LightOJ 1282

题意：http://www.lightoj.com/volume_showproblem.php?problem=1282

n^k = a.bc * 10.0^m；等式两边同时加上log10
k*log10(n) = log10(a.bc) + m;
m为k * log10(n)的整数部分，log10(a.bc)为k * lg(n)的小数部分;
x = log10(a.bc) = k*log10(n) - m = k*log10(n) - (int)k*log10(n);
x = pow(10.0, x);

double x=k*log10(n)-(LL)(k*log10(n));
int y=(int)(pow(10,x)*100);

注意后三位  不够前补0  %03d

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<vector>
#include<math.h>
#include<string>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long
#define N 10000009
#define Lson rood<<1
#define Rson rood<<1|1
int poww(int a,int b)
{
    int c=1;
    while(b)
    {
        if(b&1)
            c=c*a%1000;
        b=b/2;
        a=a*a%1000;
    }
    return c;
}
int main()
{
    int T,t=1;
    scanf("%d",&T);
    while(T--)
    {
        int n,k;
        scanf("%d %d",&n,&k);
        double x=k*log10(n)-(int)(k*log10(n));
        int y=(int)(pow(10,x)*100);
        printf("Case %d: %d %03d
",t++,y,poww(n%1000,k));
    }
    return 0;
}