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题意：n个人的小组，最少m个可以组成一组，问共有多少种方法。
思路：就是求C(n,k) + C(n,k+1) +...+ C(n,n);但是数据量超大，我们可以转换一下
2^n - ( C(n,0) + C(n,1) + ... + C(n,k-1) )而C(n,k) = ( n*(n-1)*..(n-k+1) ) / ( k! );因为要除以k的阶乘，所以就用到了费马小定理求逆元，除以一个很大的数就是乘以它的逆元。最好是把逆元预处理一下就不会超时了。
用费马小定理求逆元
费马小定理公式：a^(p-1) ≡ 1（mod p）
逆元公式：H * B ≡ 1 (mod M) -->这里假设M为质数，由公式显然B与M互质
其中M = p，B = a
联立可得H * a ≡ a^(p - 1)（mod p）
同时除以a（显然a mod p 不为0）
最终得B（或者a）关于p的逆元
H ≡ a^(p - 2)（mod p）
AC代码：
/***
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* │ Ctrl│ │Alt │ Space │ Alt│ │ │Ctrl│ │ ← │ ↓ │ → │ │ 0 │ . │←─┘│
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*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
#define PI acos(-1)
const int mod=1e9+7;
const int M=1e5 + 10;
typedef long long ll;
ll a[M]={0};
ll F(ll a, ll b)
{
ll ans = 1;
a %= mod;
while(b){
if(b&1)
ans = ((a%mod) * ans) % mod;
a = (a * a) % mod;
b >>= 1;
}
return ans % mod;
}
void init(){//预处理逆元
ll p = 1;
for (ll i = 1; i <= 1e5;i++){
p = (p * i) % mod;
a[i] = F(p, mod - 2) % mod;
if(a[i]<0)
a[i] += mod;
}
}
int main()
{
ll T, n, k;
init();
scanf("%lld", &T);
for (int kk = 1; kk <= T; kk++)
{
scanf("%lld%lld", &n,&k);
ll ans = F(2, n) - 1;//先减去C(n,0)的情况
ll p = 1, q = 1;
for (ll i = 1; i < k; i++){
p = ((p % mod) * (n - i + 1)) % mod;
q = (q * i) % mod;
ans = ans - (p % mod * (a[i] % mod)) % mod;
if (ans < 0)
ans += mod;
}
printf("Case #%d: %lld\n", kk, ans%mod);
}
return 0;
}