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.

先将N^N对数化，取10为底，又N^N可表示为a*10^x,其中9>=a>=1,故x为N^N的位数-1，eg：100=1*10^2 故x为2

lg(N^N) = lg(a*10^x) ;

化简        N*lg(N)  = lg(a) + x;

继续化       N*lg(N) - x = lg(a)

            a = 10^(N*lg(N) - x);

问题转化成求x：

因为1<=a<=9，所以lg（a）<1;

故x=（int） N*lg(N) 向下取整

最后（int）a就是最后的结果

 

#include<iostream> 
#include<cmath> 
#include<cstdio>
using namespace std;
int main() 
{ 
    int n; 
    __int64 m;
     cin>>n; 
    while(n--) 
    { 
        scanf("%I64d",&m);
        double t = m*log10(m*1.0); 
        t -= (__int64)t; 
        __int64 ans = pow((double)10, t); 
        printf("%I64d\n",ans); 
    } 
    return 0; 
}