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求n^k的前三位和后三位
- 后三位快速幂即可得到
- 对于前三位
- a.bc * 10^t = n^k
- log_{10}{a.bc} + t = klog_{10|{n}
- 所以log_{10}{a.bc} 是klog_{10|{n}的小数部分
- a.bc = 1000 * klog_{10|{n} 的整数部分
#include<bits/stdc++.h>
using namespace std;
#define fst first
#define sec second
#define sci(num) scanf("%d",&num)
#define scl(num) scanf("%lld",&num)
#define mem(a,b) memset(a,b,sizeof a)
#define cpy(a,b) memcopy(a,b,sizeof b)
typedef long long LL;
typedef pair<int,int> P;
const int MAX_N = 510;
const int MAX_M = 10000;
int mpow(int a,int n,int mod) {
int ans = 1;
while (n) {
if (n & 1) ans = ans * a % mod;
a = a * a % mod;
n >>= 1;
}
return ans;
}
int main() {
int T;
cin >> T;
for (int cs =1; cs <= T;cs++) {
LL N,K;
cin >> N >> K;
double x = K* log10(N) -(LL) (K* log10(N));
//cout << pow(10,x) << endl;
LL ans1 = 100 * pow(10,x);
LL ans2 =mpow(N % 1000,K,1000);
cout << "Case " << cs << ": ";
printf("%lld %03lld\n",ans1,ans2);
}
return 0;
}