Re: Finding all divisors of an integer

[ysth]

Here's an implementation of gri6507's suggestion (using ordinary perl integers, not bigints):

use warnings;
use strict;
use integer;

$|=1;
print "Enter number: ";
chomp(my $x = <STDIN>);
die "That's not my kind of number!" if $x=~/\D/ || $x/1 ne $x;

if ($x <= 1) { print "$x\n"; exit }

# get prime factors
my %fact;
my $fact = 2;
do {
    ++$fact while $x/$fact*$fact != $x;
    ++$fact{$fact};
    $x /= $fact;
} while ($x > 1);

# apply all combinations
my @fact = 1;
while (($fact,my $count) = each %fact) {
    my $index;
    @fact = map $_ * $fact**(++$index/@fact % ($count+1)),
                (@fact) x ($count+1);
}
print "@{[sort {$a<=>$b} @fact]}\n";

[Anonymous Monk]

oi? Newbie programer here.. dividing the number for primefactorization? sounds like this subroutine i cooked up to do my homework...

sub primefactor ($) {
 my $number=shift;
 my @factors=();
 for (my $i=2; $i<=$number;) {
  unless ($number%$i) {
   $number/=$i; push @factors, $i;
  } else {$i++}
 }
 @factors&&return(@factors);
 return($number);
}