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:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const ll mod=1e9+7; 5 6 ll n,m,k; 7 ll qpow(ll a, ll b){//快速幂 8 ll res=1; 9 while( b ){ 10 if( b&1 ) res = res*a%mod; 11 a = a*a%mod; 12 b>>=1; 13 } 14 return res; 15 } 16 17 ll inv(ll x){//求逆元 18 return qpow(x,mod-2); 19 } 20 21 int main() 22 { 23 int t,cas=0; 24 scanf("%d",&t); 25 while( t-- ){ 26 scanf("%lld%lld",&n,&k); 27 ll sum=1,pre=1; 28 for(ll i=1;i<k;i++){ 29 pre=pre*(n-i+1)%mod*inv(i)%mod;//递推公式 30 sum=(sum+pre)%mod; 31 } 32 ll ans=qpow(2,n); 33 ans = (ans-sum+mod)%mod; 34 cas++; 35 printf("Case #%d: %lld\n",cas,ans); 36 } 37 return 0; 38 }
原文：https://www.cnblogs.com/wsy107316/p/13298594.html