Factors of an integer: Difference between revisions

===Using the prime library===

<lang ruby>

require 'prime'

def factors m

return [1] if 1==m

primes, powers = Prime.prime_division(m).transpose

ranges = powers.map{|n| (0..n).to_a}

ranges[0].product( *ranges[1..-1] ).

map{|es| primes.zip(es).map{|p,e| p**e}.reduce :*}.

sort

end

[1, 7, 45, 100].each{|n| p factors n}

</lang>

Output:

<pre>

[1]

[1, 7]

[1, 3, 5, 9, 15, 45]

[1, 2, 4, 5, 10, 20, 25, 50, 100]

</pre>