hdoj 1060 Leftmost Digit 幂指函数取对数

最新推荐文章于 2019-09-20 20:11:27 发布

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最新推荐文章于 2019-09-20 20:11:27 发布

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分类专栏：数论

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本文介绍了一种高效算法，用于求解给定正整数N的N^N结果中最左侧的数字。通过数学转换，利用对数性质减少数值计算复杂度，避免了直接计算N^N带来的巨大数值运算问题。

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Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12268 Accepted Submission(s): 4689

Problem Description

Given a positive integer N, you should output the leftmost digit of N^N.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output

For each test case, you should output the leftmost digit of N^N.

Sample Input

Sample Output

  
  
   
   2
2


   
   
    
    
     
     Hint
    
    
In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2.
In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.

Descriptipn: 计算N^N结果的最左边的数
How to Do: 数学题由sum=N^N,两边对10取对数,log10(sum)=Nlog10(N),有sum=10^(Nlog10(N));
由于10的整数次幂首位均为1，则仅需考虑Nlog10(N)的结果的小数部分即可

易错点： 1、数值太大，类型需要用long long.

2、对数函数的调用。

代码：

#include <iostream>
#include <stdio.h>
#include <math.h>

using namespace std;

int main()
{
long long t,n;
long long x,m;
double m1;
cin>>t;
while(t--)
{
cin>>n;
m1=n*log10(double(n));
x=(long long)(m1);
m1-=x;
m=(long long)(pow(10.0,m1));
cout<<m<<endl;
}
return 0;
}