hdu 1060——Leftmost Digit

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

   
   
   
   

    
    
    
    2
3
4
   
   
   
   

 

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. 
    
    
    
    
     
   
   
   
   

 

Author

Ignatius.L

 

 

水题，算是为数学题目开头吧

算法思想：

由于算的是 n^n 的第一位数，刚开始有点犯难，但是后来释然了， n^n = 10 ^ (log10(n^n)) = 10 ^ (n * log10(n))

然后我们可以观察到： n^n = 10 ^ (N + s) 其中，N 是一个整数 s 是一个小数。

好了，结果出来了

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        double m=n*log10(n);
        m=m-(__int64)(m);
        m=pow(10.0,m);
        printf("%d\n",(int)(m));
    }
    return 0;
}