# Five weekends

**Five weekends**

You are encouraged to solve this task according to the task description, using any language you may know.

The month of October in 2010 has five Fridays, five Saturdays, and five Sundays.

**The 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).

- See also

## [edit] 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

## [edit] AppleScript

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:

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

## [edit] 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%

)

## [edit] 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

## [edit] ALGOL 68

Note: This specimen retains the original Fortran coding style. difffive_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

## [edit] 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

## [edit] 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

## [edit] 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:

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

## [edit] 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

## [edit] C#

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

## [edit] 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: " ,))

## [edit] 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.

### [edit] 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

## [edit] 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.

## [edit] 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]

## [edit] 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

## [edit] 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

## [edit] 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

## [edit] 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

## [edit] 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 "nr of month for 1900 to 2100 that has five weekends";total

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

' empty keyboard buffer

While InKey <> "" : Var _key_ = 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

## [edit] 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" ]

## [edit] 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

## [edit] 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

## [edit] 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

## [edit] Haskell

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

## [edit] 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.

## [edit] Icon and Unicon

link datetime,printf

procedure main(A) # five weekends

printf( "There are %d months from %d to %d with five full weekends.\n",

*(L := fiveweekends(s := 1900, f := 2100)), s,f)

printf("The first and last five such months are:\n")

every printf("%s\n",L[1 to 5]|"..."|L[-4 to 0])

printf( "There are %d years without such months as follows:\n",

*(M := Bonus(s,f,L)))

every printf("%s\n",!M)

end

procedure fiveweekends(start,finish)

L := [] # months years with five weekends FRI-SUN

every year := start to finish & month := 1 to 12 do

if month = (2|4|6|9|11) then next

else if julian(month,1,year) % 7 = 4 then

put(L,sprintf("%d-%d-1",year,month))

return L

end

procedure Bonus(start,finish,fwe)

every insert(Y := set(), start to finish)

every insert(F := set(), integer(!fwe ? tab(find("-"))))

return sort(Y--F)

end

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

## [edit] 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

## [edit] 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];

//dreizig tage habt september...

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:

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

## [edit] 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

## [edit] JavaScript

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

### [edit] Alternative Version

This 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

## [edit] 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.

## [edit] Julia

isdefined(:Date) || using Dates

const wday = Dates.Fri

const lo = 1900

const hi = 2100

const showres = 5

mfive = recur(Date(lo, 1):Month(1):Date(hi, 12)) do m

Dates.daysinmonth(m) == 31 && Dates.dayofweek(m) == wday

end

println("Considering the years from ", lo, " to ", hi, ".\n")

println("There are ", length(mfive), " months having 5 3-day weekends.")

println("The first ", showres, " such months are:")

for m in mfive[1:showres]

println(" ", Dates.monthname(m), " ", Dates.year(m))

end

println("\nThe last ", showres, " such months are:")

for m in mfive[end-showres+1:end]

println(" ", Dates.monthname(m), " ", Dates.year(m))

end

print("\nThere are ", length(filter(y -> !(y in year(mfive)), lo:hi)))

println(" years that have no such months.")

This code uses the `Dates`

module, which is being incorporated into Julian's standard library with the current development version (`0.4`). I've used `isdefined`

to make this code good for the current stable version (`0.3`) as well as for future releases. If `Dates`

is not installed on your instance of Julian try `Pkg.add("Dates")`

from the `REPL`.

- Output:

Considering the years from 1900 to 2100. There are 201 months having 5 3-day weekends. The first 5 such months are: March 1901 August 1902 May 1903 January 1904 July 1904 The last 5 such months are: March 2097 August 2098 May 2099 January 2100 October 2100 There are 29 years that have no such months.

## [edit] 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

## [edit] 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,

## [edit] Mathematica

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}

## [edit] 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

## [edit] 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]]

## [edit] MUMPS

Usage:

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

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

## [edit] 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

## [edit] Nim

import times

const LongMonths = {mJan, mMar, mMay, mJul, mAug, mOct, mDec}

var timeinfo = getLocalTime getTime()

timeinfo.monthday = 1

var sumNone = 0

for year in 1900..2100:

var none = true

for month in LongMonths:

timeinfo.year = year

timeinfo.month = month

if getLocalTime(timeInfoToTime timeinfo).weekday == dFri:

echo month," ",year

none = false

if none: inc sumNone

echo "Years 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

## [edit] 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

## [edit] Oforth

func: 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]]

## [edit] 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

};

## [edit] Pascal

See Delphi

## [edit] 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

## [edit] 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

## [edit] Phix

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)=6 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}

## [edit] 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)

## [edit] 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,

## [edit] 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

## [edit] 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

## [edit] 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

## [edit] Python

from datetime import timedelta, date

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 fiveweekendspermonth(start=START, stop=STOP):

'Compute months with five weekends between dates'

when = start

lastmonth = weekenddays = 0

fiveweekends = []

while when < stop:

year, mon, _mday, _h, _m, _s, wday, _yday, _isdst = when.timetuple()

if mon != lastmonth:

if weekenddays >= 15:

fiveweekends.append(when - DAY)

weekenddays = 0

lastmonth = mon

if wday in WEEKEND:

weekenddays += 1

when += DAY

return fiveweekends

dates = fiveweekendspermonth()

indent = ' '

print('There are %s months of which the first and last five are:' % len(dates))

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:]))

print('\nThere are %i years in the range that do not have months with five weekends'

% len(set(range(START.year, STOP.year)) - {d.year for d in dates}))

**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 fiveweekendspermonth2(start=START, stop=STOP):

return [date(yr, month, 31)

for yr in range(START.year, STOP.year)

for month in LONGMONTHS

if date(yr, month, 31).timetuple()[6] == 6 # Sunday

]

dates2 = fiveweekendspermonth2()

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

## [edit] 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 = ", "))

## [edit] 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 " weeks 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 weeks 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

## [edit] REXX

### [edit] version for newer REXXes

This version uses the latest enhancements to the **DATE** built-in function (which some older REXX interpreters don't support).

Programming justification note: the inclusion of leapyear checking is present even though it wasn't needed as

February can be excluded from the criteria of having five weekends. Having the leap year routine included

allows for a general-purpose stratagem of solving these types of problems without taking shortcuts to the

solution (although shortcuts are generally desirable). If there is a change in the criteria, the support for

more (or less) checks can be supported easily, as well as supporting more criteria.

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 with 5 weekends in them (given a date range)*/

month. =31 /*month days; Feb. is done later.*/

month.4=30; month.6=30; month.9=30; month.11=30 /*30-day months*/

parse arg yStart yStop . /*get the "start" & "stop" years.*/

if yStart=='' then yStart=1900 /*if not specified, use default. */

if yStop =='' then yStop =2100 /* " " " " " */

years=yStop-yStart+1 /*calculate the # of yrs in range*/

haps=0 /*num of five weekends happenings*/

!.=0 /*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, each year. */

if m==2 then month.2=28+leapyear(y) /*handle #days in Feb*/

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 1st 2 chars of weekday*/

end /*d*/ /*WD.su = # of Sundays in a month*/

if wd.su\==5 | wd.fr\==5 | wd.sa\==5 then iterate /*5 W.E.s?*/

say 'There are five weekends in' y date('M', dat_, "I")

haps=haps+1; !.y=1 /*bump ctr; indicate yr has 5 WEs*/

end /*m*/

end /*y*/

say

say 'There were ' haps " occurrence"s(haps) 'of five-weekend months 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 five-weekend months."

end /*y*/

say

say "There are " # ' year's(#) "that haven't any five─weekend months in year"s(years) yStart'──►'yStop

exit /*stick a fork in it, we're done.*/

/*──────────────────────────────────LEAPYEAR subroutine─────────────────*/

leapyear: procedure; parse arg y /*year could be: Y, YY, YYY, YYYY*/

if length(y)==2 then y=left(right(date(),4),2)y /*adjust for YY year.*/

if y//4\==0 then return 0 /* not ÷ by 4? Not a leap year.*/

return y//100\==0 | y//400==0 /*apply 100 and 400 year rule. */

/*──────────────────────────────────S subroutine────────────────────────*/

s: if arg(1)==1 then return arg(3); return word(arg(2) 's',1) /*plural*/

- Output:

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

### [edit] version for older REXXes

This version will work with any version of a REXX interpreter.

/*REXX program finds months with 5 weekends in them (given a date range)*/

month. =31 /*month days; Feb. is done later.*/

month.4=30; month.6=30; month.9=30; month.11=30 /*30-day months*/

@months='January February March April May June July August September October November December'

parse arg yStart yStop . /*get the "start" & "stop" years.*/

if yStart=='' then yStart=1900 /*if not specified, use default. */

if yStop =='' then yStop =2100 /* " " " " " */

years=yStop-yStart+1 /*calculate the # of yrs in range*/

haps=0 /*num of five weekends happenings*/

!.=0 /*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 in each year*/

if m==2 then month.2=28+leapyear(y) /*handle # days in Feb.*/

do d=1 for month.m

?=dow(m,d,y) /*get day-of-week for mm/dd/yyyy.*/

wd.?=wd.?+1 /*?: 1=Sun, 2=Mon, ∙∙∙ 7=Sat */

end /*d*/

if wd.1\==5 | wd.6\==5 | wd.7\==5 then iterate /*5 WEs ? */

say 'There are five weekends in' y word(@months,m)

haps=haps+1; !.y=1 /*bump ctr; indicate yr has 5 WEs*/

end /*m*/

end /*y*/

say

say 'There were ' haps " occurrence"s(haps) 'of five-weekend months 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 five─weekend months in year"s(years) yStart'──►'yStop

exit /*stick a fork in it, we're done.*/

/*──────────────────────────────────DOW─────────────────────────────────*/

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*/

/*──────────────────────────────────LEAPYEAR subroutine─────────────────*/

leapyear: procedure; parse arg y /*year could be: Y, YY, YYY, YYYY*/

if length(y)==2 then y=left(right(date(),4),2)y /*adjust for YY year.*/

if y//4\==0 then return 0 /* not ÷ by 4? Not a leap year.*/

return y//100\==0 | y//400==0 /*apply 100 and 400 year rule. */

/*──────────────────────────────────S subroutine────────────────────────*/

s: if arg(1)==1 then return arg(3); return word(arg(2) 's',1) /*plural*/

**output** is identical to the first version.

### [edit] 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

### [edit] shorter version

/*REXX program finds months with 5 weekends in them (given a date range)*/

month. =31 /*month days; Feb. is skipped. */

month.4=30; month.6=30; month.9=30; month.11=30 /*30-day months*/

yStart=1900; yStop=2100 /*define start and stop years. */

haps=0 /*num of five weekends happenings*/

!.=0 /*if a year has any five-weekends*/

do y=yStart to yStop /*process the years specified. */

do m=1 for 12; wd.=0 /*each month except Feb, each yr.*/

if m==2 then iterate /*if month is February, skip it. */

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 1st 2 chars of tge weekday*/

end /*d*/ /*WD.su=# of Sundays in the month*/

if wd.su\==5 | wd.fr\==5 | wd.sa\==5 then iterate /*5 W.E.s?*/

say 'There are five weekends in' y date('M', dat_, "I")

haps=haps+1; !.y=1 /*bump ctr; indicate yr has 5 WEs*/

end /*m*/

end /*y*/

say

say 'There were ' haps " occurrences of five-weekend months in 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 five-weekend months."

end /*y*/

say

say "There are " # " years that haven't any five─weekend months in years" yStart'──►'yStop

/*stick a fork in it, we're done.*/

**output** is identical to the first REXX version.

### [edit] 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 with 5 weekends in them (given a date range)*/

month.=31; yStart=1900; yStop=2100 /*month days; range of years. */

month.2=0; month.4=0; month.6=0; month.9=0; month.11=0 /*¬31 day months*/

haps=0 /*num of five weekends happenings*/

!.=0 /*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 /*test 31-day mons*/

dat_=y"-"right(m,2,0)'-01' /*get date in the proper format. */

if left(date('W',dat_,"I"),2)\=='Fr' then iterate /*Friday?*/

say 'There are five weekends in' y date('M', dat_, "I")

haps=haps+1; !.y=1 /*bump ctr; indicate yr has 5 WEs*/

end /*m*/

end /*y*/

say

say 'There were ' haps " occurrences of five-weekend months in 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 five-weekend months."

end /*y*/

say

say "There are " # " years that haven't any five─weekend months in years" yStart'──►'yStop

/*stick a fork in it, we're done.*/

**output** is identical to the first REXX version.

## [edit] 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.

dates = []

1900.upto(2100) do |year|

1.upto(12) do |month|

d = Date.new(year, month, -1) # -1 is last day of month

dates << d if d.sunday? && d.day == 31

end

end

puts "There are #{dates.size} months with 5 weekends from 1900 to 2100:"

puts dates.first(5).map { |d| d.strftime("%b %Y") }

puts "..."

puts dates.last(5).map { |d| d.strftime("%b %Y") }

years_with_5w = dates.map(&:year)

years = (1900..2100).to_a - years_with_5w

puts "There are #{years.size} years without months with 5 weekends:"

puts years.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

## [edit] 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

## [edit] 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

}

}

## [edit] 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

## [edit] 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[0..4].join("\n");

say happymonths[-5..-1].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

## [edit] 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

## [edit] 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

## [edit] 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

Else

i = i + 1

@(i) = y

EndIf

Next

Print "Number of months from 1900 to 2100 that have five weekends: ";t

Print i;" years don't have months with five weekends:"

For j = 1 To i

Print @(j); " ";

If (j % 8) = 0 Then Print

Next

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

## [edit] 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

## [edit] 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"

## [edit] 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

## [edit] 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,...)

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