UVA 11714 - Blind Sorting(推理贪心)

I I U P C   2 0 0 9
Problem B: Blind Sorting
I am a polar bear. But I am not just an ordinary polar bear. Yes I am extra ordinary! I love to play with numbers. One day my very good friend Mr. Panda came to me, and challenged me to solve a puzzle. He blindfolded me, and said that I have n distinct numbers. What I can ask is whether a’th number is larger than b’th number and he will answer me properly. What I have to do is to find out the largest and second largest number. I thought for a while and said “Come on, I will do it in minimum number of comparison.”
Input
There will be a non-negative integer, n in each of the line of input where n is as described above. n will be less than any 10 digit prime number and not less than the smallest prime.
Output
For each n, output number of questions that I have to ask Mr. Panda in the worst case.
Sample Input	Output for Sample Input
2 4	1 4

题意：有n个不同的数，你可以询问a，b哪个大，会得到答案，然后问最少要几次保证能挑选出最大和第二大的数。

思路：n个数，先以打擂台的方式，两两比较出最大的，n - 1次，然后在由被最大PK下去的数字中，比较出最大的，有log(n)个数，需要进行log(n) - 1次，注意是向上取整。

代码：

#include <stdio.h>
#include <math.h>
#include <string.h>

int n;

int main() {
while (~scanf("%d", &n)) {
printf("%d\n", n - 1 + (int)(ceil(log(n)/log(2)) + 1e-9) - 1);
}
return 0;
}