Last Friday of each month: Difference between revisions
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2012-11-30
2012-12-28</pre>
=={{header|AWK}}==
<lang AWK>
# syntax: GAWK -f LAST_FRIDAY_OF_EACH_MONTH.AWK year
# converted from Fortran
BEGIN {
split("31,28,31,30,31,30,31,31,30,31,30,31",daynum_array,",") # days per month in non leap year
year = ARGV[1]
if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0)) {
daynum_array[2] = 29
}
y = year - 1
k = 44 + y + int(y/4) + int(6*(y/100)) + int(y/400)
for (m=1; m<=12; m++) {
k += daynum_array[m]
d = daynum_array[m] - (k%7)
printf("%04d-%02d-%02d\n",year,m,d)
}
exit(0)
}
</lang>
<p>output:</p>
<pre>
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28
</pre>
=={{header|C}}==
Doesn't work with Julian calendar (then again, you probably don't need to plan your weekends for middle ages).
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