poor-pigs

java

python

python3

mysql

mssql

oraclesql

csharp

javascript

typescript

bash

swift

kotlin

dart

golang

ruby

scala

rust

racket

erlang

elixir

pythondata

postgresql

cangjie

Algorithms

There are <code>buckets</code> buckets of liquid, where exactly one of the buckets is poisonous. To figure out which one is poisonous, you feed some number of (poor) pigs the liquid to see whether they will die or not. Unfortunately, you only have <code>minutesToTest</code> minutes to determine which bucket is poisonous.

You can feed the pigs according to these steps:

<ol>
	<li>Choose some live pigs to feed.</li>
	<li>For each pig, choose which buckets to feed it. The pig will consume all the chosen buckets simultaneously and will take no time. Each pig can feed from any number of buckets, and each bucket can be fed from by any number of pigs.</li>
	<li>Wait for <code>minutesToDie</code> minutes. You may not feed any other pigs during this time.</li>
	<li>After <code>minutesToDie</code> minutes have passed, any pigs that have been fed the poisonous bucket will die, and all others will survive.</li>
	<li>Repeat this process until you run out of time.</li>
</ol>

Given <code>buckets</code>, <code>minutesToDie</code>, and <code>minutesToTest</code>, return the minimum number of pigs needed to figure out which bucket is poisonous within the allotted time.

&nbsp;
Example 1:

<pre>
Input: buckets = 4, minutesToDie = 15, minutesToTest = 15
Output: 2
Explanation: We can determine the poisonous bucket as follows:
At time 0, feed the first pig buckets 1 and 2, and feed the second pig buckets 2 and 3.
At time 15, there are 4 possible outcomes:
- If only the first pig dies, then bucket 1 must be poisonous.
- If only the second pig dies, then bucket 3 must be poisonous.
- If both pigs die, then bucket 2 must be poisonous.
- If neither pig dies, then bucket 4 must be poisonous.
</pre>

Example 2:

<pre>
Input: buckets = 4, minutesToDie = 15, minutesToTest = 30
Output: 2
Explanation: We can determine the poisonous bucket as follows:
At time 0, feed the first pig bucket 1, and feed the second pig bucket 2.
At time 15, there are 2 possible outcomes:
- If either pig dies, then the poisonous bucket is the one it was fed.
- If neither pig dies, then feed the first pig bucket 3, and feed the second pig bucket 4.
At time 30, one of the two pigs must die, and the poisonous bucket is the one it was fed.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= buckets &lt;= 1000</code></li>
	<li><code>1 &lt;=&nbsp;minutesToDie &lt;=&nbsp;minutesToTest &lt;= 100</code></li>
</ul>

poorPigs

Poor Pigs

Math

Dynamic Programming

Combinatorics

有<code> buckets</code> 桶液体，其中 正好有一桶&nbsp;含有毒药，其余装的都是水。它们从外观看起来都一样。为了弄清楚哪只水桶含有毒药，你可以喂一些猪喝，通过观察猪是否会死进行判断。不幸的是，你只有&nbsp;<code>minutesToTest</code> 分钟时间来确定哪桶液体是有毒的。

喂猪的规则如下：

<ol>
	<li>选择若干活猪进行喂养</li>
	<li>可以允许小猪同时饮用任意数量的桶中的水，并且该过程不需要时间。</li>
	<li>小猪喝完水后，必须有 <code>minutesToDie</code> 分钟的冷却时间。在这段时间里，你只能观察，而不允许继续喂猪。</li>
	<li>过了 <code>minutesToDie</code> 分钟后，所有喝到毒药的猪都会死去，其他所有猪都会活下来。</li>
	<li>重复这一过程，直到时间用完。</li>
</ol>

给你桶的数目 <code>buckets</code> ，<code>minutesToDie</code> 和 <code>minutesToTest</code> ，返回&nbsp;在规定时间内判断哪个桶有毒所需的 最小 猪数&nbsp;。

&nbsp;

示例 1：

<pre>
输入：buckets = 1000, minutesToDie = 15, minutesToTest = 60
输出：5
</pre>

示例 2：

<pre>
输入：buckets = 4, minutesToDie = 15, minutesToTest = 15
输出：2
</pre>

示例 3：

<pre>
输入：buckets = 4, minutesToDie = 15, minutesToTest = 30
输出：2
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= buckets &lt;= 1000</code></li>
	<li><code>1 &lt;=&nbsp;minutesToDie &lt;=&nbsp;minutesToTest &lt;= 100</code></li>
</ul>