# hannnnah_j’s Biological Test

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 681    Accepted Submission(s): 235

Problem Description
hannnnah_j is a teacher in WL High school who teaches biology.

One day, she wants to test m students, thus she arranges n different seats around a round table.

In order to prevent cheating, she thinks that there should be at least k empty seats between every two students.

hannnnah_j is poor at math, and she wants to know the sum of the solutions.So she turns to you for help.Can you help her? The answer maybe large, and you need to mod 1e9+7.

Input
First line is an integer T(T≤1000).
The next T lines were given n, m, k, respectively.
0 < m < n < 1e6, 0 < k < 1000

Output
For each test case the output is only one integer number ans in a line.

Sample Input
2 4 2 6 5 2 1

Sample Output
0 5

Source

a1+...+an = n-m....1
ai>=k                ....2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const long long  p =  1e9+7;
typedef long long LL;
LL pow_mod(LL a,LL n)
{
LL ans=1;
while(n)
{
if(n&1) ans=ans*a%p;
a=a*a%p;
n>>=1;
}
return ans;
}
LL cm(LL n,LL m,LL mod)
{
if(m>n) return 0;
LL i,ans=1,a,b;
for(i=0; i<m; i++)
{
a=(n-i)%mod;
b=(m-i)%mod;
ans=ans*( a*pow_mod(b,mod-2)%mod )%mod;
}
return ans;
}
LL lucas(LL n,LL m,LL p)
{
if(m==0) return 1;
return ( cm(n%p,m%p,p)*lucas(n/p,m/p,p) )%p;
}

int main()
{
int T;
long long n,m,k;
scanf("%d",&T);
while(T--)
{
scanf("%lld %lld %lld",&n,&m,&k);
LL a = lucas(n-m*k-1,m-1,p);
LL c = pow_mod(m,p-2)%p;
printf("%lld\n",((a*n)%p*c)%p);

}
return 0;
}

posted @ 2016-09-20 23:15  樱花庄的龙之介大人  阅读(...)  评论(...编辑  收藏