feat: add biweekly contest 160

solution/3600-3699/3602.Hexadecimal and Hexatrigesimal Conversion/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3602.Hexadecimal%20and%20Hexatrigesimal%20Conversion/README.md
+---
+
+<!-- problem:start -->
+
+# [3602. 十六进制和三十六进制转化](https://leetcode.cn/problems/hexadecimal-and-hexatrigesimal-conversion)
+
+[English Version](/solution/3600-3699/3602.Hexadecimal%20and%20Hexatrigesimal%20Conversion/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个整数 <code>n</code>。</p>
+
+<p>返回 <code>n<sup>2</sup></code> 的&nbsp;<strong>十六进制表示</strong> 和 <code>n<sup>3</sup></code> 的&nbsp;<strong>三十六进制表示</strong> 拼接成的字符串。</p>
+
+<p><strong>十六进制&nbsp;</strong>数定义为使用数字 <code>0 – 9</code> 和大写字母 <code>A - F</code> 表示 0 到 15 的值。</p>
+
+<p><strong>三十六进制&nbsp;</strong>数定义为使用数字 <code>0 – 9</code> 和大写字母 <code>A - Z</code> 表示 0 到 35 的值。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">n = 13</span></p>
+
+<p><strong>输出：&nbsp;</strong><span class="example-io">"A91P1"</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><code>n<sup>2</sup> = 13 * 13 = 169</code>。在十六进制中，它转换为 <code>(10 * 16) + 9 = 169</code>，对应于 <code>"A9"</code>。</li>
+	<li><code>n<sup>3</sup> = 13 * 13 * 13 = 2197</code>。在三十六进制中，它转换为 <code>(1 * 36<sup>2</sup>) + (25 * 36) + 1 = 2197</code>，对应于 <code>"1P1"</code>。</li>
+	<li>连接两个结果得到 <code>"A9" + "1P1" = "A91P1"</code>。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">n = 36</span></p>
+
+<p><strong>输出：</strong><span class="example-io">"5101000"</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><code>n<sup>2</sup> = 36 * 36 = 1296</code>。在十六进制中，它转换为 <code>(5 * 16<sup>2</sup>) + (1 * 16) + 0 = 1296</code>，对应于 <code>"510"</code>。</li>
+	<li><code>n<sup>3</sup> = 36 * 36 * 36 = 46656</code>。在三十六进制中，它转换为 <code>(1 * 36<sup>3</sup>) + (0 * 36<sup>2</sup>) + (0 * 36) + 0 = 46656</code>，对应于 <code>"1000"</code>。</li>
+	<li>连接两个结果得到 <code>"510" + "1000" = "5101000"</code>。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 1000</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3600-3699/3602.Hexadecimal and Hexatrigesimal Conversion/README_EN.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3602.Hexadecimal%20and%20Hexatrigesimal%20Conversion/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3602. Hexadecimal and Hexatrigesimal Conversion](https://leetcode.com/problems/hexadecimal-and-hexatrigesimal-conversion)
+
+[中文文档](/solution/3600-3699/3602.Hexadecimal%20and%20Hexatrigesimal%20Conversion/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an integer <code>n</code>.</p>
+
+<p>Return the concatenation of the <strong>hexadecimal</strong> representation of <code>n<sup>2</sup></code> and the <strong>hexatrigesimal</strong> representation of <code>n<sup>3</sup></code>.</p>
+
+<p>A <strong>hexadecimal</strong> number is defined as a base-16 numeral system that uses the digits <code>0 &ndash; 9</code> and the uppercase letters <code>A - F</code> to represent values from 0 to 15.</p>
+
+<p>A <strong>hexatrigesimal</strong> number is defined as a base-36 numeral system that uses the digits <code>0 &ndash; 9</code> and the uppercase letters <code>A - Z</code> to represent values from 0 to 35.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">n = 13</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;A91P1&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><code>n<sup>2</sup> = 13 * 13 = 169</code>. In hexadecimal, it converts to <code>(10 * 16) + 9 = 169</code>, which corresponds to <code>&quot;A9&quot;</code>.</li>
+	<li><code>n<sup>3</sup> = 13 * 13 * 13 = 2197</code>. In hexatrigesimal, it converts to <code>(1 * 36<sup>2</sup>) + (25 * 36) + 1 = 2197</code>, which corresponds to <code>&quot;1P1&quot;</code>.</li>
+	<li>Concatenating both results gives <code>&quot;A9&quot; + &quot;1P1&quot; = &quot;A91P1&quot;</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">n = 36</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;5101000&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><code>n<sup>2</sup> = 36 * 36 = 1296</code>. In hexadecimal, it converts to <code>(5 * 16<sup>2</sup>) + (1 * 16) + 0 = 1296</code>, which corresponds to <code>&quot;510&quot;</code>.</li>
+	<li><code>n<sup>3</sup> = 36 * 36 * 36 = 46656</code>. In hexatrigesimal, it converts to <code>(1 * 36<sup>3</sup>) + (0 * 36<sup>2</sup>) + (0 * 36) + 0 = 46656</code>, which corresponds to <code>&quot;1000&quot;</code>.</li>
+	<li>Concatenating both results gives <code>&quot;510&quot; + &quot;1000&quot; = &quot;5101000&quot;</code>.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 1000</code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3600-3699/3603.Minimum Cost Path with Alternating Directions II/README.md

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+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3603.Minimum%20Cost%20Path%20with%20Alternating%20Directions%20II/README.md
+---
+
+<!-- problem:start -->
+
+# [3603. 交替方向的最小路径代价 II](https://leetcode.cn/problems/minimum-cost-path-with-alternating-directions-ii)
+
+[English Version](/solution/3600-3699/3603.Minimum%20Cost%20Path%20with%20Alternating%20Directions%20II/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你两个整数 <code>m</code> 和 <code>n</code>，分别表示网格的行数和列数。</p>
+
+<p>进入单元格 <code>(i, j)</code> 的成本定义为 <code>(i + 1) * (j + 1)</code>。</p>
+
+<p>另外给你一个二维整数数组 <code>waitCost</code>，其中 <code>waitCost[i][j]</code> 定义了在该单元格&nbsp;<strong>等待&nbsp;</strong>的成本。</p>
+
+<p>你从第 1 秒开始在单元格 <code>(0, 0)</code>。</p>
+
+<p>每一步，你都遵循交替模式：</p>
+
+<ul>
+	<li>在&nbsp;<strong>奇数秒&nbsp;</strong>，你必须向&nbsp;<strong>右&nbsp;</strong>或向&nbsp;<strong>下&nbsp;</strong>移动到&nbsp;<strong>相邻&nbsp;</strong>的单元格，并支付其进入成本。</li>
+	<li>在&nbsp;<strong>偶数秒&nbsp;</strong>，你必须原地&nbsp;<strong>等待&nbsp;</strong>，并支付 <code>waitCost[i][j]</code>。</li>
+</ul>
+
+<p>返回到达 <code>(m - 1, n - 1)</code> 所需的&nbsp;<strong>最小&nbsp;</strong>总成本。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">m = 1, n = 2, waitCost = [[1,2]]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">3</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>最佳路径为：</p>
+
+<ul>
+	<li>从第 1 秒开始在单元格 <code>(0, 0)</code>，进入成本为 <code>(0 + 1) * (0 + 1) = 1</code>。</li>
+	<li><strong>第 1 秒</strong>：向右移动到单元格 <code>(0, 1)</code>，进入成本为 <code>(0 + 1) * (1 + 1) = 2</code>。</li>
+</ul>
+
+<p>因此，总成本为 <code>1 + 2 = 3</code>。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">m = 2, n = 2, waitCost = [[3,5],[2,4]]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">9</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>最佳路径为：</p>
+
+<ul>
+	<li>从第 1 秒开始在单元格 <code>(0, 0)</code>，进入成本为 <code>(0 + 1) * (0 + 1) = 1</code>。</li>
+	<li><strong>第 1 秒</strong>：向下移动到单元格 <code>(1, 0)</code>，进入成本为 <code>(1 + 1) * (0 + 1) = 2</code>。</li>
+	<li><strong>第 2 秒</strong>：在单元格 <code>(1, 0)</code> 等待，支付 <code>waitCost[1][0] = 2</code>。</li>
+	<li><strong>第 3 秒</strong>：向右移动到单元格 <code>(1, 1)</code>，进入成本为 <code>(1 + 1) * (1 + 1) = 4</code>。</li>
+</ul>
+
+<p>因此，总成本为 <code>1 + 2 + 2 + 4 = 9</code>。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">m = 2, n = 3, waitCost = [[6,1,4],[3,2,5]]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">16</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>最佳路径为：</p>
+
+<ul>
+	<li>从第 1 秒开始在单元格 <code>(0, 0)</code>，进入成本为 <code>(0 + 1) * (0 + 1) = 1</code>。</li>
+	<li><strong>第 1 秒</strong>：向右移动到单元格 <code>(0, 1)</code>，进入成本为 <code>(0 + 1) * (1 + 1) = 2</code>。</li>
+	<li><strong>第 2 秒</strong>：在单元格 <code>(0, 1)</code> 等待，支付 <code>waitCost[0][1] = 1</code>。</li>
+	<li><strong>第 3 秒</strong>：向下移动到单元格 <code>(1, 1)</code>，进入成本为 <code>(1 + 1) * (1 + 1) = 4</code>。</li>
+	<li><strong>第 4 秒</strong>：在单元格 <code>(1, 1)</code> 等待，支付 <code>waitCost[1][1] = 2</code>。</li>
+	<li><strong>第 5 秒</strong>：向右移动到单元格 <code>(1, 2)</code>，进入成本为 <code>(1 + 1) * (2 + 1) = 6</code>。</li>
+</ul>
+
+<p>因此，总成本为 <code>1 + 2 + 1 + 4 + 2 + 6 = 16</code>。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
+	<li><code>2 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
+	<li><code>waitCost.length == m</code></li>
+	<li><code>waitCost[0].length == n</code></li>
+	<li><code>0 &lt;= waitCost[i][j] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->