CodeForces - 131C The World is a Theatre 求组合数 DP

本文探讨了一个具体的组合数学问题:给定一定数量的男孩和女孩,如何计算组成特定人数剧团的不同方式的数量,且剧团中至少包含4名男孩和1名女孩。文章通过使用动态规划的方法来解决这个问题,并提供了完整的C++代码实现。

There are n boys and m girls attending a theatre club. To set a play "The Big Bang Theory", they need to choose a group containing exactly t actors containing no less than 4 boys and no less than one girl. How many ways are there to choose a group? Of course, the variants that only differ in the composition of the troupe are considered different.

Perform all calculations in the 64-bit type: long long for С/С++, int64 for Delphi and long for Java.

Input

The only line of the input data contains three integers nmt (4 ≤ n ≤ 30, 1 ≤ m ≤ 30, 5 ≤ t ≤ n + m).

Output

Find the required number of ways.

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.

Examples
Input
5 2 5
Output
10
Input
4 3 5
Output
3

c[i][j]=c[i-1][j]+c[i-1][j-1]:即第j个数在前i-1个数选与不选的情况(注意i>=j)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstdlib>
#include<cctype>
using namespace std;
#define ll long long
typedef pair<ll,int>P;
const int len=1e2+5;
ll c[31][31];
int main()
{
	for(int i=0;i<=30;++i)c[i][0]=1;
	for(int i=0;i<=30;++i)c[i][i]=1;
	for(int i=1;i<=30;++i)
		for(int j=1;j<=i;++j)c[i][j]=c[i-1][j]+c[i-1][j-1];
	ll a,b,t;
	while(cin>>a>>b>>t)
	{
		ll ans=0;
		for(int i=4;i<t;++i)
			if(i<=a&&t-i<=b)
				ans+=c[a][i]*c[b][t-i];
		cout<<ans<<endl;
	}
}

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