快速幂

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long LL;
const int MOD = 1000000007;
LL quick_pow(LL a,LL b)
{
    a = a % MOD;
    LL ans = 1;
    while(b)
    {
        if(b & 1)ans = ans * a % MOD;//如果是1在这位上，就乘
        b >>= 1; 		//进位
        a = a * a % MOD;//乘以一位的2
    }
    return ans;
}
int main()
{
    LL a,b;
    int n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%I64d%I64d",&a,&b);
        printf("%I64d\n",quick_pow(a,b));
    }
    return 0;
}