It is said that a dormitory with 6 persons has 7 chat groups ^_^. But the number can be even larger: since every 3 or more persons could make a chat group, there can be 42 different chat groups.
Given N persons in a dormitory, and every K or more persons could make a chat group, how many different chat groups could there be?
INPUT
The input starts with one line containing exactly one integer T which is the number of test cases.
Each test case contains one line with two integers N and K indicating the number of persons in a dormitory and the minimum number of persons that could make a chat group.
1 ≤ T ≤ 100.
1 ≤ N ≤ 109.
3 ≤ K ≤ 105.
OUTPUT
For each test case, output one line containing “Case #x: y” where x is the test case number (starting from 1) and y is the number of different chat groups modulo 1000000007.
Example
Input
1
6 3
Output
Case #1: 42
对于此题来讲，有几个重点。首先要知道快速幂，其次知道排列组合的递推公式，最后要知道取模后除法和减法会出现问题，需要用费马小定理来解决和加模数再取模来解决。
除法不满足取模，那么我们可以求模的逆元来进行求结果，既然乘法可以取模，如果这个数除以一个数，那么我们可以让他乘以一个数使得和除以那个数的结果相同，那么乘的这个数就是那个数的乘法逆元
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int M=1e9+7;
typedef long long ll;
ll quick(ll a,ll b) //快速幂
{
ll ans=1;
a=a%M;
while(b)
{
if(b&1)
ans=ans*a%M;
b>>=1;
a=a*a%M;
}
return ans;
}
ll Cab(ll n,ll k) //求C(n,k)
{
int temp=1;
for(int i=1; i<=k; i++)
{
temp=(temp%M*(n-i+1)%M*quick(i,M-2))%M;
}
return temp%M;
}
int main()
{
int ca=0;
int T;
cin>>T;
while(T--)
{
ll N,K,temp,ans;
scanf("%I64d%I64d",&N,&K);
if(K>=N/2)
{
ans=temp=Cab(N,K);
for(ll i=K+1; i<=N; i++)
{
temp=(temp%M*(N-i+1)%M*quick(i,M-2))%M; //此处用了费马小定理
ans=(ans+temp)%M;
}
}
else
{
ll result=quick(2,N);
ans=temp=1;
for(ll i=1; i< K; i++)
{
temp=(temp%M*(N-i+1)%M*quick(i,M-2))%M;
ans=(ans+temp)%M;
}
ans=(M+result-ans)%M; //注意减法取模问题
}
printf("Case #%d: %I64d\n",++ca,ans);
}
return 0;
}