快速幂
#include <iostream>
using namespace std;
typedef long long ll;
ll pow(ll a, ll b, ll p){
ll ans = 1, base = a;
while (b){
if (b & 1) ans = ans * base % p;
base = base * base % p;
b >>= 1;
}
return ans;
}
int main(){
ll a, b, p;
cin >> a >> b >> p;
cout << pow(a, b, p);
return 0;
}