LightOJ

题意：

求解n^k的后三位和前三位，后三位要有前导0，比如123001，前三位是123后三位是001。后三位用快速幂取模就行，前三位比较麻烦，推导，看了别人的博客之后的。

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

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).

Output

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.

Sample Input

5

123456 1

123456 2

2 31

2 32

29 8751919

Sample Output

Case 1: 123 456

Case 2: 152 936

Case 3: 214 648

Case 4: 429 296

Case 5: 665 669

#include <stdio.h>#include <ctime>#include <math.h>#include <string>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <ctype.h>#include <string.h>using namespace std;typedef long long ll;int pow_mod(int a,int n,int m){    if(n==0)        return 1;    int x=pow_mod(a,n/2,m);    long long ans=(long long)x*x%m;    if(n%2==1)        ans=ans*a%m;    return (int)ans;}int main(){    int t;    int n,k;    cin>>t;    for(int i=1;i<=t;i++)    {        cin>>n>>k;        double x=pow(10.0,fmod(k*log10(1.0*n),1));         x=x*100.0;        printf("Case %d: %d %03d\n",i,(int)x,pow_mod(n,k,1000));            }    return 0;}