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
234
 

Sample Output
22
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;}