Leftmost Digit（数学）

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. 


题目意思：求n^n结果的第一位。
解题思路：我们可以将其看成幂指函数，那么我们对其处理也就顺其自然了，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。由于10的任何整数次幂首位一定为1,所以首位只和s(小数部分)有关。 

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