标签：时间 ike .com within inter eal nat 情况 out
There are 1000 buckets, one and only one of them contains poison, the rest are filled with water. They all look the same. If a pig drinks that poison it will die within 15 minutes. What is the minimum amount of pigs you need to figure out which bucket contains the poison within one hour.
Answer this question, and write an algorithm for the follow-up general case.
Follow-up:
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 "poison" bucket within p minutes? There is exact one bucket with poison.
我草，说是简单题目，简单你妹啊！出题人脑壳有问题！！！明显是数学题，太尼玛打击老子了想了一个晚上没有搞定。
摘录别人的解答：
解答的思路：
1. 一只猪在一小时内最多能验多少桶？
一次喝一个桶的，15分钟后没挂再喝第二桶，一小时60分钟内可以喝 60/15 = 4 次，如果有5桶水，那个只要喝前4桶就只能第5桶是否有毒。
因此一只小猪在一小时可以验5桶水
2. 两只呢？
既然一只能验5桶，那么用二维的思路，2只猪应该可以验5*5桶：
猪A负责行，猪B负责列，每15分钟试喝一行/一列的所有5桶水，通过2只猪上天的时间能推断出毒水在几行几列。
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With 2 pigs, poison killing in 15 minutes, and having 60 minutes, we can find the poison in up to 25 buckets in the following way. Arrange the buckets in a 5×5 square:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
Now use one pig to find the row (make it drink from buckets 1, 2, 3, 4, 5, wait 15 minutes, make it drink from buckets 6, 7, 8, 9, 10, wait 15 minutes, etc). Use the second pig to find the column (make it drink 1, 6, 11, 16, 21, then 2, 7, 12, 17, 22, etc).
Having 60 minutes and tests taking 15 minutes means we can run four tests. If the row pig dies in the third test, the poison is in the third row. If the column pig doesn‘t die at all, the poison is in the fifth column (this is why we can cover five rows/columns even though we can only run four tests).
With 3 pigs, we can similarly use a 5×5×5 cube instead of a 5×5 square and again use one pig to determine the coordinate of one dimension (one pig drinks layers from top to bottom, one drinks layers from left to right, one drinks layers from front to back). So 3 pigs can solve up to 125 buckets.
In general, we can solve up to (?minutesToTest / minutesToDie? + 1)pigs buckets this way, so just find the smallest sufficient number of pigs for example like this:
def poorPigs(self, buckets, minutesToDie, minutesToTest):
pigs = 0
while (minutesToTest / minutesToDie + 1) ** pigs < buckets:
pigs += 1
return pigs
Or with logarithm like I‘ve seen other people do it. That‘s also where I got the idea from (I didn‘t really try solving this problem on my own because the judge‘s solution originally was wrong and I was more interested in possibly helping to make it right quickly).
还是觉得
minutesToTest / minutesToDie + 1 中+1有问题？因为60分钟，只能测试4次，也就是最后的25个桶实际上是喝不到的。 1 2 3 4 5
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标签：时间 ike .com within inter eal nat 情况 out
原文地址：https://www.cnblogs.com/bonelee/p/9065189.html
