Last Friday of each month: Difference between revisions

=={{header|Fortran}}==

Algorithm: compute day of week for last day of month, then subtract just enough to get to the preceding friday. Do this for each month. To simplify computations further, we only need to compute day of week of january 1st (the others are found by adding month lengths). Since day of week need only be known modulo 7, we do not compute modulo at all except once when subtracting.

<lang fortran>program fridays

integer :: days(1:12) = (/31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31/)

integer :: year, k, y

read *, year

if (mod(year, 400) == 0 .or. (mod(year, 4) == 0 .and. mod(year, 100) /= 0)) days(2) = 29

y = year - 1

k = 41 + y + y/4 + 6*(y/100) + y/400

do m = 1, 12

k = k + days(m)

print "(I4,A1,I2.2,A1,I2)", year, '-', m, '-', days(m) - mod(k + 3, 7)

end do

end program</lang>