E - The World is a Theatre

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

10

Input

4 3 5

Output

3

小结:

        大概的想法有两种：DP，以及排列组合，两者都可以...基本属于水题范畴，为了方便理解，我两种想法的AC代码如下：

排列组合：

#include<stdio.h>#include<string.h>int min(int a,int b){    return a<b?a:b;}long long solve(int a,int b){    if(a>b)    return 0;    int temp=min(b-a,a);    long long ans=1;    for(int i=1;i<=temp;i++)    {        ans*=(b--);        ans/=i;    }    return ans;}int main(){    int m,n,t;    scanf("%d%d%d",&m,&n,&t);    long long ans=0;    for(int i=4;i<t;i++)    {        ans+=solve(i,m)*solve(t-i,n);    }    printf("%lld\n",ans);    return 0;}

DP：

#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;}