LightOj1282(阶乘前三位和后三位）

You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of n^k.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing two integers: n (2 ≤ n < 2³¹) and k (1 ≤ k ≤ 10⁷).

Output

For each case, print the case number and the three leading digits (most significant) and three trailing digits (least significant). You can assume that the input is given such that n^k contains at least six digits.

Sample Input

5

123456 1

123456 2

2 31

2 32

29 8751919

Sample Output

Case 1: 123 456

Case 2: 152 936

Case 3: 214 648

Case 4: 429 296

Case 5: 665 669

           

           在hdu上已经做过类似一样的了 ，基本就是取对数加上快速幂就可以得出答案。

       

#include<stdio.h>  #include<math.h>  #define ll long long ll Mod(ll a,ll b,ll mod) { //快速幂算法      ll ans=1;      if(!b) return 1;      ans=Mod(a*a%mod,b/2,mod);      if(b&1)ans=ans*a%mod;      return ans;  }    int main() {      int t,cont=0;      scanf("%d",&t);      while(t--){          ll n,k;          scanf("%lld%lld",&n,&k);          int strat=(int)pow(10.0,2.0+fmod(k*log10(n*1.0),1.0));//前三位          int end=(int)Mod(n,k,1000);//后三位          printf("Case %d: %d %lld\n",++cont,strat,end);// ll 输出就溢出！       }//忽略了后三位的首位是 0 的情况      return 0;  }