题目

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.

分析：

对一个数num可写为 num=a*10^n , 即科学计数法，使a的整数部分即为num的最高位数字
k^k=a*10^n
如何求a的值？
a=k^k/10^n
如何求n的值？
把n提出来，lga=k*lgk-n
n是整数，而且0 < k* lgk-n< 1, 这说明n是k * lgk的整数部分，因此n=（int）k*lgk
因此a=10^(k*lgk-(int)k *lgk)
实际上，由于k*lgk小数点后有许多数，而int 只有32 位，那么取整用（long long）

代码

#include<stdio.h>
#include<math.h>
int main(){int n;double x,a;scanf("%d",&n);while(n--){scanf("%lf",&a);x=a*log10(a);printf("%d\n",(int)pow(10.0,x-(long long)x));}
}