LightOJ - 1282

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int power(int a,int b,int c)
{
	int res=1;
	a%=c;
	while(b)
	{
		if(b&1)
		res=(res*a)%c;
		a=(a*a)%c;
		b>>=1;
	}
	return res;
}
int main()
{
	int t,ca=1;
	cin>>t;
	while(t--)
	{
		int n,k,x,y;
		scanf("%d%d",&n,&k);
		//公式推导，得前三位求法 
		//设n^k=a.bc*10^m(除开前三位，其余的约掉了)
		//lg(n^k) = lg(a.bc * 10^m)
		//k * lg(n) = lg(a.bc) + lg(10^m) = lg(a.bc) + m
		//m为k * lg(n)的整数部分，lg(a.bc)为k * lg(n)的小数部分
		//x = lg(a.bc) = k * lg(n) - m = k * lg(n) - (int)(k * lg(n))
		//a.bc = pow(10, x);
		//abc = a.bc * 100;
		double m=k*log10(n)-(int)(k*log10(n));
		m=pow(10,m);
		x=m*100;
		//------------------------------------------------------------ 
		y=power(n,k,1000);
		printf("Case %d: %d %03d\n",ca++,x,y);
	}
}