P1226 【模板】快速幂||取余运算

link

#include <iostream>
# define LL long long
using namespace std;

LL b;
LL p,k;

int main(){
    scanf("%lld%lld%lld", &b, &p, &k);
    LL d=p;
    LL ans=1;
    LL base=b;
    while(p>0){
        if(p&1){
            ans*=base;
            ans%=k;
        }
        base*=base;
        base%=k;
        p>>=1;
    }
    printf("%lld^%lld mod %lld=%lld", b,d,k,ans%k);
    return 0;
}