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 题意是求n的k次方前三位和后三位 后3位快速幂 前3位：c=klog10(n)-(int)(klog10(n)); c=pow(10,c); 设10p=nk,。10p=10y*10x(科学计数法),所以log10(n)=x+y，我们设x为(int)log10(n).这样10x就是科学计数法后面的10几次方，10y就是前面的小数部分
#include<stdio.h> #include<string.h> #include<cmath> #include<algorithm> using namespace std; typedef long long ll; int mod=1000; long long rpow(ll n,ll p) { long long ans=1; while(p) { if(p%2) ans=ans*n%mod; n=n*n%mod; p/=2; } return ans%mod; } int main() { int t,q=1; scanf("%d",&t); while(t--) { ll n,k; scanf("%lld%lld",&n,&k); double c=k*log10(n)-(int)(k*log10(n)); ll b=rpow(n,k); c=pow(10,c); ll a=c*100; printf("Case %d: %lld lld\n",q++,a,b); } return 0; }