There are N buildings standing in a straight line in the City, numbered from 1 to N. The heights of all the buildings are distinct and between 1 and N. You can see F buildings when you standing in front of the first building and looking forward, and B buildings when you are behind the last building and looking backward. A building can be seen if the building is higher than any building between you and it.
Now, given N, F, B, your task is to figure out how many ways all the buildings can be.

First line of the input is a single integer T (T<=100000), indicating there are T test cases followed.
Next T lines, each line consists of three integer N, F, B, (0<N, F, B<=2000) described above.

For each case, you should output the number of ways mod 1000000007(1e9+7).

2
3 2 2
3 2 1

2
1

2012 Multi-University Training Contest 8

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ac代码

#include<stdio.h>
#include<string.h>
#define mod 1000000007
__int64 s[2010][2010],c[2010][2010],n,f,b;
void fun()
{
    int i,j;
    for(i=0;i<2010;i++)
    {
        c[i][0]=1;
        c[i][i]=1;
        s[i][0]=0;
        s[i][i]=1;
        for(j=0;j<i;j++)
        {
            c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
            s[i][j]=(s[i-1][j]*(i-1)%mod+s[i-1][j-1])%mod;
        }
    }
}
int main()
{
    int t;
    fun();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d%I64d",&n,&f,&b);
        printf("%I64d\n",c[b+f-2][f-1]*s[n-1][f+b-2]%mod);
    }
}