E - Chat Group Gym - 101775A——组合数累加

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

才知道n的所有组合数累加是2n$2^n$，难怪做不出来，那么只要减去k之前的所有累加就好了。
先预处理阶乘的逆元，否则好像会爆

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=1e9+7;
ll ny[100005];
ll n,m;
ll qpow(ll a,ll b)
{
    ll ret=a;
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*ret)%mod;
        ret=(ret*ret)%mod;
        b>>=1;
    }
    return ans;
}
void init()
{
    for(ll i=1;i<=100000;i++)
        ny[i]=qpow(i,mod-2);
}
int main()
{
    int t;
    scanf("%d",&t);
    init();
    int cas=0;
    while(t--)
    {
        scanf("%lld%lld",&n,&m);
        ll ans=qpow(2,n);
        ll cut=1;
        for(int i=0;i<m;i++)
        {
            ans=(ans+mod-cut)%mod;
            cut=(cut*(n-i)%mod*ny[i+1]%mod)%mod;
        }
        printf("Case #%d: %lld\n",++cas,ans);
    }
    return 0;
}