I have no idea what language is that, but I assume its using some big
integer data type which makes slow, but you don't have to do % 100003 at the
end

using this two properties of modular arithmetic the values never grow big
(x + y) % M == (x % M + y % M) % M
(x * y) % M == (x % M * y % M) % M

for the choose, it wont work as is, if it is still slow you can use a choose
of the form

choose(n,k) = chose(n-1,k-1) + choose(n-1,k)

you can apply modular arithmetic to that formulation, and even cache. I'll
let you figure out the base cases.


Carlos Guía


On Sat, May 22, 2010 at 9:18 PM, Chris Carton <[email protected]> wrote:

> I missed this question during the competition because I mis-read it - doh!
> Then after I figured out what it was asking I came up with a solution that
> solves the small case but it runs too slow on the large set.  What
> optimization I am missing?
>
>
>
> def XchooseN(x,n):
>         return factorial(x) / ( factorial(n) * factorial(x-n) )
>
> cache = {}
>
> def solve(N, K):
>         global cache
>         if K < 1: return 0
>         if K == 1 or K == 2: return 1
>         if (K == (N-1)): return 1
>         if (N,K) in cache: return cache[(N,K)]
>         r = sum((solve(K,Kp) * XchooseN(N-K-1,K-Kp-1) for Kp in
> xrange(2*K-N, K)))
>         cache[(N,K)] = r
>         return r
>
>
> if __name__ == '__main__':
>     while test_case < test_case_count:
>         r = sum((solve(N,K) for K in xrange(1,N)))
>         print "Case #%d: %s" % (test_case, r % 100003)
>
>
>
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