Doomsday rule: Difference between revisions

=={{header|BASIC}}==

Note that 1/6/1800 is actually a Monday, not a Sunday. As far as I can tell this is actually the correct day.

<lang BASIC>10 DIM D$(7): FOR I=1 TO 7: READ D$(I): NEXT I

20 DIM D(12,1): FOR I=0 TO 1: FOR J=1 TO 12: READ D(J,I): NEXT J,I

30 READ Y: IF Y=0 THEN END ELSE READ M,D

40 PRINT USING "##/##/####: ";M;D;Y;

50 C=Y\100: R=Y MOD 100

60 S=R\12: T=R MOD 12

70 A=(5*(C AND 3)+2) MOD 7

80 B=(S+T+(T\4)+A) MOD 7

90 PRINT D$((B+D-D(M,-(Y MOD 4=0 AND (Y MOD 100<>0 OR Y MOD 400=0)))+7) MOD 7+1)

100 GOTO 30

110 DATA Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday

120 DATA 3,7,7,4,2,6,4,1,5,3,7,5

130 DATA 4,1,7,4,2,6,4,1,5,3,7,5

140 DATA 1800,1,6, 1875,3,29, 1915,12,7, 1970,12,23

150 DATA 2043,5,14, 2077,2,12, 2101,4,2, 0</lang>

{{out}}

<pre> 1/ 6/1800: Monday

3/29/1875: Monday

12/ 7/1915: Tuesday

12/23/1970: Wednesday

5/14/2043: Thursday

2/12/2077: Friday

4/ 2/2101: Saturday</pre>

=={{header|BCPL}}==

Note that 1/6/1800 is actually a Monday, not a Sunday. As far as I can tell this is actually the correct day.

<lang bcpl>get "libhdr"

let dayname(n) =

n = 0 -> "Sunday",

n = 1 -> "Monday",

n = 2 -> "Tuesday",

n = 3 -> "Wednesday",

n = 4 -> "Thursday",

n = 5 -> "Friday",

n = 6 -> "Saturday",

dayname(n rem 7)

let leap(year) = year rem 4 = 0 & (year rem 100 ~= 0 | year rem 400 = 0)

let weekday(y, m, d) = valof

$( let leapdoom = table 4,1,7,4,2,6,4,1,5,3,7,5

let normdoom = table 3,7,7,4,2,6,4,1,5,3,7,5

let c = y / 100 and r = y rem 100

let s = r / 12 and t = r rem 12

let an = (5 * (c rem 4) + 2) rem 7

let doom = (s + t + (t/4) + an) rem 7

let anchor = (leap(y) -> leapdoom, normdoom)!(m-1)

resultis (doom + d - anchor + 7) rem 7

$)

let start() be

$( writef("January 6, 1800 was on a %S.*N", dayname(weekday(1800, 1, 6)))

writef("March 29, 1875 was on a %S.*N", dayname(weekday(1875, 3, 29)))

writef("December 7, 1915 was on a %S.*N", dayname(weekday(1915, 12, 7)))

writef("December 23, 1970 was on a %S.*N", dayname(weekday(1970, 12, 23)))

writef("May 14, 2043 will be on a %S.*N", dayname(weekday(2043, 5, 14)))

writef("February 12, 2077 will be on a %S.*N", dayname(weekday(2077, 2, 12)))

writef("April 2, 2101 will be on a %S.*N", dayname(weekday(2101, 4, 2)))

$)</lang>

{{out}}

<pre>January 6, 1800 was on a Monday.

March 29, 1875 was on a Monday.

December 7, 1915 was on a Tuesday.

December 23, 1970 was on a Wednesday.

May 14, 2043 will be on a Thursday.

February 12, 2077 will be on a Friday.

April 2, 2101 will be on a Saturday.</pre>