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.

Examples
Input
5 2 5
Output
10
Input
4 3 5
Output
3
挺水的這道題，但是一次做對也不是很容易。首先打表求出組合數。其次數組不要開得太小，dp[i][j]，如果j>i的時候，就是0。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=100; int n,m,t; ll dp[maxx][maxx];inline void init() {memset(dp,0,sizeof(dp));for(int i=0;i<=30;i++){dp[0][i]=0;dp[i][0]=1ll;}for(int i=1;i<=30;i++){for(int j=1;j<=i;j++){dp[i][j]=(dp[i-1][j]+dp[i-1][j-1]);}} } int main() {init();while(~scanf("%d%d%d",&n,&m,&t)){if(t<5||n<4||m<1) puts("0");else{ll sum=0;for(int i=4;i<=n&&t-i>=1;i++){sum+=dp[n][i]*dp[m][t-i];//單純的排列組合}printf("%I64d\n",sum);}} }

努力加油a啊，(^o)/~

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