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

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).

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

  
  

   
   2
3
4
  
  

  
  

   
   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.

   
   
   
    
  
  

Ignatius.L

思路：这道题目是参考网络上大牛们写的，想法就是N*N=M,那么M=10^ ( N*log10 (N) ),【取一次对数再取10的对数次方】

     又由于10^(a*b)【a为整数，b为小数】的最左一位数仅仅与b有关，===> 10^( c.d )[这c.d=a*b]，也就是10^c*10^0.d次方，所以最高位仅 仅        受0.d的影响，所以有ans=ans-(int)ans , ans=N*log10(N), 再求( int ) 10^ans，就是答案了。膜拜数学大牛。。

AC代码如下：

#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;


int main(){

    double a,ans,tmp;
    long long ttmp;
    int t;
    cin>>t;
    while(t--){
        cin>>a;
        tmp=a*log10(a);
        ttmp=(long long)tmp;
        ans=tmp-ttmp;
        cout<<(int)pow(10.0,ans)<<endl;
    }


    return 0;
}