/*
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);
}
}
转载于:https://www.cnblogs.com/zsben991126/p/10387083.html
