（HDOJ 1060）Leftmost Digit

 Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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. 
        
      
         
     
       

 

  Sample code：

   #include<stdio.h>

#include < math.h >
int  main()
{
      int  n,a;
     scanf( " %d " , & n);
      while (n -- )
     {
       scanf( " %d " , & a);
       double  m = a * log10(a);
       long   long  z = ( long   long )m;
       double  p = m - z;
       int  r = pow( 10 ,p);
      printf( " %d\n " ,r);
     }
     return   0 ;
}