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;

}