hdu_1060_Leftmost Digit_201311071827-2

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11305    Accepted Submission(s): 4329

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

 

 //x ^ x= 10^(x*lg(x))=10^(整数部分+小数部分) ；则x ^ x的最高位是由小数部分决定的（因为10的整数次幂不会影响最高位，只在最末位加0）。
 //去掉整数部分（floor函数向下去整）得到10^(小数部分)最高位即是所求……

//x ^ x= 10^(x*lg(x))=10^(整数部分+小数部分) ；
#include <stdio.h>
#include <math.h>

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,t;
        double a;
        scanf("%d",&n);
        a = log10((double)n);
        a -= (int)a;
        //n*a的结果可能很大，所以先减去整数部分再乘 
        a = n*a;
        t = (int)a;
        a -= t;
        t = (int)pow(10.0,a);
        printf("%d
",t);
    }
    return 0;
}

链接：http://www.cnblogs.com/hxsyl/archive/2012/09/04/2671068.html

总结：数学题，log的运用，掌握的不行，需多练习