洛谷P3414 SAC#1

链接

　　https://www.luogu.org/problem/show?pid=3414

题解

　　

ans=C0n+C2n+C4n+...=C0n+C1n+C2n+...+Cnn2=C0n1n10+C1n1n−111+C2n1n−211+...+Cnn101n2=(1+1)n2=2n−1

　　快速幂

代码

//二项式定理#include <cstdio>#include <algorithm>#define ll long long#define mod 6662333using namespace std;ll N;ll pow(ll a, ll b, ll p){    ll ans, t;    for(ans=1,t=a;b;b>>=1,t=t*t%p)if(b&1)ans=ans*t%p;    return ans;}int main(){    scanf("%lld",&N);    printf("%lld",pow(2,N-1,mod));    return 0;}