2013 - ECJTU 暑期训练赛第三场-problem-D

D - D

Submit Status Practice POJ 3070

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

10

Input

4 3 5

Output

3

组合数学问题，记住一个公式，这道题就迎刃而解了，Cn(m)=C(n-1)(m-1)+C(n-1)m

AC代码：

#include
#define MAX 61
__int64 s[MAX][MAX];
int main()
{
	int i,j,n,m,t;
	__int64 sum;
		s[1][1]=1;
		for(i=60;i>=1;i--)//初始化，因为Cn(1)=n
		s[1][i]=i;
		for(i=60;i>=1;i--)//同样，因为Cn(n)=1
		s[i][i]=1;
		for(i=2;i<=60;i++)
		{
			for(j=2;j<=60;j++)
			{
				if(j>=i)
				s[i][j]=s[i-1][j-1]+s[i][j-1];//Cm(n)=C(m-1)(n-1)+Cm(n-1),我这里反过来了本来是Cn(m)=C(n-1)(m-1)+C(n-1)m
			}
		}
	while(scanf("%d%d%d",&n,&m,&t)!=EOF)
	{
		sum=0;
		for(i=4;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
			 if((i+j)==t)
			 {
			  sum+=s[i][n]*s[j][m];
			 }
			}
		}
		printf("%I64d\n",sum);
	}
	return 0;
}