%mod compatible ways of generating Binomial Coeffi

问题:

I would like to optimize a part of my program where I'm calculating the sum of Binomial Coefficients up to K. i.e.

C(N,0) + C(N,1) + ... + C(N,K)

Since the values go beyond the data type (long long) can support, I'm to calculate values mod M and was looking for procedures to do that.

Currently, I've done it with Pascal's Triangle but it seems to be taking a bit of load. so, I was wondering if there's any other efficient way to do this. I've considered Lucas' Theorem, although M I have is already large enough so that C(N,k) goes out of hand!

Any pointers as how can I do this differently, maybe calculate the whole sum altogether with some other neat expression of teh sum. If not I'll leave it with the Pascal's Triangle method itself.

Thank you,

Here is what I have so far O(N^2) :

#define MAX 1000000007
long long NChooseK_Sum(int N, int K){
    vector<long long> prevV, V;
    prevV.push_back(1);     prevV.push_back(1);
    for(int i=2;i<=N;++i){
            V.clear();
            V.push_back(1);
            for(int j=0;j<(i-1);++j){
                    long long val = prevV[j] + prevV[j+1];
                    if(val >= MAX)
                            val %= MAX;
                    V.push_back(val);
            }
            V.push_back(1);
            prevV = V;
    }
    long long res=0;
    for(int i=0;i<=K;++i){
            res+=V[i];
            if(res >= MAX)
                    res %= MAX;
    }
    return res;
}

回答1:

An algorithm that performs a linear number of arithmetic bignum operations is

def binom(n):
    nck = 1
    for k in range(n + 1):  # 0..n
        yield nck
        nck = (nck * (n - k)) / (k + 1)

This uses division, but modulo a prime p, you can accomplish much the same thing by multiplying by the solution i to the equation i * (k + 1) = 1 mod p. The value i can be found in a logarithmic number of arithmetic ops via the extended Euclidean algorithm.