lightoj 1282 取对数的操作

/*
前三位 
len=log10n^k(乘积的长度) 
len=klog10n
n^k=x*10^(len-1)
x=n^k/10^(len-1)
log10x = k*log10n - (len-1)
x=pow(10,k*log10n - (len-1))

后三位
快速幂解决 
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll n,k,x;
ll power(ll a,ll n){
    ll res=1;
    while(n){
        if(n%2)res=res*a%1000;
        a=a*a%1000;
        n>>=1;
    }
    return res%1000;
}
int main(){
    int T;
    cin>>T;
    for(int tt=1;tt<=T;tt++){
        cin>>n>>k;
        ll len=k*log10((double)n);
        double x=pow((double)10,(double)k*log10((double)n)-(len-1));
        while(x<100)x*=10;
        printf("Case %d: %d ",tt,(int)x);
        n%=1000;
        ll y=power(n,k);//后三位用快速幂
        if(y==0)printf("000\n");
        else if(y<10)printf("00%d\n",y);
        else if(y<100)printf("0%d\n",y);
        else printf("%d\n",y); 
    }    
}