🚀 go-pugleaf

Bruce Kaskel wrote:
> 

> So after you have finished the n-th iteration, you have multiplied your
> number by n, removed the trailing zeros, and subtracted 10000 until the
> number is not greater than 10000. Are you claiming that you end up with
> a number with the same last digit as n! ?
> 
> Strictly speaking this is not correct.
> 
> Take the 3125-th iteration. Your algorithm yields 995.
> But the rightmost non-zero digit of 3125! is not 5.
> 
> --Bruce Kaskel

Duh...uh..well, sorry to shoot you down but:

The original USACO problem (where this idea came from...and whoever
posted it to this group needs a serious course in ethics) specified that
you only needed to test up to 1000!....So, his idea works within the
program specs 

BTW-I didn't enter the competition, but I also solved it that particular
way...with a little more clumsy approach using strings instead of
longints, but the basic concept is the same - you only need to save the
last 4 non-zero digits, because you are only ever multiplying the result
by something less than 1000.

Tim 
tass@armory.com