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You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of n^k.

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing two integers: n (2 ≤ n < 2³¹) and k (1 ≤ k ≤ 10⁷).

For each case, print the case number and the three leading digits (most significant) and three trailing digits (least significant). You can assume that the input is given such that n^k contains at least six digits.

5

123456 1

123456 2

2 31

2 32

29 8751919

Case 1: 123 456

Case 2: 152 936

Case 3: 214 648

Case 4: 429 296

Case 5: 665 669

题意，求出n的k次幂前三位和后三位。

后三位由快速幂，前三位可由数学公式推导。

n的k次幂我们用科学记数法表示则a*10的m次方，两边同时取对数

得k*log10（n）=log10（a）+m。这里可以发现m为k*log10（n)得整数部分（因为a大于0小于10，取对数后为0.xxx形式），

所以可求得log10（a），这里求前三位故加2（相当于乘100），算10得次方既可。

code：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define mod 1000
LL quick_pow(LL x,int k)
{
    LL ans=1;
    while(k>0)
    {
        if(k&1)
        {
            ans*=x;
            ans%=mod;
        }
        x*=x;
        x%=mod;
        k>>=1;
    }
    return ans;
}


int main()
{
    int T;
    scanf("%d",&T);
    for(int c= 1; c<=T; c++)
    {
        int k,i;
        LL n;
        scanf("%lld %d",&n,&k);
        double temp=k*log10(n);
       temp=fmod(temp,1);
       int res=(int)(pow(10,temp)*100);
        printf("Case %d: %03d %03lld\n",c,res,quick_pow(n,k));
    }
}