458. Poor Pigs
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.
我尝试利用0和1编码解决，解出的答案不是最小值，因此查看了最高赞的答案（
StefanPochmann)
def poorPigs(self, buckets, minutesToDie, minutesToTest):
pigs = 0
while (minutesToTest / minutesToDie + 1) ** pigs < buckets:
pigs += 1
return pigs作者给出了清晰的解释：
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:
有两个地方
- 把每只猪作为一个维度的坐标轴
- 每个坐标轴长度是5，不是4.因为虽然只能测4次，但是根据排除法，可以推断第五次的情况。
自己用C++实现了一下
int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
int numOfTest=floor(minutesToTest/minutesToDie)+1;
int pigs=0;
while(pow(numOfTest,pigs)
}
return pigs;
}