in reply to Finding divisors from factors

danaj:

For my version, I took the powerset version, used tail recursion to turn it into a loop, and used a hash to remove duplicates continuously. I don't expect that it's all that fast with the extra hash operations and a sort at the end, but it's simple enough:

#!/usr/bin/perl use strict; use warnings; my @all_factors = (1, 2, 2, 3, 5, 11, 277412413); my @divisors = build_factors(@all_factors); print join(", ", @divisors),"\n"; sub build_factors { my @orig = @_; my %new = (1=>0); while (my $factor = shift @orig) { @new{map{$factor*$_} keys %new}=0; } return sort {$a<=>$b} keys %new; }

...roboticus

When your only tool is a hammer, all problems look like your thumb.