Both approaches will save time, but the first one is very prone to integer overflow. Approach 1: This approach will generate result in shortest time (in at most n/2 iterations), and the possibility of overflow can be reduced by doing the multiplications carefully: long long C(int n, int r) { if(r > n – r) … Read more
So, here is how you can solve your problem. Of course you know the formula: comb(n,k) = n!/(k!*(n-k)!) = (n*(n-1)*…(n-k+1))/k! (See http://en.wikipedia.org/wiki/Binomial_coefficient#Computing_the_value_of_binomial_coefficients) You know how to compute the numerator: long long res = 1; for (long long i = n; i > n- k; –i) { res = (res * i) % p; } Now, … Read more
Your formula is totally wrong, it’s supposed to be fact(n)/fact(r)/fact(n-r), but that is in turn a very inefficient way to compute it. See Fast computation of multi-category number of combinations and especially my comments on that question. (Oh, and please reopen that question also so I can answer it properly) The single-split case is actually … Read more