题解 | 【模板】快速幂

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

ll fast_pow(ll a,ll b,ll p){
	ll ans = 1;
	a = a % p;
	
	while(b){
		if(b & 1) ans = (ans * a) % p;
		
		a = (a*a) % p;
		
		b = b >> 1;
	}
	
	return ans;
}

int main(){
	int t;  cin >> t;
	
	while(t--){
		ll a,b,p;  cin >> a >> b >> p;
		
		cout << fast_pow(a,b,p) << endl;
	}
	
	return 0;
}