hdu1060

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10997    Accepted Submission(s): 4209

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.

 

分析：

m=n^n,两边分别对10取对数得 log10(m)=n*log10(n),
得m=10^(n*log10(n)),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

 

#include<stdio.h>
#include<math.h>
int main()
{
    int n,t;
    double a,a1,a2;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d",&n);
        a=n*log10(n*1.0);
        a1=a-(__int64)a;
        a2=pow(10.0,a1);
        printf("%d ", (int)a2);
    }
    return 0;
}