CodeForces - 131C The World is a Theatre（组合数）

题意：已知有n个男生，m个女生。现在要选t个人，要求有至少4个男生，至少1个女生，求有多少种选法。

分析：

1、展开，将分子中的m！与分母中n！相约，即可推出函数C。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e5 + 10;
const int MAXT = 10000 + 10;
using namespace std;
LL C(LL n, LL m){
    LL ans = 1;
    for(LL i = 1; i <= m; ++i){
        ans *= n - i + 1;
        ans /= i;
    }
    return ans;
}
int main(){
    LL n, m, t;
    while(scanf("%I64d%I64d%I64d", &n, &m, &t) == 3){
        LL ans = 0;
        for(LL i = 4; i < t; ++i){
            ans += C(n, i) * C(m, t - i);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}

2、递推求组合数。

高中学的组合数公式：C(n, m) = C(n - 1, m - 1) + C(n - 1, m)。

注意m <= n。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 30 + 10;
const int MAXT = 10000 + 10;
using namespace std;
LL c[MAXN][MAXN];
void init(){
    c[0][0] = 1;
    for(int i = 1; i <= 30; ++i){
        c[i][0] = c[i][i] = 1;
        for(int j = 1; j < i; ++j){
            c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
        }
    }
}
int main(){
    LL n, m, t;
    init();
    while(scanf("%I64d%I64d%I64d", &n, &m, &t) == 3){
        LL ans = 0;
        for(LL i = 4; i < t; ++i){
            if(t - i <= m && i <= n) ans += c[n][i] * c[m][t - i];
        }
        printf("%I64d\n", ans);
    }
    return 0;
}