逆元应用求组合数

例题：here

\(求C_{n}^{k}+C_{n}^{k+1}+\cdots +C_{n}^{n}\)

\(C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}+\cdots +C_{n}^{n}=2^{n}\)

\(则C_{n}^{k}+C_{n}^{k+1}+\cdots +C_{n}^{n}=2^{n}-\left ( C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\cdots +C_{n}^{k-1}\right )\)

\(2^{n}用快速幂算，C_{n}^{k}=\frac{\left ( n-k+1\right )}{k}*C_{n}^{k-1}\left ( 组合数递推公式\right )\)

\(\frac{\left ( n-k+1\right )}{k}除法改为乘法，所以求逆元\)

模板：AC_Code:

 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;

ll n,m,k;
ll qpow(ll a, ll b){//快速幂
    ll res=1;
    while( b ){
        if( b&1 ) res = res*a%mod;
        a = a*a%mod;
        b>>=1;
    }
    return res;
}

ll inv(ll x){//求逆元
    return qpow(x,mod-2);
}

int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while( t-- ){
        scanf("%lld%lld",&n,&k);
        ll sum=1,pre=1;
        for(ll i=1;i<k;i++){
            pre=pre*(n-i+1)%mod*inv(i)%mod;//递推公式
            sum=(sum+pre)%mod;
        }
        ll ans=qpow(2,n);
        ans = (ans-sum+mod)%mod;
        cas++;
        printf("Case #%d: %lld\n",cas,ans);
    }
    return 0;
}