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Here's a solution with no limits (beyond the memory used by memoize). It answers largest(1000, 1001, 1002, 1003, 1004, 1005) in 19 secs (the answer is 199,999 by the way). If you turn off memoize it takes much longer, although probably just days rather than centuries, It's based on a previous perlmonks challenge: How to generate restricted partitions of an integer.

It's definitely not golf! Is there a golfish answer that runs in similar time?

#! /usr/bin/perl use strict; use warnings; my @all; use Memoize; memoize('breakup'); print largest(@ARGV)."\n"; sub largest { my $small = shift; my %need; @all = reverse @_; my $highest = 0; for my $m_target (1..($small-1)) { # find the smallest number ($target) which can be made from the in +put # numbers and which gives remainder $m_target when divided by $sma +ll. my $target = $m_target; while ($target += $small) { if (0) # change to 1 to see details { my @b = breakup($target, 0); nice($target, @b); } last if breakup($target, 0); } $target -= $small; print "$target == $m_target (mod $small)\n"; $highest = $target if $highest < $target; } return $highest; } sub nice { my $target = shift; print "---------------\n\ntarget = $target\n"; foreach my $sol (@_) { print join(", ", @$sol)."\n"; } } sub breakup { my ($target, $sub) = @_; # $count++; my $cur = $all[$sub]; my @solns; if ($target == $cur) { push(@solns, [$cur]); } if ($target >= $cur) { push(@solns, map {[$cur, @$_]} (breakup($target - $cur, $sub))); } push(@solns, breakup($target, $sub + 1)) unless ($sub == $#all); return @solns; }

I'm going to bed now so I won't explain in detail. Hopefully someone will have posted a more golfish solution by the time I get to work tomorrow and I won't have to bother.

In reply to Re^3: Golf: Buying with exact change by fergal
in thread Golf: Buying with exact change by blokhead

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