E - 取模运算

本题就是求(A1^B1+
A2^B2+ … +AH^BH)mod M.

由于数太大
我们也需要每次运算都对m取余
因为大于m的部分不会影响结果！

#include<iostream>
using namespace std;
#define ll long long
ll quickpow(ll x, ll n,ll h) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res * x%h;
		x = (x * x)%h;
		n = (n >> 1);
	}

	return res;
}

int main() {
	int m;
	cin >> m;
	for (int i = 0; i < m; ++i) {
		ll n;
		cin >> n;
		int h;
		cin >> h;
		const int t = h;
		int a[1000][2];
		ll sum = 0;
		for (int j = 0; j < h; ++j) {
			cin >> a[j][0] >> a[j][1];
			sum += quickpow(a[j][0], a[j][1],n);
		}
		cout << sum % n<<endl;
	}
	return 0;
}