HDUOJ1060Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11600    Accepted Submission(s): 4430

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

 

一个数n^n=p= x*10^m  所以log(p)=n*log10(n);

而这和我们要求的最高位i的那个数有什么关系乐 比如求1234的最高位为1，而log10(1234)=log10(1.234*10^3)=log10(1.234)+3;

而最高位即为其pow(10,1og10(1.234))=1.234的整数1；

=log10(n^n)=n*log10(n);所以同样的道理，就可以求这个了.....

代码：

#include<stdio.h>
#include<math.h>
int main()
{
    int n,m;
    scanf("%d",&m);
    while(m--)
    {
        scanf("%d",&n);
        double a=n*log10(1.0*n);
        a-=(__int64)a;
        printf("%d\n",(int)pow(10,a));
    }
    return 0;
}