//硬币（除法运算用逆元代替） //2 1 0 #include <iostream> #include <algorithm> #include <functional> using namespace std; const int mod = 1e9 + 7; long long power(int x, int y, const int mod) {     int t = 1;     for (x %= mod; y > 0; x = (long long)x * x % mod, y /= 2) {         if (y % 2 != 0) {             t = (long long)t * x % mod;         }     }     return t; } int Cmn(int m, int n, const int mod) {     int t = 1;     for (int i = 0; i < m; ++i) { t = ((t * (n - i) % mod) * power((i + 1), mod - 2, mod)) % mod; }    //t = t * (n - i) / (i + 1);     return t; } int main(void) {     int n, p, q;     cin >> n >> p >> q;     //double e = pow(0.5, n);     int u = 0, v = 0;     for (; p + q <= n; ++p) {         u = (u + p * Cmn(p, n, mod) % mod) % mod;    //u += p * Cmn(p, n) * e;         v = (v + Cmn(p, n, mod) % mod) % mod;    //v += Cmn(p, n) * e;     }     cout << (float)u / v << endl;     cout << (u * power(v, mod - 2, mod)) % mod << endl;     getchar();     getchar();     return 0; } 和楼上的方法一样，用逆元代替除法运算，费马小定理