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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 挺水的这道题，但是一次做对也不是很容易。首先打表求出组合数。其次数组不要开得太小，dp[i][j]，如果j>i的时候，就是0。 代码如下：

#include
   
    #define ll long longusing 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];//单纯的排列组合			}			printf("%I64d\n",sum);		}	}}
   

努力加油a啊，(^o)/~

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