit)
- Output:
1901: 1 ( Mar ) 1902: 1 ( Aug ) 1903: 1 ( May ) 1904: 2 ( Jan Jul ) 1905: 1 ( Dec ) ... 2095: 1 ( Jul ) 2097: 1 ( Mar ) 2098: 1 ( Aug ) 2099: 1 ( May ) 2100: 2 ( Jan Oct ) Total months with five weekends: 201 Years with no five weekends months: 29
Nim
import times
const LongMonths = {mJan, mMar, mMay, mJul, mAug, mOct, mDec}
var sumNone = 0
for year in 1900..2100:
var none = true
for month in LongMonths:
if initDateTime(1, month, year, 0, 0, 0).weekday == dFri:
echo month, " ", year
none = false
if none: inc sumNone
echo "\nYears without a 5 weekend month: ", sumNone
- Output:
March 1901 August 1902 May 1903 January 1904 July 1904 December 1905 [...] March 2097 August 2098 May 2099 January 2100 October 2100 Years without a 5 weekend month: 29
OCaml
open CalendarLib
let list_first_five = function
| x1 :: x2 :: x3 :: x4 :: x5 :: _ -> [x1; x2; x3; x4; x5]
| _ -> invalid_arg "list_first_five"
let () =
let months = ref [] in
for year = 1900 to 2100 do
for month = 1 to 12 do
let we = ref 0 in
let num_days = Date.days_in_month (Date.make_year_month year month) in
for day = 1 to num_days - 2 do
let d0 = Date.day_of_week (Date.make year month day)
and d1 = Date.day_of_week (Date.make year month (day + 1))
and d2 = Date.day_of_week (Date.make year month (day + 2)) in
if (d0, d1, d2) = (Date.Fri, Date.Sat, Date.Sun) then incr we
done;
if !we = 5 then months := (year, month) :: !months
done;
done;
Printf.printf "Number of months with 5 weekends: %d\n" (List.length !months);
print_endline "First and last months between 1900 and 2100:";
let print_month (year, month) = Printf.printf "%d-%02d\n" year month in
List.iter print_month (list_first_five (List.rev !months));
List.iter print_month (List.rev (list_first_five !months));
;;
- Output:
$ ocaml unix.cma str.cma -I +calendar calendarLib.cma five_we.ml Number of months with 5 weekends: 201 First and last months between 1900 and 2100: 1901-03 1902-08 1903-05 1904-01 1904-07 2097-03 2098-08 2099-05 2100-01 2100-10
Oforth
: fiveWeekEnd(y1, y2)
| y m |
ListBuffer new
y1 y2 for: y [
Date.JANUARY Date.DECEMBER for: m [
Date.DaysInMonth(y, m) 31 ==
[ y, m, 01 ] asDate dayOfWeek Date.FRIDAY == and
ifTrue: [ [ y, m ] over add ]
]
]
dup size println dup left(5) println right(5) println ;
- Output:
>fiveWeekEnd(1900, 2100) 201 [[1901, 3], [1902, 8], [1903, 5], [1904, 1], [1904, 7]] [[2097, 3], [2098, 8], [2099, 5], [2100, 1], [2100, 10]]
PARI/GP
fiveWeekends()={
my(day=6); \\ 0 = Friday; this represents Thursday for March 1, 1900.
my(ny=[31,30,31,30,31,31,30,31,30,31,31,28],ly=ny,v,s);
ly[12]=29;
for(year=1900,2100,
v=if((year+1)%4,ny,ly); \\ Works for 1600 to 2398
for(month=1,12,
if(v[month] == 31 && !day,
if(month<11,
print(year" "month+2)
,
print(year+1" 1")
);
s++
);
day = (day + v[month])%7
)
);
s
};
Pascal
See Delphi
PascalABC.NET
uses system;
const
startYear = 1900;
endYear = 2100;
begin
var query := new List<datetime>;
foreach var year in (startyear..endyear) do
foreach var month in (1..12) do
if DateTime.DaysInMonth(year, month) = 31 then
query.Add(new DateTime(year, month, 1));
query := query.where(d -> d.dayofweek = dayofweek.Friday).ToList;
println('Count:', query.Count);
println;
println('First five:');
foreach var date in query.Take(5) do
writeln(date.Year, '-', date.Month);
println;
Println('Last five:');
foreach var date in query.Skip(query.Count - 5) do
writeln(date.Year, '-', date.Month);
println;
println('Years without 5 weekends:');
(startyear..endyear).except(query.Select(d -> d.year)).Println
end.
- Output:
Count: 201 First five: 1901-3 1902-8 1903-5 1904-1 1904-7 Last five: 2097-3 2098-8 2099-5 2100-1 2100-10 Years without 5 weekends: 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
Perl
#!/usr/bin/perl -w
use DateTime ;
my @happymonths ;
my @workhardyears ;
my @longmonths = ( 1 , 3 , 5 , 7 , 8 , 10 , 12 ) ;
my @years = 1900..2100 ;
foreach my $year ( @years ) {
my $countmonths = 0 ;
foreach my $month ( @longmonths ) {
my $dt = DateTime->new( year => $year ,
month => $month ,
day => 1 ) ;
if ( $dt->day_of_week == 5 ) {
$countmonths++ ;
my $yearfound = $dt->year ;
my $monthfound = $dt->month_name ;
push ( @happymonths , "$yearfound $monthfound" ) ;
}
}
if ( $countmonths == 0 ) {
push ( @workhardyears, $year ) ;
}
}
print "There are " . @happymonths . " months with 5 full weekends!\n" ;
print "The first 5 and the last 5 of them are:\n" ;
foreach my $i ( 0..4 ) {
print "$happymonths[ $i ]\n" ;
}
foreach my $i ( -5..-1 ) {
print "$happymonths[ $i ]\n" ;
}
print "No long weekends in the following " . @workhardyears . " years:\n" ;
map { print "$_\n" } @workhardyears ;
- Output:
There are 201 months with 5 full weekends! The first 5 and the last 5 of them are: 1901 March 1902 August 1903 May 1904 January 1904 July 2097 March 2098 August 2099 May 2100 January 2100 October No 5 weekends per month in the following 29 years: 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
Phix
with javascript_semantics sequence m31 = {"January",0,"March",0,"May",0,"July","August",0,"October",0,"December"} integer y,m, nmonths = 0 string months sequence res = {}, none = {} for y=1900 to 2100 do months = "" for m=1 to 12 do if string(m31[m]) and day_of_week(y,m,1,true)="Friday" then if length(months)!=0 then months &= ", " end if months &= m31[m] nmonths += 1 end if end for if length(months)=0 then none = append(none,y) else res = append(res,sprintf("%d : %s\n",{y,months})) end if end for printf(1,"Found %d months with five full weekends\n",nmonths) res[6..-6] = {" ...\n"} puts(1,join(res,"")) printf(1,"Found %d years with no month having 5 weekends:\n",{length(none)}) none[6..-6] = {".."} ?none
- Output:
Found 201 months with five full weekends 1901 : March 1902 : August 1903 : May 1904 : January, July 1905 : December ... 2095 : July 2097 : March 2098 : August 2099 : May 2100 : January, October Found 29 years with no month having 5 weekends: {1900,1906,1917,1923,1928,"..",2068,2074,2085,2091,2096}
Picat
go ?=>
println("Months with five weekends:"),
FiveWeekends = [ [Year,Month] : Year in 1900..2100, Month in [1,3,5,7,8,10,12], dow(Year,Month,1) == 5],
WLen = FiveWeekends.len,
println(take(FiveWeekends,5)),
println("..."),
println(drop(FiveWeekends,WLen-5)),
println(len=WLen),
nl,
println("Years w/o five weekends:"),
FiveWeekendYears = [Year : [Year,_] in FiveWeekends].remove_dups,
NoHitYears = [Year : Year in 1900..2100, not member(Year,FiveWeekendYears)],
NHLen = NoHitYears.len,
println(take(NoHitYears,5)),
println("..."),
println(drop(NoHitYears,NHLen-5)),
println(len=NHLen),
nl.
go => true.
% Day of week, Sakamoto's method
dow(Y, M, D) = R =>
T = [0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4],
if M < 3 then
Y := Y - 1
end,
R = (Y + Y // 4 - Y // 100 + Y // 400 + T[M] + D) mod 7.
% Days in a month.
max_days_in_month(Year,Month) = Days =>
if member(Month, [1,3,5,7,8,10,12]) then
Days = 31
elseif member(Month,[4,6,9,11]) then
Days = 30
else
if leap_year(Year) then
Days = 29
else
Days = 28
end
end.
leap_year(Year) =>
(Year mod 4 == 0, Year mod 100 != 0)
;
Year mod 400 == 0.
- Output:
Months with five weekends: [[1901,3],[1902,8],[1903,5],[1904,1],[1904,7]] ... [[2097,3],[2098,8],[2099,5],[2100,1],[2100,10]] len = 201 Years w/o five weekends: [1900,1906,1917,1923,1928] ... [2068,2074,2085,2091,2096] len = 29
PicoLisp
(setq Lst
(make
(for Y (range 1900 2100)
(for M (range 1 12)
(and
(date Y M 31)
(= "Friday" (day (date Y M 1)))
(link (list (get *Mon M) Y)) ) ) ) ) )
(prinl "There are " (length Lst) " months with five weekends:")
(mapc println (head 5 Lst))
(prinl "...")
(mapc println (tail 5 Lst))
(prinl)
(setq Lst (diff (range 1900 2100) (uniq (mapcar cadr Lst))))
(prinl "There are " (length Lst) " years with no five-weekend months:")
(println Lst)
- Output:
There are 201 months with five weekends: (Mar 1901) (Aug 1902) (May 1903) (Jan 1904) (Jul 1904) ... (Mar 2097) (Aug 2098) (May 2099) (Jan 2100) (Oct 2100) There are 29 years with no five-weekend months: (1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096)
Pike
int(0..1) weekends(object day)
{
return (<5,6,7>)[day->week_day()];
}
int(0..1) has5(object month)
{
return sizeof(filter(month->days(), weekends))==15;
}
object range = Calendar.Year(1900)->distance(Calendar.Year(2101));
array have5 = filter(range->months(), has5);
write("found %d months:\n%{%s\n%}...\n%{%s\n%}",
sizeof(have5), have5[..4]->format_nice(), have5[<4..]->format_nice());
array rest = range->years() - have5->year();
write("%d years without any 5 weekend month:\n %{%d,%}\n", sizeof(rest), rest->year_no());
- Output:
found 201 months: March 1901 August 1902 May 1903 January 1904 July 1904 ... March 2097 August 2098 May 2099 January 2100 October 2100 29 years without any 5 weekend month: 1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096,
PL/I
weekends: procedure options (main); /* 28/11/2011 */
declare tally fixed initial (0);
declare (d, dend, dn) fixed (10);
declare (date_start, date_end) picture '99999999';
declare Leap fixed (1);
date_start = '01011900';
do date_start = date_start to '01012100';
d = days(date_start, 'DDMMYYYY');
date_end = date_start + 30110000;
dend = days(date_end, 'DDMMYYYY');
Leap = dend-d-364;
do dn = d, d+59+Leap, d+120+Leap, d+181+Leap, d+212+Leap,
d+273+Leap, d+334+Leap;
if weekday(dn) = 6 then
do;
put skip list (daystodate(dn, 'MmmYYYY') || ' has 5 weekends' );
tally = tally + 1;
end;
end;
end;
put skip list ('Total number of months having 3-day weekends =', tally);
end weekends;
- Output:
Mar1901 has 5 weekends Aug1902 has 5 weekends May1903 has 5 weekends Jan1904 has 5 weekends Jul1904 has 5 weekends Dec1905 has 5 weekends ..... Jan2094 has 5 weekends Oct2094 has 5 weekends Jul2095 has 5 weekends Mar2097 has 5 weekends Aug2098 has 5 weekends May2099 has 5 weekends Jan2100 has 5 weekends Oct2100 has 5 weekends Total number of months having 3-day weekends = 201
Part 2: Years not having any month of 5 weekends:
weekends: procedure options (main); declare tally fixed initial (0); declare (d, dend, dn) fixed (10); declare (date_start, date_end) picture '99999999'; declare Leap fixed (1); declare Long_weekend bit (1); date_start = '01011900'; do date_start = date_start to '01012100'; d = days(date_start, 'DDMMYYYY'); date_end = date_start + 30110000; dend = days(date_end, 'DDMMYYYY'); Leap = dend-d-364; long_weekend = '0'b; do dn = d, d+59+Leap, d+120+Leap, d+181+Leap, d+212+Leap, d+273+Leap, d+334+Leap; if weekday(dn) = 6 then long_weekend = '1'b; end; if ^long_weekend then put skip list (daystodate(dn, 'YYYY') || ' has no month of 5 weekends'); end; end weekends;
- Output:
1900 has no month of 5 weekends 1906 has no month of 5 weekends 1917 has no month of 5 weekends 1923 has no month of 5 weekends 1928 has no month of 5 weekends 1934 has no month of 5 weekends 1945 has no month of 5 weekends 1951 has no month of 5 weekends 1956 has no month of 5 weekends 1962 has no month of 5 weekends 1973 has no month of 5 weekends 1979 has no month of 5 weekends 1984 has no month of 5 weekends 1990 has no month of 5 weekends 2001 has no month of 5 weekends 2007 has no month of 5 weekends 2012 has no month of 5 weekends 2018 has no month of 5 weekends 2029 has no month of 5 weekends 2035 has no month of 5 weekends 2040 has no month of 5 weekends 2046 has no month of 5 weekends 2057 has no month of 5 weekends 2063 has no month of 5 weekends 2068 has no month of 5 weekends 2074 has no month of 5 weekends 2085 has no month of 5 weekends 2091 has no month of 5 weekends 2096 has no month of 5 weekends
PowerShell
$fiveWeekends = @()
$yearsWithout = @()
foreach ($y in 1900..2100) {
$hasFiveWeekendMonth = $FALSE
foreach ($m in @("01","03","05","07","08",10,12)) {
if ((Get-Date "$y-$m-1").DayOfWeek -eq "Friday") {
$fiveWeekends += "$y-$m"
$hasFiveWeekendMonth = $TRUE
}
}
if ($hasFiveWeekendMonth -eq $FALSE) {
$yearsWithout += $y
}
}
Write-Output "Between the years 1900 and 2100, inclusive, there are $($fiveWeekends.count) months with five full weekends:"
Write-Output "$($fiveWeekends[0..4] -join ","),...,$($fiveWeekends[-5..-1] -join ",")"
Write-Output ""
Write-Output "Extra Credit: these $($yearsWithout.count) years have no such month:"
Write-Output ($yearsWithout -join ",")
- Output:
Between the years 1900 and 2100, inclusive, there are 201 months with five full weekends: 1901-03,1902-08,1903-05,1904-01,1904-07,...,2097-03,2098-08,2099-05,2100-01,2100-10 Extra Credit: these 29 years have no such month: 1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063 ,2068,2074,2085,2091,2096
Prolog
Works with SWI-Prolog;
main() :-
weekends(1900, 2100, FiveWeekendList, RemainderWeekendList),
length(FiveWeekendList, FiveLen),
maplist(write, ["Total five weekend months:", FiveLen, '\n']),
slice(FiveWeekendList, 5, FirstFiveList),
maplist(write, ["First five {year,month} pairs:", FirstFiveList, '\n']),
slice(FiveWeekendList, -5, LastFiveList),
maplist(write, ["Last five {year,month} pairs:", LastFiveList, '\n']),
maplist(take_year, FiveWeekendList, FiveYearList),
list_to_set(FiveYearList, FiveYearSet),
maplist(take_year, RemainderWeekendList, RemainderYearList),
list_to_set(RemainderYearList, RemainderYearSet),
subtract(RemainderYearSet, FiveYearSet, NonFiveWeekendSet),
length(NonFiveWeekendSet, NonFiveWeekendLen),
maplist(write, ["Total years with no five weekend months:", NonFiveWeekendLen, '\n']),
writeln(NonFiveWeekendSet).
weekends(StartYear, EndYear, FiveWeekendList, RemainderWeekendList) :-
numlist(StartYear, EndYear, YearList),
numlist(1, 12, MonthList),
pair(YearList, MonthList, YearMonthList),
partition(has_five_weekends, YearMonthList, FiveWeekendList, RemainderWeekendList).
has_five_weekends({Year, Month}) :-
long_month(Month),
starts_on_a_friday(Year, Month).
starts_on_a_friday(Year, Month) :-
Date = date(Year, Month, 1),
day_of_the_week(Date, DayOfTheWeek),
DayOfTheWeek == 5.
take_year({Year, _}, Year).
long_month(1).
long_month(3).
long_month(5).
long_month(7).
long_month(8).
long_month(10).
long_month(12).
% Helpers
% https://stackoverflow.com/a/7739806
pair(L1, L2, Pairs):-findall({A,B}, (member(A, L1), member(B, L2)), Pairs).
slice(_, 0, []).
slice(List, N, NList):-
N < 0,
N1 is abs(N),
last_n_elements(List, N1, NList).
slice(List, N, NList):-
N > 0,
first_n_elements(List, N, NList).
first_n_elements(List, N, FirstN):-
length(FirstN, N),
append(FirstN, _, List).
last_n_elements(List, N, LastN) :-
length(LastN, N),
append(_, LastN, List).
- Output:
?- main. Total five weekend months:201 First five {year,month} pairs:[{1901,3},{1902,8},{1903,5},{1904,1},{1904,7}] Last five {year,month} pairs:[{2097,3},{2098,8},{2099,5},{2100,1},{2100,10}] Total years with no five weekend months:29 [1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096] true .
PureBasic
Procedure DateG(year.w, month.b, day)
;Returns the number of days before or after the earliest reference date
;in PureBasic's Date Library (1 Jan 1970) based on an assumed Gregorian calendar calculation
Protected days
days = (year) * 365 + (month - 1) * 31 + day - 1 - 719527 ;DAYS_UNTIL_1970_01_01 = 719527
If month >= 3
days - Int(0.4 * month + 2.3)
Else
year - 1
EndIf
days + Int(year/4) - Int(year/100) + Int(year/400)
ProcedureReturn days
EndProcedure
Procedure startsOnFriday(year, month)
;0 is Sunday, 1 is Monday, ... 5 is Friday, 6 is Saturday
Protected referenceDay = DayOfWeek(Date(1970, 1, 1, 0, 0, 0)) ;link to the first day in the PureBasic's date library
Protected resultDay = (((DateG(year, month, 1) + referenceDay) % 7) + 7) % 7
If resultDay = 5
ProcedureReturn #True
EndIf
EndProcedure
Procedure has31Days(month)
Select month
Case 1, 3, 5, 7 To 8, 10, 12
ProcedureReturn #True
EndSelect
EndProcedure
Procedure checkMonths(year)
Protected month, count
For month = 1 To 12
If startsOnFriday(year, month) And has31Days(month)
count + 1
PrintN(Str(year) + " " + Str(month))
EndIf
Next
ProcedureReturn count
EndProcedure
Procedure fiveWeekends()
Protected startYear = 1900, endYear = 2100, year, monthTotal, total
NewList yearsWithoutFiveWeekends()
For year = startYear To endYear
monthTotal = checkMonths(year)
total + monthTotal
;extra credit
If monthTotal = 0
AddElement(yearsWithoutFiveWeekends())
yearsWithoutFiveWeekends() = year
EndIf
Next
PrintN("Total number of months: " + Str(total) + #CRLF$)
PrintN("Years with no five-weekend months: " + Str(ListSize(yearsWithoutFiveWeekends())) )
EndProcedure
If OpenConsole()
fiveWeekends()
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf
- Sample output:
1901 3 1902 8 1903 5 1904 1 1904 7 . .(output clipped) . 2097 3 2098 8 2099 5 2100 1 2100 10 Total number of months: 201 Years with no five-weekend months: 29
Python
from datetime import (date,
timedelta)
DAY = timedelta(days=1)
START, STOP = date(1900, 1, 1), date(2101, 1, 1)
WEEKEND = {6, 5, 4} # Sunday is day 6
FMT = '%Y %m(%B)'
def five_weekends_per_month(start: date = START,
stop: date = STOP) -> list[date]:
"""Compute months with five weekends between dates"""
current_date = start
last_month = weekend_days = 0
five_weekends = []
while current_date < stop:
if current_date.month != last_month:
if weekend_days >= 15:
five_weekends.append(current_date - DAY)
weekend_days = 0
last_month = current_date.month
if current_date.weekday() in WEEKEND:
weekend_days += 1
current_date += DAY
return five_weekends
dates = five_weekends_per_month()
indent = ' '
print(f"There are {len(dates)} months of which the first and last five are:")
print(indent + ('\n' + indent).join(d.strftime(FMT) for d in dates[:5]))
print(indent + '...')
print(indent + ('\n' + indent).join(d.strftime(FMT) for d in dates[-5:]))
years_without_five_weekends_months = (STOP.year - START.year
- len({d.year for d in dates}))
print(f"\nThere are {years_without_five_weekends_months} years in the "
f"range that do not have months with five weekends")
Alternate Algorithm
The condition is equivalent to having a thirty-one day month in which the last day of the month is a Sunday.
LONGMONTHS = (1, 3, 5, 7, 8, 10, 12) # Jan Mar May Jul Aug Oct Dec
def five_weekends_per_month2(start: date = START,
stop: date = STOP) -> list[date]:
return [last_day
for year in range(start.year, stop.year)
for month in LONG_MONTHS
if (last_day := date(year, month, 31)).weekday() == 6] # Sunday
dates2 = five_weekends_per_month2()
assert dates2 == dates
- Sample output:
There are 201 months of which the first and last five are: 1901 03(March) 1902 08(August) 1903 05(May) 1904 01(January) 1904 07(July) ... 2097 03(March) 2098 08(August) 2099 05(May) 2100 01(January) 2100 10(October) There are 29 years in the range that do not have months with five weekends
Quackery
[ over 3 < if [ 1 - ]
dup 4 / over +
over 100 / -
swap 400 / +
swap 1 -
[ table
0 3 2 5 0 3
5 1 4 6 2 4 ]
+ + 7 mod ] is dayofweek ( day month year --> weekday )
[ 1 -
[ table
$ "January" $ "February"
$ "March" $ "April"
$ "May" $ "June"
$ "July" $ "August"
$ "September" $ "October"
$ "November" $ "December" ]
do ] is monthname ( monthnumber --> $ )
[] [] temp put
201 times
[ true temp put
i^ 1900 +
' [ 1 3 5 7 8 10 12 ]
witheach
[ 2dup swap
1 unrot dayofweek
5 = iff
[ false temp replace
over join nested
swap dip join ]
else drop ]
temp take iff
[ temp take
swap join
temp put ]
else drop ]
temp take swap
say "There are "
dup size echo
say " months with five weekends."
cr cr
5 split swap
say "Five weekends: "
witheach
[ do swap
monthname echo$
sp echo
i if say ", " ]
cr say " ..."
cr space 15 of echo$
-5 split nip
witheach
[ do swap
monthname echo$
sp echo
i if say ", " ]
cr cr
say "Years without five weekends: "
witheach
[ echo
i if say ", "
i^ 8 mod 7 = if
[ cr space 29 of echo$ ] ]
- Output:
There are 201 months with five weekends. Five weekends: March 1901, August 1902, May 1903, January 1904, July 1904 ... March 2097, August 2098, May 2099, January 2100, October 2100 Years without five weekends: 1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096
R
ms = as.Date(sapply(c(1, 3, 5, 7, 8, 10, 12),
function(month) paste(1900:2100, month, 1, sep = "-")))
ms = format(sort(ms[weekdays(ms) == "Friday"]), "%b %Y")
message("There are ", length(ms), " months with five weekends.")
message("The first five: ", paste(ms[1:5], collapse = ", "))
message("The last five: ", paste(tail(ms, 5), collapse = ", "))
Racket
#lang racket
(require srfi/19)
(define long-months '(1 3 5 7 8 10 12))
(define days #(sun mon tue wed thu fri sat))
(define (week-day date)
(vector-ref days (date-week-day date)))
(define (five-weekends-a-month start end)
(for*/list ([year (in-range start (+ end 1))]
[month long-months]
[date (in-value (make-date 0 0 0 0 31 month year 0))]
#:when (eq? (week-day date) 'sun))
date))
(define weekends (five-weekends-a-month 1900 2100))
(define count (length weekends))
(displayln (~a "There are " count " months with five weekends."))
(displayln "The first five are: ")
(for ([w (take weekends 5)])
(displayln (date->string w "~b ~Y")))
(displayln "The last five are: ")
(for ([w (drop weekends (- count 5))])
(displayln (date->string w "~b ~Y")))
- Output:
There are 201 months with five weekends. The first five are: Mar 1901 Aug 1902 May 1903 Jan 1904 Jul 1904 The last five are: Mar 2097 Aug 2098 May 2099 Jan 2100 Oct 2100
Raku
(formerly Perl 6)
# A month has 5 weekends iff it has 31 days and starts on Friday.
my @years = 1900 .. 2100;
my @has31 = 1, 3, 5, 7, 8, 10, 12;
my @happy = ($_ when *.day-of-week == 5 for (@years X @has31).map(-> ($y, $m) { Date.new: $y, $m, 1 }));
say 'Happy month count: ', +@happy;
say 'First happy months: ' ~ @happy[^5];
say 'Last happy months: ' ~ @happy[*-5 .. *];
say 'Dreary years count: ', @years - @happy».year.squish;
- Output:
Happy month count: 201 First happy months: 1901-03-01 1902-08-01 1903-05-01 1904-01-01 1904-07-01 Last happy months: 2097-03-01 2098-08-01 2099-05-01 2100-01-01 2100-10-01 Dreary years count: 29
REXX
version for newer REXXes
This version uses the latest enhancements to the DATE built-in function (which some older REXX interpreters don't support).
This version was written in such a way that it counts the number of days-of-the-week for each month, thereby
generalizing the problem without taking shortcuts specific to the task at hand.
Changing the task's requirements to find how many extended weekends (Fr-Sa-Su or Sa-Su-Mo) there are for
each month would entail changing only a single if statement.
/*REXX program finds months that contain five weekends (given a date range). */
month. =31; month.2=0 /*month days; February is skipped. */
month.4=30; month.6=30; month.9=30; month.11=30 /*all the months with thirty-days. */
parse arg yStart yStop . /*get the "start" and "stop" years.*/
if yStart=='' | yStart=="," then yStart= 1900 /*Not specified? Then use the default.*/
if yStop =='' | yStop =="," then yStop = 2100 /* " " " " " " */
years=yStop - yStart + 1 /*calculate the number of yrs in range.*/
haps=0 /*number of five weekends happenings. */
!.=0; @5w= 'five-weekend months' /*flag if a year has any five-weekends.*/
do y=yStart to yStop /*process the years specified. */
do m=1 for 12; wd.=0 /*process each month and also each year*/
do d=1 for month.m; dat_= y"-"right(m,2,0)'-'right(d,2,0)
parse upper value date('W', dat_, "I") with ? 3
wd.?=wd.?+1 /*? is the first two chars of weekday.*/
end /*d*/ /*WD.su = number of Sundays in a month.*/
if wd.su\==5 | wd.fr\==5 | wd.sa\==5 then iterate /*five weekends ?*/
say 'There are five weekends in' y date('M', dat_, "I")
haps=haps+1; !.y=1 /*bump counter; indicate yr has 5 WE's.*/
end /*m*/
end /*y*/
say
say "There were " haps ' occurrence's(haps) "of" @5w 'in year's(years) yStart"──►"yStop
say; #=0
do y=yStart to yStop; if !.y then iterate /*skip if OK.*/
#=#+1; say 'Year ' y " doesn't have any" @5wem'.'
end /*y*/
say
say "There are " # ' year's(#) "that haven't any" @5w 'in year's(years) yStart'──►'yStop
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
s: if arg(1)==1 then return arg(3); return word(arg(2) 's',1) /*pluralizer.*/
output when using the default inputs:
There are five weekends in 1901 March There are five weekends in 1902 August There are five weekends in 1903 May There are five weekends in 1904 January There are five weekends in 1904 July There are five weekends in 1905 December There are five weekends in 1907 March There are five weekends in 1908 May There are five weekends in 1909 January There are five weekends in 1909 October There are five weekends in 1910 July There are five weekends in 1911 December There are five weekends in 1912 March There are five weekends in 1913 August There are five weekends in 1914 May There are five weekends in 1915 January There are five weekends in 1915 October There are five weekends in 1916 December There are five weekends in 1918 March There are five weekends in 1919 August There are five weekends in 1920 October There are five weekends in 1921 July There are five weekends in 1922 December There are five weekends in 1924 August There are five weekends in 1925 May There are five weekends in 1926 January There are five weekends in 1926 October There are five weekends in 1927 July There are five weekends in 1929 March There are five weekends in 1930 August There are five weekends in 1931 May There are five weekends in 1932 January There are five weekends in 1932 July There are five weekends in 1933 December There are five weekends in 1935 March There are five weekends in 1936 May There are five weekends in 1937 January There are five weekends in 1937 October There are five weekends in 1938 July There are five weekends in 1939 December There are five weekends in 1940 March There are five weekends in 1941 August There are five weekends in 1942 May There are five weekends in 1943 January There are five weekends in 1943 October There are five weekends in 1944 December There are five weekends in 1946 March There are five weekends in 1947 August There are five weekends in 1948 October There are five weekends in 1949 July There are five weekends in 1950 December There are five weekends in 1952 August There are five weekends in 1953 May There are five weekends in 1954 January There are five weekends in 1954 October There are five weekends in 1955 July There are five weekends in 1957 March There are five weekends in 1958 August There are five weekends in 1959 May There are five weekends in 1960 January There are five weekends in 1960 July There are five weekends in 1961 December There are five weekends in 1963 March There are five weekends in 1964 May There are five weekends in 1965 January There are five weekends in 1965 October There are five weekends in 1966 July There are five weekends in 1967 December There are five weekends in 1968 March There are five weekends in 1969 August There are five weekends in 1970 May There are five weekends in 1971 January There are five weekends in 1971 October There are five weekends in 1972 December There are five weekends in 1974 March There are five weekends in 1975 August There are five weekends in 1976 October There are five weekends in 1977 July There are five weekends in 1978 December There are five weekends in 1980 August There are five weekends in 1981 May There are five weekends in 1982 January There are five weekends in 1982 October There are five weekends in 1983 July There are five weekends in 1985 March There are five weekends in 1986 August There are five weekends in 1987 May There are five weekends in 1988 January There are five weekends in 1988 July There are five weekends in 1989 December There are five weekends in 1991 March There are five weekends in 1992 May There are five weekends in 1993 January There are five weekends in 1993 October There are five weekends in 1994 July There are five weekends in 1995 December There are five weekends in 1996 March There are five weekends in 1997 August There are five weekends in 1998 May There are five weekends in 1999 January There are five weekends in 1999 October There are five weekends in 2000 December There are five weekends in 2002 March There are five weekends in 2003 August There are five weekends in 2004 October There are five weekends in 2005 July There are five weekends in 2006 December There are five weekends in 2008 August There are five weekends in 2009 May There are five weekends in 2010 January There are five weekends in 2010 October There are five weekends in 2011 July There are five weekends in 2013 March There are five weekends in 2014 August There are five weekends in 2015 May There are five weekends in 2016 January There are five weekends in 2016 July There are five weekends in 2017 December There are five weekends in 2019 March There are five weekends in 2020 May There are five weekends in 2021 January There are five weekends in 2021 October There are five weekends in 2022 July There are five weekends in 2023 December There are five weekends in 2024 March There are five weekends in 2025 August There are five weekends in 2026 May There are five weekends in 2027 January There are five weekends in 2027 October There are five weekends in 2028 December There are five weekends in 2030 March There are five weekends in 2031 August There are five weekends in 2032 October There are five weekends in 2033 July There are five weekends in 2034 December There are five weekends in 2036 August There are five weekends in 2037 May There are five weekends in 2038 January There are five weekends in 2038 October There are five weekends in 2039 July There are five weekends in 2041 March There are five weekends in 2042 August There are five weekends in 2043 May There are five weekends in 2044 January There are five weekends in 2044 July There are five weekends in 2045 December There are five weekends in 2047 March There are five weekends in 2048 May There are five weekends in 2049 January There are five weekends in 2049 October There are five weekends in 2050 July There are five weekends in 2051 December There are five weekends in 2052 March There are five weekends in 2053 August There are five weekends in 2054 May There are five weekends in 2055 January There are five weekends in 2055 October There are five weekends in 2056 December There are five weekends in 2058 March There are five weekends in 2059 August There are five weekends in 2060 October There are five weekends in 2061 July There are five weekends in 2062 December There are five weekends in 2064 August There are five weekends in 2065 May There are five weekends in 2066 January There are five weekends in 2066 October There are five weekends in 2067 July There are five weekends in 2069 March There are five weekends in 2070 August There are five weekends in 2071 May There are five weekends in 2072 January There are five weekends in 2072 July There are five weekends in 2073 December There are five weekends in 2075 March There are five weekends in 2076 May There are five weekends in 2077 January There are five weekends in 2077 October There are five weekends in 2078 July There are five weekends in 2079 December There are five weekends in 2080 March There are five weekends in 2081 August There are five weekends in 2082 May There are five weekends in 2083 January There are five weekends in 2083 October There are five weekends in 2084 December There are five weekends in 2086 March There are five weekends in 2087 August There are five weekends in 2088 October There are five weekends in 2089 July There are five weekends in 2090 December There are five weekends in 2092 August There are five weekends in 2093 May There are five weekends in 2094 January There are five weekends in 2094 October There are five weekends in 2095 July There are five weekends in 2097 March There are five weekends in 2098 August There are five weekends in 2099 May There are five weekends in 2100 January There are five weekends in 2100 October There were 201 occurrences of five-weekend months in years 1900──►2100 Year 1900 doesn't have any five-weekend months. Year 1906 doesn't have any five-weekend months. Year 1917 doesn't have any five-weekend months. Year 1923 doesn't have any five-weekend months. Year 1928 doesn't have any five-weekend months. Year 1934 doesn't have any five-weekend months. Year 1945 doesn't have any five-weekend months. Year 1951 doesn't have any five-weekend months. Year 1956 doesn't have any five-weekend months. Year 1962 doesn't have any five-weekend months. Year 1973 doesn't have any five-weekend months. Year 1979 doesn't have any five-weekend months. Year 1984 doesn't have any five-weekend months. Year 1990 doesn't have any five-weekend months. Year 2001 doesn't have any five-weekend months. Year 2007 doesn't have any five-weekend months. Year 2012 doesn't have any five-weekend months. Year 2018 doesn't have any five-weekend months. Year 2029 doesn't have any five-weekend months. Year 2035 doesn't have any five-weekend months. Year 2040 doesn't have any five-weekend months. Year 2046 doesn't have any five-weekend months. Year 2057 doesn't have any five-weekend months. Year 2063 doesn't have any five-weekend months. Year 2068 doesn't have any five-weekend months. Year 2074 doesn't have any five-weekend months. Year 2085 doesn't have any five-weekend months. Year 2091 doesn't have any five-weekend months. Year 2096 doesn't have any five-weekend months. There are 29 years that haven't any five─weekend months in years 1900──►2100
version for older REXXes
This version will work with any version of a REXX interpreter.
/*REXX program finds months that contain five weekends (given a date range). */
month. =31; month,2=0 /*month days; February is skipped. */
month.4=30; month.6=30; month.9=30; month.11=30 /*all the months with thirty-days. */
@months='January February March April May June July August September October November December'
parse arg yStart yStop . /*get the "start" and "stop" years.*/
if yStart=='' | yStart=="," then yStart= 1900 /*Not specified? Then use the default.*/
if yStop =='' | yStop =="," then yStop = 2100 /* " " " " " " */
years=yStop - yStart + 1 /*calculate the number of yrs in range.*/
haps=0 /*number of five weekends happenings. */
!.=0; @5w= 'five-weekend months' /*flag if a year has any five-weekends.*/
do y=yStart to yStop /*process the years specified. */
do m=1 for 12; wd.=0 /*process each month and also each year*/
do d=1 for month.m
?=dow(m,d,y) /*get the day-of-week for mm/dd/yyyy*/
wd.?=wd.?+1 /*?: 1=Sun, 2=Mon, 3=Tue ∙∙∙ 7=Sat.*/
end /*d*/
if wd.1\==5 | wd.6\==5 | wd.7\==5 then iterate /*not a weekend ? */
say 'There are five weekends in' y word(@months, m)
haps=haps+1; !.y=1 /*bump counter; indicate yr has 5 WE's.*/
end /*m*/
end /*y*/
say
say 'There were ' haps " occurrence"s(haps) 'of' @5w "in year"s(years) yStart'──►'yStop
#=0; say
do y=yStart to yStop; if !.y then iterate /*skip if OK.*/
#=#+1
say 'Year ' y " doesn't have any five-weekend months."
end /*y*/
say
say "There are " # ' year's(#) "that haven't any" @5w 'in year's(years) yStart'──►'yStop
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
dow: procedure; parse arg m,d,y; if m<3 then do; m=m+12; y=y-1; end
yL=left(y,2); yr=right(y,2); w=(d+(m+1)*26%10+yr+yr%4+yL%4+5*yL) // 7
if w==0 then w=7; return w /*Sunday=1, Monday=2, ... Saturday=7 */
/*──────────────────────────────────────────────────────────────────────────────────────*/
s: if arg(1)==1 then return arg(3); return word(arg(2) 's',1) /*pluralizer.*/
output is identical to the 1^{st} REXX version.
version short and focussed at the task description
/* REXX ***************************************************************
* Short(er) solution focussed at the task's description
* Only 7 months can have 5 full weekends
* and it's enough to test if the 1st day of the month is a Friday
* 30.08.2012 Walter Pachl
**********************************************************************/
Numeric digits 20
nr5fwe=0
years_without_5fwe=0
mnl='Jan Mar May Jul Aug Oct Dec'
ml='1 3 5 7 8 10 12'
Do j=1900 to 2100
year_has_5fwe=0
Do mi=1 To words(ml)
m=word(ml,mi)
jd=greg2jul(j m 1)
IF jd//7=4 Then Do /* 1st m j is a Friday */
nr5fwe=nr5fwe+1
year_has_5fwe=1
If j<=1905 | 2095<=j Then
Say word(mnl,mi) j 'has 5 full weekends'
End
End
If j=1905 Then Say '...'
if year_has_5fwe=0 Then years_without_5fwe=years_without_5fwe+1
End
Say ' '
Say nr5fwe 'occurrences of 5 full weekends in a month'
Say years_without_5fwe 'years without 5 full weekends'
exit
greg2jul: Procedure
/***********************************************************************
* Converts a Gregorian date to the corresponding Julian day number
* 19891101 Walter Pachl REXXified algorithm published in CACM
* (Fliegel & vanFlandern, CACM Vol.11 No.10 October 1968)
* 19891125 PA copy leapyear test into this to avoid the dependency
***********************************************************************/
numeric digits 12
Parse Arg yy mm d
If mm<1 | 12<mm Then Call err 'month ('mm') not within 1 to 12'
mdl='31' (28+leapyear(yy)) '31 30 31 30 31 31 30 31 30 31'
md=word(mdl,mm)
If d<1 | md<d Then Call err 'day ('d') not within 1 to' md
/***********************************************************************
* The published formula:
* res=d-32075+1461*(yy+4800+(mm-14)%12)%4+,
* 367*(mm-2-((mm-14)%12)*12)%12-3*((yy+4900+(mm-14)%12)%100)%4
***********************************************************************/
mma=(mm-14)%12
yya=yy+4800+mma
result=d-32075+1461*yya%4+367*(mm-2-mma*12)%12-3*((yya+100)%100)%4
Return result /* return the result */
leapyear: Return ( (arg(1)//4=0) & (arg(1)//100<>0) ) | (arg(1)//400=0)
- Output:
Mar 1901 has 5 full weekends Aug 1902 has 5 full weekends May 1903 has 5 full weekends Jan 1904 has 5 full weekends Jul 1904 has 5 full weekends Dec 1905 has 5 full weekends ... Jul 2095 has 5 full weekends Mar 2097 has 5 full weekends Aug 2098 has 5 full weekends May 2099 has 5 full weekends Jan 2100 has 5 full weekends Oct 2100 has 5 full weekends 201 occurrences of 5 full weekends in a month 29 years without 5 full weekends
shorter version
/*REXX program finds months that contain five weekends (given a date range). */
month. =31; month.2=0 /*month days; February is skipped. */
month.4=30; month.6=30; month.9=30; month.11=30 /*all the months with thirty-days. */
parse arg yStart yStop . /*get the "start" and "stop" years.*/
if yStart=='' | yStart=="," then yStart= 1900 /*Not specified? Then use the default.*/
if yStop =='' | yStop =="," then yStop = 2100 /* " " " " " " */
years=yStop - yStart + 1 /*calculate the number of yrs in range.*/
haps=0 /*number of five weekends happenings. */
!.=0; @5w= 'five-weekend months' /*flag if a year has any five-weekends.*/
do y=yStart to yStop /*process the years specified. */
do m=1 for 12; wd.=0 /*process each month and also each year*/
do d=1 for month.m; dat_= y"-"right(m, 2, 0)'-'right(d, 2, 0)
parse upper value date('W', dat_, "I") with ? 3
wd.?=wd.?+1 /*?: 1=Sun, 2=Mon, 3=Tue ∙∙∙ 7=Sat.*/
end /*d*/ /*WD.su=number of Sundays in the month.*/
if wd.su\==5 | wd.fr\==5 | wd.sa\==5 then iterate /*is this a weekend ? */
say 'There are five weekends in' y date('M', dat_, "I")
haps=haps+1; !.y=1 /*bump counter; indicate yr has 5 WE's.*/
end /*m*/
end /*y*/
say
say 'There were ' haps " occurrence"s(haps) 'of' @5w "in year"s(years) yStart'──►'yStop
#=0; say
do y=yStart to yStop; if !.y then iterate /*skip if OK.*/
#=#+1
say 'Year ' y " doesn't have any five-weekend months."
end /*y*/
say
say "There are " # ' year's(#) "that haven't any" @5w 'in year's(years) yStart'──►'yStop
/*stick a fork in it, we're all done. */
output is identical to the 1^{st} REXX version.
shorter and more focused
This REXX version takes advantage that a month with five full weekends must start on a Friday and have 31 days.
/*REXX program finds months that contain five weekends (given a date range). */
month. =31 /*days in "all" the months. */
month.2=0; month.4=0; month.6=0; month.9=0; month.11=0 /*not 31 day months.*/
month.4=30; month.6=30; month.9=30; month.11=30 /*all the months with thirty-days. */
parse arg yStart yStop . /*get the "start" and "stop" years.*/
if yStart=='' | yStart=="," then yStart= 1900 /*Not specified? Then use the default.*/
if yStop =='' | yStop =="," then yStop = 2100 /* " " " " " " */
years=yStop - yStart + 1 /*calculate the number of yrs in range.*/
haps=0 /*number of five weekends happenings. */
!.=0; @5w= 'five-weekend months' /*flag if a year has any five-weekends.*/
do y=yStart to yStop /*process the years specified. */
do m=1 for 12; if month.m==0 then iterate /*only test 31-day months.*/
dat_= y"-"right(m,2,0)'-01' /*get the date in the desired format. */
if left(date('W',dat_,"I"),2)\=='Fr' then iterate /*isn't not a Friday? */
say 'There are five weekends in' y date('M', dat_, "I")
haps=haps+1; !.y=1 /*bump counter; indicate yr has 5 WE's.*/
end /*m*/
end /*y*/
say
say 'There were ' haps " occurrence"s(haps) 'of' @5w "in year"s(years) yStart'──►'yStop
#=0; say
do y=yStart to yStop; if !.y then iterate /*skip if OK.*/
#=#+1
say 'Year ' y " doesn't have any five-weekend months."
end /*y*/
say
say "There are " # ' year's(#) "that haven't any" @5w 'in year's(years) yStart'──►'yStop
output is identical to the 1^{st} REXX version.
Ring
sum = 0
month = list(12)
mo = [4,0,0,3,5,1,3,6,2,4,0,2]
mon = [31,28,31,30,31,30,31,31,30,31,30,31]
mont = ["January","February","March","April","May","June",
"July","August","September","October","November","December"]
for year = 1900 to 2100
if year < 2100 leap = year - 1900 else leap = year - 1904 ok
m = ((year-1900)%7) + floor(leap/4) % 7
oldsum = sum
for n = 1 to 12
month[n] = (mo[n] + m) % 7
x = (month[n] + 1) % 7
if x = 2 and mon[n] = 31 sum += 1 see "" + year + "-" + mont[n] + nl ok
next
if sum = oldsum see "" + year + "-" + "(none)" + nl ok
next
see "Total : " + sum + nl
Output:
1900-(none) 1901-March 1902-August 1903-May 1904-July 1905-December 1906-(none) ... 2093-May 2094-January 2094-October 2095-July 2096-(none) 2097-March 2098-August 2099-May 2100-January 2100-October Total : 201
RPL
RPL code | Comment |
---|---|
≪ OVER 3 < - R→B DUP 4 / OVER 100 / - OVER 400 / + + B→R { 0 3 2 5 0 3 5 1 4 6 2 4 } ROT GET + + 7 MOD ≫ 'DAYoW' STO ≪ {} (0,0) 1900 2100 FOR y 1 CF 1 12 FOR m IF { 1 3 5 7 8 10 12 } m POS THEN IF 1 m y DAYoW 5 == THEN IF DUP RE 5 < OVER RE 195 > OR THEN y m R→C ROT SWAP + SWAP END 1 + 1 SF END END NEXT IF 1 FC? THEN (0,1) + END NEXT ≫ 'MW5WE' STO |
DAYoW ( j mm aaaa -- day ) if ( m < 3 ) { y -= 1; } return (y + y/4 - y/100 + y/400 + t[m-1] + d) % 7 ; MW5WE ( -- {(first)..(last)} (months,years) ) For each year in range reset flag scan long months if month starts on Fridays if in first 5 or last 5 ones store as (yyyy,mm) increase month counter, set flag if flag not set, count the year |
- Output:
2: { (1901,3) (1902,8) (1903,5) (1904,1) (1904,7) (2097,3) (2098,8) (2099,5) (2100,1) (2100,10) } 1: (201,29)
Ruby
require 'date'
# The only case where the month has 5 weekends is when the last day
# of the month falls on a Sunday and the month has 31 days.
LONG_MONTHS = [1,3,5,7,8,10,12]
YEARS = (1900..2100).to_a
dates = YEARS.product(LONG_MONTHS).map{|y, m| Date.new(y,m,31)}.select(&:sunday?)
years_4w = YEARS - dates.map(&:year)
puts "There are #{dates.size} months with 5 weekends from 1900 to 2100:"
puts dates.first(5).map {|d| d.strftime("%b %Y") }, "..."
puts dates.last(5).map {|d| d.strftime("%b %Y") }
puts "There are #{years_4w.size} years without months with 5 weekends:"
puts years_4w.join(", ")
Output
There are 201 months with 5 weekends from 1900 to 2100: Mar 1901 Aug 1902 May 1903 Jan 1904 Jul 1904 ... Mar 2097 Aug 2098 May 2099 Jan 2100 Oct 2100 There are 29 years without months with 5 weekends: 1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096
Run BASIC
preYear = 1900
for yyyy = 1900 to 2100
for mm = 1 to 12 ' go thru all 12 months
dayOne$ = mm;"-01-";yyyy ' First day of month
n = date$(dayOne$) ' Days since 1700
dow = 1 + (n mod 7) ' Day of Week month begins
m1 = mm '
n1 = n + 27 ' find end of month starting with 27th day
while m1 = mm ' if month changes we have the end of the month
n1 = n1 + 1
n$ = date$(n1)
m1 = val(left$(n$,2))
wend
mmDays = n1 - n ' Days in the Month
if dow = 4 and mmDays = 31 then ' test for 5 weeks
count = count + 1
print using("###",count);" ";yyyy;"-";left$("0";mm,2)
end if
next mm
if preCount = count then
noCount = noCount + 1 ' count years that have none
print yyyy;" has none ";noCount
end if
preCount = count
next yyyy
- Output:
1900 has none 1 1 1901-03 2 1902-08 3 1903-05 4 1904-01 5 1904-07 6 1905-01 1906 has none 2 7 1907-03 ........ 196 2095-07 2096 has none 29 197 2097-03 198 2098-08 199 2099-05 200 2100-01 201 2100-01
Rust
extern crate chrono;
use chrono::prelude::*;
/// Months with 31 days
const LONGMONTHS: [u32; 7] = [1, 3, 5, 7, 8, 10, 12];
/// Get all the tuples (year, month) in wich there is five Fridays, five Saturdays and five Sundays
/// between the years start and end (inclusive).
fn five_weekends(start: i32, end: i32) -> Vec<(i32, u32)> {
let mut out = vec![];
for year in start..=end {
for month in LONGMONTHS.iter() {
// Five weekends if a 31-days month starts with a Friday.
if Local.ymd(year, *month, 1).weekday() == Weekday::Fri {
out.push((year, *month));
}
}
}
out
}
fn main() {
let out = five_weekends(1900, 2100);
let len = out.len();
println!(
"There are {} months of which the first and last five are:",
len
);
for (y, m) in &out[..5] {
println!("\t{} / {}", y, m);
}
println!("...");
for (y, m) in &out[(len - 5..)] {
println!("\t{} / {}", y, m);
}
}
#[test]
fn test() {
let out = five_weekends(1900, 2100);
assert_eq!(out.len(), 201);
}
Scala
import java.util.Calendar._
import java.util.GregorianCalendar
import org.scalatest.{FlatSpec, Matchers}
class FiveWeekends extends FlatSpec with Matchers {
case class YearMonth[T](year: T, month: T)
implicit class CartesianProd[T](val seq: Seq[T]) {
def x(other: Seq[T]) = for(s1 <- seq; s2 <- other) yield YearMonth(year=s1,month=s2)
def -(other: Seq[T]): Seq[T] = seq diff other
}
def has5weekends(ym: { val year: Int; val month: Int}) = {
val date = new GregorianCalendar(ym.year, ym.month-1, 1)
date.get(DAY_OF_WEEK) == FRIDAY && date.getActualMaximum(DAY_OF_MONTH) == 31
}
val expectedFirstFive = Seq(
YearMonth(1901,3), YearMonth(1902,8), YearMonth(1903,5), YearMonth(1904,1), YearMonth(1904,7))
val expectedFinalFive = Seq(
YearMonth(2097,3), YearMonth(2098,8), YearMonth(2099,5), YearMonth(2100,1), YearMonth(2100,10))
val expectedNon5erYears = Seq(1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962,
1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035,
2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096)
"Five Weekend Algorithm" should "match specification" in {
val months = (1900 to 2100) x (1 to 12) filter has5weekends
months.size shouldBe 201
months.take(5) shouldBe expectedFirstFive
months.takeRight(5) shouldBe expectedFinalFive
(1900 to 2100) - months.map(_.year) shouldBe expectedNon5erYears
}
}
Seed7
$ include "seed7_05.s7i";
include "time.s7i";
const proc: main is func
local
var integer: months is 0;
var time: firstDayInMonth is time.value;
begin
for firstDayInMonth.year range 1900 to 2100 do
for firstDayInMonth.month range 1 to 12 do
if daysInMonth(firstDayInMonth) = 31 and dayOfWeek(firstDayInMonth) = 5 then
writeln(firstDayInMonth.year <& "-" <& firstDayInMonth.month lpad0 2);
incr(months);
end if;
end for;
end for;
writeln("Number of months:" <& months);
end func;
- Output:
1901-03 1902-08 1903-05 1904-01 1904-07 1905-12 ... 2095-07 2097-03 2098-08 2099-05 2100-01 2100-10 Number of months:201
Sidef
require('DateTime');
var happymonths = [];
var workhardyears = [];
var longmonths = [1, 3, 5, 7, 8, 10, 12];
range(1900, 2100).each { |year|
var countmonths = 0;
longmonths.each { |month|
var dt = %s'DateTime'.new(
year => year,
month => month,
day => 1
);
if (dt.day_of_week == 5) {
countmonths++;
var yearfound = dt.year;
var monthfound = dt.month_name;
happymonths.append(join(" ", yearfound, monthfound));
}
}
if (countmonths == 0) {
workhardyears.append(year);
}
}
say "There are #{happymonths.len} months with 5 full weekends!";
say "The first 5 and the last 5 of them are:";
say happymonths.first(5).join("\n");
say happymonths.last(5).join("\n");
say "No long weekends in the following #{workhardyears.len} years:";
say workhardyears.join(",");
- Output:
There are 201 months with 5 full weekends! The first 5 and the last 5 of them are: 1901 March 1902 August 1903 May 1904 January 1904 July 2097 March 2098 August 2099 May 2100 January 2100 October No long weekends in the following 29 years: 1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096
Simula
! TRANSLATION OF FREEBASIC VERSION ;
BEGIN
INTEGER PROCEDURE WD(M, D, Y); INTEGER M, D, Y;
BEGIN
! ZELLERISH
! 0 = SUNDAY, 1 = MONDAY, 2 = TUESDAY, 3 = WEDNESDAY
! 4 = THURSDAY, 5 = FRIDAY, 6 = SATURDAY
;
IF M < 3 THEN ! IF M = 1 OR M = 2 THEN ;
BEGIN
M := M + 12;
Y := Y - 1
END;
WD := MOD(Y + (Y // 4)
- (Y // 100)
+ (Y // 400)
+ D + ((153 * M + 8) // 5), 7)
END WD;
! ------=< MAIN >=------
! ONLY MONTHS WITH 31 DAY CAN HAVE FIVE WEEKENDS
! THESE MONTHS ARE: JANUARY, MARCH, MAY, JULY, AUGUST, OCTOBER, DECEMBER
! IN NR: 1, 3, 5, 7, 8, 10, 12
! THE 1E DAY NEEDS TO BE ON A FRIDAY (= 5)
;
TEXT PROCEDURE MONTHNAMES(M); INTEGER M;
MONTHNAMES :- IF M = 1 THEN "JANUARY"
ELSE IF M = 2 THEN "FEBRUARY"
ELSE IF M = 3 THEN "MARCH"
ELSE IF M = 4 THEN "APRIL"
ELSE IF M = 5 THEN "MAY"
ELSE IF M = 6 THEN "JUNE"
ELSE IF M = 7 THEN "JULY"
ELSE IF M = 8 THEN "AUGUST"
ELSE IF M = 9 THEN "SEPTEMBER"
ELSE IF M = 10 THEN "OCTOBER"
ELSE IF M = 11 THEN "NOVEMBER"
ELSE IF M = 12 THEN "DECEMBER"
ELSE NOTEXT;
INTEGER M, YR, TOTAL, I, J;
INTEGER ARRAY YR_WITHOUT(1:200);
TEXT ANSWER;
FOR YR := 1900 STEP 1 UNTIL 2100 DO ! GREGORIAN CALENDAR ;
BEGIN
ANSWER :- NOTEXT;
FOR M := 1 STEP 2 UNTIL 12 DO
BEGIN
IF M = 9 THEN M := 8;
IF WD(M, 1, YR) = 5 THEN
BEGIN
ANSWER :- ANSWER & MONTHNAMES(M) & ", ";
TOTAL := TOTAL + 1
END
END;
IF ANSWER =/= NOTEXT THEN
BEGIN
OUTIMAGE;
OUTINT(YR, 4); OUTTEXT(" | ");
OUTTEXT(ANSWER.SUB(1, ANSWER.LENGTH - 2)) ! GET RID OF EXTRA " ," ;
END
ELSE
BEGIN
I := I + 1;
YR_WITHOUT(I) := YR
END
END;
OUTIMAGE;
OUTTEXT("NR OF MONTH FOR 1900 TO 2100 THAT HAS FIVE WEEKENDS ");
OUTINT(TOTAL, 0);
OUTIMAGE;
OUTIMAGE;
OUTINT(I, 0);
OUTTEXT(" YEARS DON'T HAVE MONTHS WITH FIVE WEEKENDS");
OUTIMAGE;
FOR J := 1 STEP 1 UNTIL I DO
BEGIN
OUTINT(YR_WITHOUT(J), 0); OUTCHAR(' ');
IF MOD(J, 8) = 0 THEN OUTIMAGE
END;
OUTIMAGE
END.
- Output:
1901 | MARCH 1902 | AUGUST 1903 | MAY 1904 | JANUARY, JULY 1905 | DECEMBER ... 2095 | JULY 2097 | MARCH 2098 | AUGUST 2099 | MAY 2100 | JANUARY, OCTOBER NR OF MONTH FOR 1900 TO 2100 THAT HAS FIVE WEEKENDS 201 29 YEARS DON'T HAVE MONTHS WITH FIVE WEEKENDS 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096
Stata
clear
set obs `=tm(2101m1)-tm(1900m1)'
gen month=tm(1900m1)+_n-1
format %tm month
gen day=dofm(month)
keep if dofm(month+1)-day==31 & dow(day)==5
drop day
count
201
list in f/5, noobs noheader
+--------+
| 1901m3 |
| 1902m8 |
| 1903m5 |
| 1904m1 |
| 1904m7 |
+--------+
list in -5/l, noobs noheader
+---------+
| 2097m3 |
| 2098m8 |
| 2099m5 |
| 2100m1 |
| 2100m10 |
+---------+
Tcl
package require Tcl 8.5
set months {}
set years {}
for {set year 1900} {$year <= 2100} {incr year} {
set count [llength $months]
foreach month {Jan Mar May Jul Aug Oct Dec} {
set date [clock scan "$month/01/$year" -format "%b/%d/%Y" -locale en_US]
if {[clock format $date -format %u] == 5} {
# Month with 31 days that starts on a Friday => has 5 weekends
lappend months "$month $year"
}
}
if {$count == [llength $months]} {
# No change to number of months; year must've been without
lappend years $year
}
}
puts "There are [llength $months] months with five weekends"
puts [join [list {*}[lrange $months 0 4] ... {*}[lrange $months end-4 end]] \n]
puts "There are [llength $years] years without any five-weekend months"
puts [join $years ","]
- Output:
There are 201 months with five weekends Mar 1901 Aug 1902 May 1903 Jan 1904 Jul 1904 ... Mar 2097 Aug 2098 May 2099 Jan 2100 Oct 2100 There are 29 years without any five-weekend months 1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096
TUSCRIPT
$$ MODE TUSCRIPT
LOOP year=1900,2100
LOOP month="1'3'5'7'8'10'12"
SET dayofweek=DATE (number,1,month,year,nummer)
IF (dayofweek==5) PRINT year,"-",month
ENDLOOP
ENDLOOP
- Output:
1901-3 1902-8 1903-5 1904-1 1904-7 1905-12 1907-3 1908-5 1909-1 1909-10 1910-7 1911-12 1912-3 1913-8 1914-5 1915-1 1915-10 1916-12 1918-3 1919-8 1920-10 1921-7 1922-12 1924-8 1925-5 1926-1 1926-10 1927-7 1929-3 1930-8 1931-5 1932-1 1932-7 1933-12 1935-3 1936-5 1937-1 1937-10 1938-7 1939-12 1940-3 1941-8 1942-5 1943-1 1943-10 1944-12 1946-3 1947-8 1948-10 1949-7 1950-12 1952-8 1953-5 1954-1 1954-10 1955-7 1957-3 1958-8 1959-5 1960-1 1960-7 1961-12 1963-3 1964-5 1965-1 1965-10 1966-7 1967-12 1968-3 1969-8 1970-5 1971-1 1971-10 1972-12 1974-3 1975-8 1976-10 1977-7 1978-12 1980-8 1981-5 1982-1 1982-10 1983-7 1985-3 1986-8 1987-5 1988-1 1988-7 1989-12 1991-3 1992-5 1993-1 1993-10 1994-7 1995-12 1996-3 1997-8 1998-5 1999-1 1999-10 2000-12 2002-3 2003-8 2004-10 2005-7 2006-12 2008-8 2009-5 2010-1 2010-10 2011-7 2013-3 2014-8 2015-5 2016-1 2016-7 2017-12 2019-3 2020-5 2021-1 2021-10 2022-7 2023-12 2024-3 2025-8 2026-5 2027-1 2027-10 2028-12 2030-3 2031-8 2032-10 2033-7 2034-12 2036-8 2037-5 2038-1 2038-10 2039-7 2041-3 2042-8 2043-5 2044-1 2044-7 2045-12 2047-3 2048-5 2049-1 2049-10 2050-7 2051-12 2052-3 2053-8 2054-5 2055-1 2055-10 2056-12 2058-3 2059-8 2060-10 2061-7 2062-12 2064-8 2065-5 2066-1 2066-10 2067-7 2069-3 2070-8 2071-5 2072-1 2072-7 2073-12 2075-3 2076-5 2077-1 2077-10 2078-7 2079-12 2080-3 2081-8 2082-5 2083-1 2083-10 2084-12 2086-3 2087-8 2088-10 2089-7 2090-12 2092-8 2093-5 2094-1 2094-10 2095-7 2097-3 2098-8 2099-5 2100-1 2100-10
uBasic/4tH
' ------=< MAIN >=------
' only months with 31 day can have five weekends
' these months are: January, March, May, July, August, October, December
' in nr: 1, 3, 5, 7, 8, 10, 12
' the 1e day needs to be on a friday (= 5)
For y = 1900 To 2100 ' Gregorian calendar
a = 0
For m = 1 To 12 Step 2
If m = 9 Then m = 8
If Func(_wd(m , 1 , y)) = 5 Then
If a Then
Print ", ";
Else
Print y; " | ";
a = 1
EndIf
GoSub m*10
t = t + 1
EndIf
Next
If a Then
Print
Else
i = i + 1
@(i) = y
EndIf
Next
Print
Print "Number of months from 1900 to 2100 that have five weekends: ";t
Print
Print i;" years don't have months with five weekends:"
Print
For j = 1 To i
Print @(j); " ";
If (j % 8) = 0 Then Print
Next
Print
End
_wd Param(3)
' Zellerish
' 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday
' 4 = Thursday, 5 = Friday, 6 = Saturday
If a@ < 3 Then ' If a@ = 1 Or a@ = 2 Then
a@ = a@ + 12
c@ = c@ - 1
EndIf
Return ((c@ + (c@ / 4) - (c@ / 100) + (c@ / 400) + b@ + ((153 * a@ + 8) / 5)) % 7)
10 Print "January"; : Return
20 Print "February"; : Return
30 Print "March"; : Return
40 Print "April"; : Return
50 Print "May"; : Return
60 Print "June"; : Return
70 Print "July"; : Return
80 Print "August"; : Return
90 Print "September"; : Return
100 Print "October"; : Return
110 Print "November"; : Return
120 Print "December"; : Return
- Output:
1901 | March 1902 | August 1903 | May 1904 | January, July ... 2099 | May 2100 | January, October Number of months from 1900 to 2100 that have five weekends: 201 29 years don't have months with five weekends: 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096 0 OK, 0:888
UNIX Shell
This is a 2-step-solution:
- create a file with 1-month calendars for all the required month and years
- feed this file to an awk-script to look for the months with 5 weekends.
echo "Creating cal-file..."
echo > cal.txt
for ((y=1900; y <= 2100; y++)); do
for ((m=1; m <= 12; m++)); do
#echo $m $y
cal -m $m $y >> cal.txt
done
done
ls -la cal.txt
echo "Looking for month with 5 weekends:"
awk -f 5weekends.awk cal.txt
See also: awk and Calendar.
Try it out online: compileonline.com
VBA
Option Explicit
Sub Main()
Dim y As Long, m As Long, t As String, cpt As Long, cptm As Long
For y = 1900 To 2100
t = vbNullString
For m = 1 To 12 Step 2
If m = 9 Then m = 8
If Weekday(DateSerial(y, m, 1)) = vbFriday Then
t = t & ", " & m
cptm = cptm + 1
End If
Next
If t <> "" Then
Debug.Print y & t
Else
cpt = cpt + 1
End If
Next
Debug.Print "There is " & cptm & " months with five full weekends from the year 1900 through 2100"
Debug.Print "There is " & cpt & " years which don't have months with five weekends"
End Sub
- Output:
1901, 3 1902, 8 1903, 5 1904, 1, 7 1905, 12 1907, 3 1908, 5 1909, 1, 10 1910, 7 ...... 2090, 12 2092, 8 2093, 5 2094, 1, 10 2095, 7 2097, 3 2098, 8 2099, 5 2100, 1, 10 There is 201 months with five full weekends from the year 1900 through 2100 There is 29 years which don't have months with five weekends
VBScript
For y = 1900 To 2100
For m = 1 To 12
d = DateSerial(y, m + 1, 1) - 1
If Day(d) = 31 And Weekday(d) = vbSunday Then
WScript.Echo y & ", " & MonthName(m)
i = i + 1
End If
Next
Next
WScript.Echo vbCrLf & "Total = " & i & " months"
Wren
import "./date" for Date
import "./seq" for Lst
var dates = []
var years = []
for (y in 1900..2100) {
var hasFive = false
for (m in 1..12) {
var fri = 0
var sat = 0
var sun = 0
for (d in 1..Date.monthLength(y, m)) {
var d = Date.new(y, m, d)
var dow = d.dayOfWeek
if (dow == 5) {
fri = fri + 1
} else if (dow == 6) {
sat = sat + 1
} else if (dow == 7) {
sun = sun + 1
}
}
var fd = Date.new(y, m, 1)
if (fri == 5 && sat == 5 && sun == 5) {
dates.add(fd)
hasFive = true
}
}
if (!hasFive) years.add(y)
}
Date.default = "mmm|-|yyyy"
System.print("Between 1900 and 2100:-")
System.print(" There are %(dates.count) months that have five full weekends.")
System.print(" The first 5 are:")
for (i in 0..4) System.print(" %(dates[i])")
System.print(" and the last 5 are:")
for (i in -5..-1) System.print(" %(dates[i])")
System.print("\n There are %(years.count) years that do not have at least one five-weekend month, namely:")
var chunks = Lst.chunks(years, 10)
for (i in 0...chunks.count) System.print(" %(chunks[i])")
- Output:
Between 1900 and 2100:- There are 201 months that have five full weekends. The first 5 are: Mar-1901 Aug-1902 May-1903 Jan-1904 Jul-1904 and the last 5 are: Mar-2097 Aug-2098 May-2099 Jan-2100 Oct-2100 There are 29 years that do not have at least one five-weekend month, namely: [1900, 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962] [1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035] [2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096]
XPL0
include c:\cxpl\codes; \intrinsic 'code' declarations
func WeekDay(Year, Month, Day); \Return day of week (0=Sat 1=Sun..6=Fri)
int Year, Month, Day;
[if Month<=2 then [Month:= Month+12; Year:= Year-1];
return rem((Day + (Month+1)*26/10 + Year + Year/4 + Year/100*6 + Year/400) / 7);
]; \WeekDay
int MonthTbl, Year, I, C;
[MonthTbl:= [1, 3, 5, 7, 8, 10, 12]; \months with 31 days
C:= 0;
for Year:= 1900 to 2100 do
for I:= 0 to 6 do \for all the 31-day months...
if WeekDay(Year, MonthTbl(I), 1) = 6 then \first of month is a Friday
[C:= C+1; \count this year
if C<=5 or C>201-5 then \show first 5 and last 5 years
[IntOut(0, Year); ChOut(0, ^ );
IntOut(0, MonthTbl(I)); CrLf(0);
];
];
IntOut(0, C); CrLf(0); \show number of years
\Count and show all years that don't have any 5-weekend months
C:= 0;
for Year:= 1900 to 2100 do
[for I:= 0 to 6 do \for all the 31-day months...
if WeekDay(Year, MonthTbl(I), 1) = 6 \Friday\ then
I:= 10; \bail out of 'for' loop
if I<10 then \'for' loop completed
[if (C&$F) = 0 then CrLf(0); \(format 16 years per line)
C:= C+1; \ without finding a 5-weekend
IntOut(0, Year); ChOut(0, ^ ); \ so show the year
];
];
CrLf(0); IntOut(0, C); CrLf(0); \show number of years
]
- Output:
1901 3 1902 8 1903 5 1904 1 1904 7 2097 3 2098 8 2099 5 2100 1 2100 10 201 1900 1906 1917 1923 1928 1934 1945 1951 1956 1962 1973 1979 1984 1990 2001 2007 2012 2018 2029 2035 2040 2046 2057 2063 2068 2074 2085 2091 2096 29
zkl
Months with five weekends:
var [const] D=Time.Date, r=L();
foreach y,m in ([1900..2100],[1..12]){
if (D.daysInMonth(y,m)==31 and 5==D.weekDay(y,m,1)) r.append(String(y,"-",m))
}
Alternatively, in a functional style using list comprehensions:
r:=[[(y,m); [1900..2100];
[1..12],{D.daysInMonth(y,m)==31 and 5==D.weekDay(y,m,1)};
{String(y,"-",m)}]];
- Output:
r.len() //-->201 r[0,5] //-->L("1901-3","1902-8","1903-5","1904-1","1904-7") r[-5,*] //-->L("2097-3","2098-8","2099-5","2100-1","2100-10")
Extra credit: Years with no five day weekends:
var [const] D=Time.Date, r=L();
foreach y in ([1900..2100]){ yes:=True;
foreach m in ([1..12]){
if (D.daysInMonth(y,m)==31 and 5==D.weekDay(y,m,1)) { yes=False; break; }
}
if (yes) r.append(y)
}
Alternatively:
var yes, r=[[(y,m);
[1900..2100];
[1..12],{if(m==1)yes=True; if(D.daysInMonth(y,m)==31 and 5==D.weekDay(y,m,1)) yes=False; True},
{if (m==12) yes else False};
{y}]]
Bit of a sleaze using a global var (yes) to hold state. First filter: On the first month, reset global, note if month has five weekends, always pass. Second filter: fail if any five day weekends and ignore all months other than December.
- Output:
r.len() //-->29 r //-->L(1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,...)