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feat: add weekly contest 451 (#4431)
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‎solution/3500-3599/3560.Find Minimum Log Transportation Cost/README.md

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---
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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3560.Find%20Minimum%20Log%20Transportation%20Cost/README.md
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---
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<!-- problem:start -->
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# [3560. 木材运输的最小成本](https://leetcode.cn/problems/find-minimum-log-transportation-cost)
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[English Version](/solution/3500-3599/3560.Find%20Minimum%20Log%20Transportation%20Cost/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给你三个整数 <code>n</code>、<code>m</code> 和 <code>k</code>。</p>
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<p>有两根长度分别为 <code>n</code> 和 <code>m</code> 单位的木材，需要通过三辆卡车运输。每辆卡车最多只能装载一根长度&nbsp;<strong>不超过</strong> <code>k</code> 单位的木材。</p>
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<p>你可以将木材切成更小的段，其中将长度为 <code>x</code> 的木材切割成长度为 <code>len1</code> 和 <code>len2</code> 的段的成本为 <code>cost = len1 * len2</code>，并且满足 <code>len1 + len2 = x</code>。</p>
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<p>返回将木材分配到卡车上的&nbsp;<strong>最小总成本&nbsp;</strong>。如果木材不需要切割，总成本为 0。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">n = 6, m = 5, k = 5</span></p>
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<p><strong>输出：</strong> <span class="example-io">5</span></p>
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<p><strong>解释：</strong></p>
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<p>将长度为 6 的木材切割成长度为 1 和 5 的两段，成本为 <code>1 * 5 == 5</code>。现在三段长度分别为 1、5 和 5 的木材可以分别装载到每辆卡车。</p>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">n = 4, m = 4, k = 6</span></p>
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<p><strong>输出：</strong> <span class="example-io">0</span></p>
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<p><strong>解释：</strong></p>
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<p>两根木材已经可以直接装载到卡车上，因此不需要切割。</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>2 &lt;= k &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= n, m &lt;= 2 * k</code></li>
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<li>输入数据保证木材总存在能被运输的方案。</li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一
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<!-- tabs:start -->
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#### Python3
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```python
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```
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#### Java
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```java
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```
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#### C++
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```cpp
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```
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#### Go
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```go
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/3500-3599/3560.Find Minimum Log Transportation Cost/README_EN.md

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---
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comments: true
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difficulty: Easy
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3560.Find%20Minimum%20Log%20Transportation%20Cost/README_EN.md
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---
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<!-- problem:start -->
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# [3560. Find Minimum Log Transportation Cost](https://leetcode.com/problems/find-minimum-log-transportation-cost)
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[中文文档](/solution/3500-3599/3560.Find%20Minimum%20Log%20Transportation%20Cost/README.md)
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## Description
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<!-- description:start -->
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<p>You are given integers <code>n</code>, <code>m</code>, and <code>k</code>.</p>
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<p>There are two logs of lengths <code>n</code> and <code>m</code> units, which need to be transported in three trucks where each truck can carry one log with length <strong>at most</strong> <code>k</code> units.</p>
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<p>You may cut the logs into smaller pieces, where the cost of cutting a log of length <code>x</code> into logs of length <code>len1</code> and <code>len2</code> is <code>cost = len1 * len2</code> such that <code>len1 + len2 = x</code>.</p>
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<p>Return the <strong>minimum total cost</strong> to distribute the logs onto the trucks. If the logs don&#39;t need to be cut, the total cost is 0.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">n = 6, m = 5, k = 5</span></p>
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<p><strong>Output:</strong> <span class="example-io">5</span></p>
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<p><strong>Explanation:</strong></p>
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<p>Cut the log with length 6 into logs with length 1 and 5, at a cost equal to <code>1 * 5 == 5</code>. Now the three logs of length 1, 5, and 5 can fit in one truck each.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">n = 4, m = 4, k = 6</span></p>
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<p><strong>Output:</strong> <span class="example-io">0</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The two logs can fit in the trucks already, hence we don&#39;t need to cut the logs.</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>2 &lt;= k &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= n, m &lt;= 2 * k</code></li>
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<li>The input is generated such that it is always possible to transport the logs.</li>
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</ul>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1
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<!-- tabs:start -->
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#### Python3
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```python
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```
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#### Java
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```java
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```
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#### C++
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```cpp
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```
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#### Go
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```go
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```
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<!-- tabs:end -->
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<!-- problem:end -->

‎solution/3500-3599/3561.Resulting String After Adjacent Removals/README.md

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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3561.Resulting%20String%20After%20Adjacent%20Removals/README.md
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---
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<!-- problem:start -->
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# [3561. 移除相邻字符](https://leetcode.cn/problems/resulting-string-after-adjacent-removals)
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[English Version](/solution/3500-3599/3561.Resulting%20String%20After%20Adjacent%20Removals/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给你一个由小写英文字母组成的字符串 <code>s</code>。</p>
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<p>你&nbsp;<strong>必须&nbsp;</strong>在字符串 <code>s</code> 中至少存在两个&nbsp;<strong>连续&nbsp;</strong>字符时，反复执行以下操作：</p>
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<ul>
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<li>移除字符串中&nbsp;<strong>最左边&nbsp;</strong>的一对按照字母表&nbsp;<strong>连续&nbsp;</strong>的相邻字符（无论是按顺序还是逆序，例如 <code>'a'</code> 和 <code>'b'</code>，或 <code>'b'</code> 和 <code>'a'</code>）。</li>
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<li>将剩余字符向左移动以填补空隙。</li>
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</ul>
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<p>当无法再执行任何操作时，返回最终的字符串。</p>
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<p><strong>注意：</strong>字母表是循环的，因此 <code>'a'</code> 和 <code>'z'</code> 也视为连续。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong> <span class="example-io">s = "abc"</span></p>
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<p><strong>输出:</strong> <span class="example-io">"c"</span></p>
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<p><strong>解释:</strong></p>
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<ul>
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<li>从字符串中移除 <code>"ab"</code>，剩下 <code>"c"</code>。</li>
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<li>无法进行进一步操作。因此，所有可能移除操作后的最终字符串为 <code>"c"</code>。</li>
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</ul>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong> <span class="example-io">s = "adcb"</span></p>
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<p><strong>输出:</strong> <span class="example-io">""</span></p>
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<p><strong>解释:</strong></p>
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<ul>
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<li>从字符串中移除 <code>"dc"</code>，剩下 <code>"ab"</code>。</li>
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<li>从字符串中移除 <code>"ab"</code>，剩下 <code>""</code>。</li>
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<li>无法进行进一步操作。因此，所有可能移除操作后的最终字符串为 <code>""</code>。</li>
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</ul>
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</div>
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<p><strong class="example">示例 3：</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong> <span class="example-io">s = "zadb"</span></p>
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<p><strong>输出:</strong> <span class="example-io">"db"</span></p>
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<p><strong>解释:</strong></p>
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<ul>
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<li>从字符串中移除 <code>"za"</code>，剩下 <code>"db"</code>。</li>
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<li>无法进行进一步操作。因此，所有可能移除操作后的最终字符串为 <code>"db"</code>。</li>
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</ul>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
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<li><code>s</code> 仅由小写英文字母组成。</li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一
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<!-- tabs:start -->
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#### Python3
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```python
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```
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#### Java
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```java
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```
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#### C++
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```cpp
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```
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#### Go
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```go
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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