LeetCode 458. Poor Pigs--智力题「小白鼠试毒」--C++,Python解法

题目地址：Poor Pigs - LeetCode

There are 1000 buckets, one and only one of them is poisonous, while the rest are filled with water. They all look identical. If a pig drinks the poison it will die within 15 minutes. What is the minimum amount of pigs you need to figure out which bucket is poisonous within one hour?

Answer this question, and write an algorithm for the general case.

General case:

If there are n buckets and a pig drinking poison will die within m minutes, how many pigs (x) you need to figure out the poisonous bucket within p minutes? There is exactly one bucket with poison.

Note:

A pig can be allowed to drink simultaneously on as many buckets as one would like, and the feeding takes no time.
After a pig has instantly finished drinking buckets, there has to be a cool down time of m minutes. During this time, only observation is allowed and no feedings at all.
Any given bucket can be sampled an infinite number of times (by an unlimited number of pigs).

这道题目是一道数学智力题，具体解法参考这2篇文章：

【数学】用信息论去解决「小白鼠试毒」问题
1000桶水，其中一桶有毒，猪喝毒水后会在15分钟内死去，想用一个小时找到这桶毒水，至少需要几头猪？

Python解法如下：

class Solution:def poorPigs(self, buckets: int, minutesToDie: int, minutesToTest: int) -> int:states = minutesToTest // minutesToDie + 1return math.ceil(math.log(buckets) / math.log(states))

C++解法如下：

class Solution {public:int poorPigs(int buckets, int minutesToDie, int minutesToTest) {int states = minutesToTest / minutesToDie + 1;return ceil(log(buckets) / log(states)); }
};

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LeetCode 458. Poor Pigs--智力题「小白鼠试毒」--C++,Python解法

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