Light oj 1282 (求一个数的前三位和后三位）

F - F 使用long long

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1282 uDebug

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 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

题意：求n^k的前三位和后三位；

思路：

求前三位时，n^k=10^(k*log10(n));

令b=k*logn10(n),求出b的小数部分b=b-(int)b;然后求pow(10,b)*100，就是前三位；

例如：123456=1.23456*10^5=10^log10(1.23456)*10^5=10^(5+log10(1.23456));

b=5+log10(1.23456),小数部分为log10(1.23456),则pow(10,log10(1.23456))=1.23456,1.23*100=123.456,转化为整形就求得了前三位；

代码：

#include<stdio.h>#include<string.h>#include<math.h>long long quick_pow(int n,int k){long long base=n,ans=1;while(k){if(k&1){ans=(ans*base)%1000;}base=(base*base)%1000;k>>=1;}return ans;}int main(){int n,k,t,mm=1;scanf("%d",&t);while(t--){scanf("%d%d",&n,&k);double b=k*1.0*log10(n*1.0);    b=b-(long long)b;//求小数部分；     long long ans1=pow(10,b)*100;//求出来的是小数，强制转换为long long整形;    long long  ans2=quick_pow(n,k);//求后三位； printf("Case %d: %lld %03lld\n",mm++,ans1,ans2);}return 0;}