LightOJ

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

求一个数的后三位大家都会求，只要 mod 1000 就可以了。其实求前三位也很简单，只需要将其变成整数位为一位的浮点型，最后乘以100取整就可以了。比如说12345，可以先变成1.2345，[1.2345*100]=123。乘方计算用快速幂。

代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>using namespace std;typedef long long LL;int getlen(long long n){    int len=0;    while(n)    {        n/=10;        len++;    }    return len;}int main(){    int t;    scanf("%d",&t);    for(int tt=1; tt<=t; tt++)    {        long long n,k;        scanf("%lld%lld",&n,&k);        double m=n/pow(10,getlen(n)-1);        long long x=n;        int res = 1,mod = 1000;        double r=1;        while(k > 0)        {            if(k & 1)            {                res = res * x % mod;                r = r * m;                while(r>=10)                {                    r /= 10;                }            }            x = x * x % mod;            m = m * m;            while(m>=10)            {                m /= 10;            }            k >>= 1;        }        printf("Case %d:",tt);        if(res<10)            printf(" %d 00%d\n",(int)(r*100),res);        else if(res<100)            printf(" %d 0%d\n",(int)(r*100),res);        else            printf(" %d %d\n",(int)(r*100),res);    }    return 0;}