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 2 4 Sample Output 1 4
问题链接：UVA11714 Blind Sorting 问题简述：（略） 问题分析： 从n个数中，找出最大的2个数，求其最少的比较次数？ 考虑分治思想，先对所有数进行两两比较，然后淘汰1个数，如同构筑一个完全二叉树。但是需要考虑细节…。其他不解释。 程序说明：（略） 参考链接：（略） 题记：（略）
AC的C++语言程序如下：
/* UVA11714 Blind Sorting */ #include <bits/stdc++.h> using namespace std; int main() { int n; while(~scanf("%d", &n)) printf("%d\n", n + (int)ceil(log2(n)) - 2); return 0; }