uva 11714 - Blind Sorting

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. nwill 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

 

Problem Setter: Mahbubul Hasan

Special Thanks: Shamim Hafiz

用最少的比较次数，找出最大的两个数。从n个数中，找最大的一个数，要比较n-1次。但再加上找第二大的呢？再用n-2次，并不是最优的。开始想用快排的思路，发现并是不是最优的方法。最优的是竞赛树，找到最大的后，只需要比较之前做比较时，比最大的数小的那些数就可以了，那些数的总数就是树的高度。所以答案是n+ceil(log2(n))-2。