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

题意：求一个数的前三位和后三位

解析：后三位直接用快速幂求出，复杂的是前三位。设x = log (n^k) = k * log10（n），
那么10^x = n^k.将x = a(整数） + b（小数）,
整数部分10^a是小数点的位置，并不影响前三位数字。
故只需要求出10^b取前三位。
使用fmod（a， 1）表示求浮点型数 a 的小数部分。
原型：extern float fmod(float x, float y)
用法：#include
功能：计算x/y的余数

代码：

 #include"stdio.h"
#include"math.h"
int main()
{
	int t,tt=1;
	scanf("%d",&t);
	while(t--)
	{
		long long n,k;
		int a,b=1;
		scanf("%lld %lld",&n,&k);
		a=(int)pow(10.0,2.0+fmod(k*log10(n*1.0),1.0));//前三位
		while(k)//后三位
		{
			if(k%2==1)
			b=b*n%1000;
			k/=2;
			n=n*n%1000;
		}
		printf("Case %d: %d %03d\n",tt++,a,b);
	}
	return 0;
}