Question

This is actually a SPOJ problem: WAYS

Now this is a very easy task what we need to do is to compute the Central binomial coefficients.

However the problem setter includes a very notorious source limit of 120 bytes, so my question is how to get past that source code limit in the languages that are allowed?

Answer 1

Assuming, that C(2n,n) = (2n)!/(n!)^2 = (2n(2n-1)/n^2) * C(2(n-1),n-1) = ((4n-2)/n)*C(2(n-1),n-1) here is function, which calculates central binomial:

int f(int n)
{
    return n==1? 2 : f(n-1)*(4*n-2)/n;
}

Edit: Here is probably shortest code:

int f(int n){return n<2?2:f(n-1)*(4*n-2)/n;}

It is only 44 characters.