From: Kevin Smith <sent@...>
Date: 2001-03-10T12:19:03+09:00
Subject: [ruby-talk:12367] Re: ...and the challenge

gotoken@math.sci.hokudai.ac.jp wrote:
>On my machine, yours takes 1.15 sec. 
>The following takes 0.55 sec here.
>
>        n=1_000_000
>        nums=[]
>        for a in 0..20
>	  a0 = 2**a
>	  for b in 0..13
>	    b0 = 3**b
>	    for c in 0..9
>	      x = a0 * b0 * 5**c
>	      nums.push x if x <= n
>	    end
>	  end
>	end
>        nums.sort!
>        for x in nums do p x end

Pure speculation: Could we speed it up a bit by 
breaking out of each loop (or at least the 
innermost loop) after "n" was exceeded? If so, it 
would probably be even faster to loop by 9, 13, 
20 instead of 20, 13, 9.

Of course, speed isn't everything. Kudos for 
creating this code which is clean and 
straightforward!

Kevin