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……
//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……
1 //x ^ x= 10^(x*lg(x))=10^(整数部分+小数部分) ； 2 #include <stdio.h> 3 #include <math.h> 4 5 int main() 6 { 7 int T; 8 scanf("%d",&T); 9 while(T--) 10 { 11 int n,t; 12 double a; 13 scanf("%d",&n); 14 a = log10((double)n); 15 a -= (int)a; 16 //n*a的结果可能很大，所以先减去整数部分再乘 17 a = n*a; 18 t = (int)a; 19 a -= t; 20 t = (int)pow(10.0,a); 21 printf("%d ",t); 22 } 23 return 0; 24 }
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