lightoj 1282 - Leading and Trailing (数学--log使用)

1282 - Leading and Trailing

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Time Limit: 2 second(s)Memory Limit: 32 MB

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

Output for Sample Input

5

123456 1

123456 2

2 31

2 32

29 8751919

Case 1: 123 456

Case 2: 152 936

Case 3: 214 648

Case 4: 429 296

Case 5: 665 669

 题意：问N^K的前三位和后三位（要补0处理）

思路：后三位：快速幂取模

前三位技巧：

N^K = X *10 ^(len-1) //len是整数的长度

两边取对数：K*log10(N)=(len-1)+log( X )//求出X

总结：tmp=k*log10(n)

tmp=tmp-[tmp]

tmp=tmp*100

代码：

#include<cstdio>#include<cmath>#define LL long longLL powmod(LL x,LL n)//快速幂 {LL res=1;while(n>0){if(n&1)res=res*x%1000;x=x*x%1000;n>>=1;}return res;}int main(){int t,kcase=1,m=100;LL n,k;LL s1,s2;scanf("%d",&t);while(t--){scanf("%lld%lld",&n,&k);s1=powmod(n,k);double tmp=k*log10(n*1.0);tmp=tmp-(long long)tmp;tmp=pow(10.0,tmp);printf("Case %d: %lld %03lld\n",kcase++,(LL)(tmp*100),s1);//补0处理}return 0;}