hdu 1060 Leftmost Digit

m=n^n;两边同取对数，得到，log10(m)=n*log10(n);再得到，m=10^(n*log10(n));
然后，对于10的整数次幂，第一位是1，所以，第一位数取决于n*log10(n)的小数部分

Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13967    Accepted Submission(s): 5339


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

代码：

#include <stdio.h>
#include <math.h>
int main()
{
    double test;
    double a;
    double c;
    __int64 b;
    __int64 d;
    int n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%lf",&test);
    
        a=test*log10(test);
        b=(__int64)a;
        c=a-b;
        d=(__int64)(pow(10,c));
        printf("%I64d\n",d);
    }
    return 0;
}