C(n,m) =n!/(m!* (n-m)!  );

o(n) 求　1-m的逆元　

o(n) 求　n的阶乘

代码实现  

https://www.cnblogs.com/linyujun/p/5194189.html

#include
     
      const int N = 200000 + 5;const int MOD = (int)1e9 + 7;int F[N], Finv[N], inv[N];//F是阶乘，Finv是逆元的阶乘 void init(){    inv[1] = 1;    for(int i = 2; i < N; i ++){        inv[i] = (MOD - MOD / i) * 1ll * inv[MOD % i] % MOD;    }    F[0] = Finv[0] = 1;    for(int i = 1; i < N; i ++){        F[i] = F[i-1] * 1ll * i % MOD;        Finv[i] = Finv[i-1] * 1ll * inv[i] % MOD;    }}int comb(int n, int m){
   //comb(n, m)就是C(n, m)     if(m < 0 || m > n) return 0;    return F[n] * 1ll * Finv[n - m] % MOD * Finv[m] % MOD;}int main(){    init();}