LightOJ 1282

Leading and Trailing

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status

Description

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

Source

Problem Setter: Shamim Hafiz

Special Thanks: Jane Alam Jan (Solution, Dataset)

/**
          题意：求a的b次幂，结果的高三位和低三位是什么
          做法：快速幂 
**/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<algorithm>
#define mod 1000
using namespace std;
int quickpow(int n, int k)
{
    int res = 1;
    int base = n%mod;
    while(k)
    {
        if(k&1)
            res *= base;
        base *= base;
        res %= mod;
        base %= mod;
        k >>= 1;
    }
    return res;
}
int solve(int   a,int   b)
{
          double s = b*log10((double)a) - (int)(b*log10((double)a));
    s = pow(10.0,s);
    return s*100;
}
int main()
{
#ifndef ONLINE_JUDGE
         freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
          int Case = 1;
          int T;
          scanf("%d",&T);
          while(T--)
          {
                    int  a,b;
                    scanf("%d %d",&a,&b);
                    int  res = quickpow(a,b);
                    int cet =pow(10, 2+fmod(b *(double)log10((double)a), 1));
                   printf("Case %d: %d %03d
",Case,cet,res);
                   Case++;
          }
          return 0;
}