组合数的求法 （n<=1e8 可以过来看）

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<cstdio>
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();
}

 

转载于:https://www.cnblogs.com/Andromeda-Galaxy/p/9882649.html