Lightoj1282（HDOJ1060）

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 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).

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.

5

123456 1

123456 2

2 31

2 32

29 8751919

Case 1: 123 456

Case 2: 152 936

Case 3: 214 648

Case 4: 429 296

Case 5: 665 669

取n的k次方的前三位和后三位

后三位直接取该数的后三位快速幂取得

用科学计数法表示n^k      22222=2.2222* 10^4

设x=log10（n^k）=k*log10（n）

那么10^x=n^k   设x=a(整数)+b(小数)

整数部分10^a是小数点的位置 不影响前三位数字

只需求出10^b的前三位(取前1位直接取整  取前三位*100再取整)

小数部分取整

如果输出结果不够3位要补0

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
long long  hou(long long x,long long y)//快速幂 
{
	 x=x%1000;
	long long res=1; 
	while(y)
	{
		if(y&1)
     	{
		res=res*x%1000;
	   }
	   y=y/2;
	   x=x*x%1000;
	}	
	return res;
}
int main()
{
	int t;
	int ans=0; 
	scanf("%d",&t);
	while(t--)
	{
		ans++;
		 long long  n,k;
		scanf("%lld%lld",&n,&k);
		printf("Case %d: ",ans);
	    double x=(log10(n))*(double)k;//科学计数法 
	          x=x-(long long)x;
		   double qian=pow(10,x);
	     long long aaa=(long long)(qian*100);
		printf("%lld ",aaa);
		long long  hs=hou(n,k);
		if(hs<10)
		{
			printf("00%lld
",hs);
		}
		else
		{
		  if(hs<100)
		  {
		  	 printf("0%lld
",hs);
		  }
	     	else
		   {
		    printf("%lld
",hs); 
		   }
		}
	
	}
	return 0;
}