E - The World is a Theatre

最新推荐文章于 2021-11-19 15:05:40 发布

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本文介绍了一个基于动态规划解决的问题——如何计算在特定条件下选择演员组合的数量。问题要求从一定数量的男孩和女孩中选出至少包含4名男孩和1名女孩的演员组合，总数为t人。

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E - The World is a Theatre

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Description

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 n, m, t (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.

Sample Input

Input

5 2 5

Output

Input

4 3 5

Output

小结：

感觉想法和划分数差不多....大概的想法是，DP[i][j]表示从i中取j个的取法，这里的i，j个物品没有区别。

然后就有状态转移方程DP[i][j]=DP[i-1][j]+DP[i-1][j-1].

以下是AC代码：

#include<stdio.h>
#include<string.h>
int main()
{
int n,m,t;
scanf("%d%d%d",&n,&m,&t);
long long dp[65][65];
memset(dp,0,sizeof(dp));
for(int i=0;i<=30;i++)
{
dp[i][0]=1;
for(int j=1;j<=i;j++)
{
dp[i][j]=dp[i-1][j]+dp[i-1][j-1];
}
}
long long ans=0;
for(int i=4;i<t;i++)
{
ans+=dp[n][i]*dp[m][t-i];
}
printf("%lld\n",ans);
return 0;
}