The word "combination" in the title brought Algorithm::Combinatorics to mind. I'm not sure if duplicate sums, e.g. the several 5+5+10+80 combinations in the third example, should all be shown but I have eliminated them. This code
use strict;
use warnings;
use feature qw{ say };
use Algorithm::Combinatorics qw{ combinations };
use List::Util qw{ sum };
my @tests = (
{
target => 100,
values => [ 1, 99, 2, 40, 50, 100, 60, 90, 3, 5, 95, 100 ],
},
{
target => 10,
values => [ 1, 3, 2, 4 ],
},
{
target => 100,
values => [ 5, 5, 5, 5, 10, 15, 80, 99 ],
},
);
foreach my $rhTest ( @tests )
{
say
qq{\nFind sums from },
join( q{, }, @{ $rhTest->{ values } } ),
qq{ making $rhTest->{ target }};
say for do {
my %seen;
grep { ! $seen{ $_ } ++ }
grep { $_ == $rhTest->{ target } }
@{ $rhTest->{ values } };
};
for my $sumsOf ( 2 .. scalar @{ $rhTest->{ values } } )
{
my $combIter = combinations( $rhTest->{ values }, $sumsOf );
my %seen;
while ( my $raComb = $combIter->next() )
{
next if $seen{ join q{+}, sort { $a <=> $b } @{ $raComb }
+} ++;
say join q{+}, @{ $raComb }
if $rhTest->{ target } == sum @{ $raComb };
}
}
}
produces
Find sums from 1, 99, 2, 40, 50, 100, 60, 90, 3, 5, 95, 100 making 100
100
1+99
40+60
5+95
2+3+95
2+90+3+5
2+40+50+3+5
Find sums from 1, 3, 2, 4 making 10
1+3+2+4
Find sums from 5, 5, 5, 5, 10, 15, 80, 99 making 100
5+15+80
5+5+10+80
5+5+5+5+80
I hope this is helpful.
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