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
#include<stdio.h> typedef long long ll; double change(double a) { while(a>=1000.0) { a=a/10.0; } return a; } int main() { int p,t=1; scanf("%d",&p); while(p--) { ll n,k; scanf("%lld %lld",&n,&k); ll sum=n,ans=1,l=k; double a,b=1.0; a=n*1.0; while(k)//后三位 { if(k%2) { ans=ans*sum%1000; } sum=sum*sum%1000; k=k/2; } while(l)//前三位 { if(l%2) { b=b*a; b=change(b); } a=a*a; l=l/2; a=change(a); } printf("Case %d: d lld\n",t++,(int)b,ans); } }