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| Time Limit: 2000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Description
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
Source
/** 题意：求a的b次幂，结果的高三位和低三位是什么 做法：快速幂 **/ #include<iostream> #include<stdio.h> #include<string.h> #include<cmath> #include<algorithm> #define mod 1000 using namespace std; int quickpow(int n, int k) { int res = 1; int base = n%mod; while(k) { if(k&1) res *= base; base *= base; res %= mod; base %= mod; k >>= 1; } return res; } int solve(int a,int b) { double s = b*log10((double)a) - (int)(b*log10((double)a)); s = pow(10.0,s); return s*100; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif // ONLINE_JUDGE int Case = 1; int T; scanf("%d",&T); while(T--) { int a,b; scanf("%d %d",&a,&b); int res = quickpow(a,b); int cet =pow(10, 2+fmod(b *(double)log10((double)a), 1)); printf("Case %d: %d %03d\n",Case,cet,res); Case++; } return 0; }
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原文地址：http://www.cnblogs.com/chenyang920/p/4445561.html
