LightOJ - 1282 Leading and Trailing

Leading and Trailing

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

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 nk 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

//求一个数的几次方的前三位有一个公式;
//head=n^k/(10^(t-3));
//fmod(double,int)是一个函数，求一个数的小数部分；
//由于任意一个数都可以写成10的几次方，只不过这个几次方可能是个小数；
//所以小数部分就是决定前几位的数是几的关键，然后再用pow()函数。
#include<stdio.h>
#include<math.h>
long long mi(long long x,long long y)
{
    x=x%1000;
    long long m=x,n=1;
    while(y)
    {
        if(y%2==1)
            n=n*m%1000;
        m=m*m%1000;
        y/=2;
    }
    n=n%1000;
    return n;
}
int main()
{
    int t,l=0;
    scanf("%d",&t);
    while(t--)
    {
        l++;
        long long a,b,c,d;
        scanf("%lld%lld",&a,&b);
        c=mi(a,b);
       d=(int)pow(10.0,2.0+fmod(b*log10(1.0*a),1));
        printf("Case %d: %03lld %03lld\n",l,d,c);
    }
}