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

[align=left]Problem Description[/align]
Given a positive integer N, you should output the leftmost digit of N^N.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, you should output the leftmost digit of N^N.

[align=left]Sample Input[/align]

2
3
4

[align=left]Sample Output[/align]

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.

[align=left]Author[/align]
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\n",t);
}
return 0;
}

链接：/article/4736587.html

总结：数学题，log的运用，掌握的不行，需多练习