| I I U P C 2 0 0 9 |
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| Problem B: Blind Sorting |
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| 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.”
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| Input |
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| 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.
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| Output |
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| For each n, output number of questions that I have to ask Mr. Panda in the worst case.
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| 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#include #include int n; int main() { while (~scanf("%d", &n)) { printf("%d\n", n - 1 + (int)(ceil(log(n)/log(2)) + 1e-9) - 1); } return 0; }