You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of nk.
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 nk 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
令x=lg(n^k)的整数部分，y=lg(n^k)的小数部分，则n^k由y决定——因为10^x是1000…0。所以10^y再乘上100取整就是前三位。 代码
#include<bits/stdc++.h> #define LL long long using namespace std; const int MAXN = 50000+10; const int MAXM = 1e5; LL qp(LL a,int n,LL mod){ LL ans=1; while(n){ if(n&1) ans=ans*a%mod; a=a*a%mod; n>>=1; } return ans; } int main() { LL n;int t,k; scanf("%d",&t);int ncase=1; while(t--){ scanf("%lld%d",&n,&k); double temp=(double)k*log10(n); temp-=floor(temp); temp=pow(10.0,temp); temp=100*temp; LL b=qp(n,k,1000); printf("Case %d: %lld lld\n",ncase++,(LL)temp,b); //注意后三位可能都是0 ，所以要加上前导零 } return 0; }