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#!perl
################################################################################
=comment
Perl Weekly Challenge 013
=========================
Challenge #1
------------
Write a script to print the date of last *Friday* of every month of a given
year. For example, if the given year is *2019* then it should print the follow-
ing:
2019/01/25
2019/02/22
2019/03/29
2019/04/26
2019/05/31
2019/06/28
2019/07/26
2019/08/30
2019/09/27
2019/10/25
2019/11/29
2019/12/27
=cut
################################################################################
#--------------------------------------#
# Copyright © 2019 PerlMonk Athanasius #
#--------------------------------------#
use strict;
use warnings;
use Const::Fast;
use DateTime;
use Regexp::Common;
const my @DAYS_IN_MONTH => ( 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 );
const my $DAYS_IN_WEEK => 7;
const my $DEFAULT_YEAR => 2019;
const my $FEBRUARY_INDEX => 1;
const my @OFFSET_1ST_FRI => ( 4, 3, 2, 1, 0, 6, 5 );
const my $USAGE => "USAGE: perl $0 <year>";
$| = 1;
MAIN:
{
my $year = get_year();
my $dt = DateTime->new(year => $year, month => 1, day => 1);
my $first = 1 + $OFFSET_1ST_FRI[ $dt->day_of_week - 1 ];
my @days = @DAYS_IN_MONTH;
++$days[ $FEBRUARY_INDEX ] if $dt->is_leap_year();
print "\nLast Fridays in each month of $year:\n\n";
for my $month (0 .. 11)
{
my $days = $days[ $month ];
my $last = $first;
$last += $DAYS_IN_WEEK until $last > $days;
$first = $last - $days;
$last -= $DAYS_IN_WEEK;
printf "%4d/%02d/%02d\n", $year, ($month + 1), $last;
}
}
sub get_year
{
scalar @ARGV <= 1
or die "\n$USAGE\n";
my $year = $ARGV[0] // $DEFAULT_YEAR;
$year =~ /^$RE{num}{int}$/
or die "\nInvalid year '$year': must be an integer\n";
return $year;
}
################################################################################
|
