快速幂取模运算

Posted 2021-08-21 DaWeiGuo

落谷p1226
理论依据

#include<iostream>
#include<cstdio>
using namespace std;
long long fun(long long a,long long b,long long c){
    long long An = 1;
    long long T = a % c;
    while(b!=0){
        if(b&1){
            An = (An * T)%c;
        }
        b >>= 1;
        T = (T*T)%c;
    }
    return An;
}
int main(){
    long long a,b,c;
    cin >> a >> b >> c;
    // cout << fun(a,b,c);
    printf("%lld^%lld mod %lld=%lld",a,b,c,fun(a,b,c));
}

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