AcWing 886 求组合数 II 题解 （求组合数）

#include<iostream>
#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e5 + 10,  mod = 1e9 + 7;

int heat[N], inheat[N];//分别储存i的阶乘和i的阶乘的逆元

int qmi(int a, int k, int p){//快速幂
    int res = 1;
    while(k){
        if(k & 1) res = (ll)res * a % mod; 
        a = (ll)a * a % mod;
        k >>= 1;
    }
    return res;
}

int main()
{
    heat[0] = inheat[0] = 1;
    for(int i = 1; i <= N; i ++ ){
        heat[i] = (ll)heat[i - 1] * i % mod;
        inheat[i] = (ll)inheat[i - 1] * qmi(i, mod - 2, mod) % mod;
    }
    
    int n;
    cin>>n;
    while(n -- ){
        int a, b;
        cin>>a>>b;
        int t = (ll)heat[a] * inheat[a - b] % mod * inheat[b] % mod;//公式
        cout<<t<<endl;
    }
    return 0;
}

标签：组合,int,题解,ll,II,inheat,heat,res,mod
来源： https://blog.csdn.net/qiaodxs/article/details/120236207