Calculating nCr mod p efficiently when n is very large

Question

I need to calculate nCr mod p efficiently. Right now, I have written this piece of code, but it exceeds the time limit. Please suggest a more optimal solution.

For my case, p = 10^9 + 7 and 1 ≤ n ≤ 100000000

I have to also make sure that there is no overflow as nCr mod p is guaranteed to fit in 32 bit integer, however n! may exceed the limit.

def nCr(n,k):
    r = min(n-k,k)
    k = max(n-k,k)
    res = 1
    mod = 10**9 + 7

    for i in range(k+1,n+1):
        res = res * i
        if res > mod:
            res = res % mod

    res = res % mod
    for i in range(1,r+1):
        res = res/i
    return res

PS : Also I think my code may not be completely correct. However, it seems to work for small n correctly. If its wrong, please point it out !

Answer 1

From http://apps.topcoder.com/wiki/display/tc/SRM+467 :

long modPow(long a, long x, long p) {
    //calculates a^x mod p in logarithmic time.
    long res = 1;
    while(x > 0) {
        if( x % 2 != 0) {
            res = (res * a) % p;
        }
        a = (a * a) % p;
        x /= 2;
    }
    return res;
}

long modInverse(long a, long p) {
    //calculates the modular multiplicative of a mod m.
    //(assuming p is prime).
    return modPow(a, p-2, p);
}
long modBinomial(long n, long k, long p) {
// calculates C(n,k) mod p (assuming p is prime).

    long numerator = 1; // n * (n-1) * ... * (n-k+1)
    for (int i=0; i<k; i++) {
        numerator = (numerator * (n-i) ) % p;
    }

    long denominator = 1; // k!
    for (int i=1; i<=k; i++) {
        denominator = (denominator * i) % p;
    }

    // numerator / denominator mod p.
    return ( numerator* modInverse(denominator,p) ) % p;
}

Notice that we use modpow(a, p-2, p) to compute the mod inverse. This is in accordance to Fermat's Little Theorem which states that (a^(p-1) is congruent to 1 modulo p) where p is prime. It thus implies that (a^(p-2) is congruent to a^(-1) modulo p).

C++ to Python conversion should be easy :)