HDOJ-1060-Leftmost Digit（求n^n的最高位）

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2924    Accepted Submission(s): 1026

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.
   
   
   
   
   
   
   
   

   
   
   
   

    
    
    
    对一个数num可写为 num=10
    
    
    
    n
    
    
    
     + a, 即科学计数法，使a的整数部分即为num的最高位数字
   
   
   
   
   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    
    num
    
    
    
    num
    
    
    
    =10
    
    
    
    n
    
    
    
     + a 这里的n与上面的n不等
   
   
   
   
   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    
    两边取对数： num*lg(num) = n + lg(a);
   
   
   
   
   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    
    因为a<10，所以0<lg(a)<1
   
   
   
   
   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    
    令x=n+lg(a); 则n为x的整数部分，lg(a)为x的小数部分
   
   
   
   
   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    
    又x=num*lg(num);
   
   
   
   
   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    
    a=10
    
    
    
    (x-n) 
    
    
    
    = 10
    
    
    
    (x-int(x)))
   
   
   
   
   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    
    再取a的整数部分即得num的最高位
   
   
   
   
   
   
   
   

   
   
   
   

   
   
   
   

    
    
    
    #include <iostream>
#include <math.h>
using namespace std;
int main()
{
    int t;
    while(cin>>t)
    {
        while(t--)
        {
            unsigned long n;
            cin>>n;
            double x=n*log10(n*1.0);
            x-=(__int64)x;
            int a=pow(10.0, x);
            cout<<a<<endl;
        }
    }
    return 0;
}