Leftmost Digit（杭电1060）（求N^N的最高位）

Leftmost Digit

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

Total Submission(s): 13574 Accepted Submission(s): 5216



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.

Author

Ignatius.L

/*本题思路： 
1,令M=N^N； 

2，分别对等式两边取对数得 log10(M)=N*log10(N),得M=10^(N*log10(N))；                  

3，令N*log10(N)=a+b,a为整数，b为小数；

4，C函数：log10()，计算对数，pow(a,b)计算a^b

5，由于10的任何整数次幂首位一定为1,所以,M的首位只和N*log10(N)的小数部分有关，

   即只用求10^b救可以了；
   
6，最后对10^b取整，输出取整的这个数就行了。（因为0<=b<1,所以1<=10^b<10对

其取整，那么的到的就是一个个位，也就是所求的数）。
*/
//hdoj系统就是坑，各种CE，各种WA，多亏康晓辉指点。 
#include<stdio.h>
#include<math.h>
int main()
{
	int test;
	double n,k,a;
	scanf("%d",&test);
	while(test--)
	{
		scanf("%lf",&n);
		a=n*log10(n);
		k=a-(__int64)a; //这里强制转换需要用int64才能A，否则会w或者CE！！！ 
		printf("%d\n",int(pow(10,k)));
	}
	return 0;
}