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Commit 0f1941e

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feat: add weekly contest 373 (#2017)
 --------- Co-authored-by: Doocs Bot <doocs-bot@outlook.com>
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# [2946. 循环移位后的矩阵相似检查](https://leetcode.cn/problems/matrix-similarity-after-cyclic-shifts)
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[English Version](/solution/2900-2999/2946.Matrix%20Similarity%20After%20Cyclic%20Shifts/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个大小为 <code>m x n</code> 的整数矩阵 <code>mat</code> 和一个整数 <code>k</code> 。请你将矩阵中的<strong> 奇数</strong> 行循环 <strong>右</strong> 移 <code>k</code> 次,<strong>偶数</strong> 行循环 <strong>左</strong> 移 <code>k</code> 次。</p>
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<p>如果初始矩阵和最终矩阵完全相同,则返回 <code>true</code> ,否则返回 <code>false</code> 。</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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<pre>
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<strong>输入:</strong>mat = [[1,2,1,2],[5,5,5,5],[6,3,6,3]], k = 2
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<strong>输出:</strong>true
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<strong>解释:</strong>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2946.Matrix%20Similarity%20After%20Cyclic%20Shifts/images/similarmatrix.png" style="width: 500px; height: 117px;" />
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初始矩阵如图一所示。
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图二表示对奇数行右移一次且对偶数行左移一次后的矩阵状态。
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图三是经过两次循环移位后的最终矩阵状态,与初始矩阵相同。
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因此,返回 true 。
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</pre>
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<p><strong class="example">示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>mat = [[2,2],[2,2]], k = 3
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<strong>输出:</strong>true
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<strong>解释:</strong>由于矩阵中的所有值都相等,即使进行循环移位,矩阵仍然保持不变。因此,返回 true 。
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</pre>
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<p><strong class="example">示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>mat = [[1,2]], k = 1
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<strong>输出:</strong>false
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<strong>解释:</strong>循环移位一次后,mat = [[2,1]],与初始矩阵不相等。因此,返回 false 。
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</pre>
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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;= mat.length &lt;= 25</code></li>
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<li><code>1 &lt;= mat[i].length &lt;= 25</code></li>
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<li><code>1 &lt;= mat[i][j] &lt;= 25</code></li>
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<li><code>1 &lt;= k &lt;= 50</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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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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### **...**
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```
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```
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<!-- tabs:end -->
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# [2946. Matrix Similarity After Cyclic Shifts](https://leetcode.com/problems/matrix-similarity-after-cyclic-shifts)
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[中文文档](/solution/2900-2999/2946.Matrix%20Similarity%20After%20Cyclic%20Shifts/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> <code>m x n</code> integer matrix <code>mat</code> and an integer <code>k</code>. You have to cyclically <strong>right</strong> shift <strong>odd</strong> indexed rows <code>k</code> times and cyclically <strong>left</strong> shift <strong>even</strong> indexed rows <code>k</code> times.</p>
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<p>Return <code>true</code> <em>if the initial and final matrix are exactly the same and </em><code>false</code> <em>otherwise.</em></p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> mat = [[1,2,1,2],[5,5,5,5],[6,3,6,3]], k = 2
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<strong>Output:</strong> true
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<strong>Explanation:</strong>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2946.Matrix%20Similarity%20After%20Cyclic%20Shifts/images/similarmatrix.png" style="width: 500px; height: 117px;" />
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Initially, the matrix looks like the first figure.
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Second figure represents the state of the matrix after one right and left cyclic shifts to even and odd indexed rows.
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Third figure is the final state of the matrix after two cyclic shifts which is similar to the initial matrix.
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Therefore, return true.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> mat = [[2,2],[2,2]], k = 3
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<strong>Output:</strong> true
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<strong>Explanation:</strong> As all the values are equal in the matrix, even after performing cyclic shifts the matrix will remain the same. Therefeore, we return true.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> mat = [[1,2]], k = 1
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<strong>Output:</strong> false
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<strong>Explanation:</strong> After one cyclic shift, mat = [[2,1]] which is not equal to the initial matrix. Therefore we return false.
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</pre>
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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>1 &lt;= mat.length &lt;= 25</code></li>
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<li><code>1 &lt;= mat[i].length &lt;= 25</code></li>
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<li><code>1 &lt;= mat[i][j] &lt;= 25</code></li>
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<li><code>1 &lt;= k &lt;= 50</code></li>
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</ul>
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## Solutions
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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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### **...**
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```
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```
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<!-- tabs:end -->
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# [2947. 统计美丽子字符串 I](https://leetcode.cn/problems/count-beautiful-substrings-i)
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[English Version](/solution/2900-2999/2947.Count%20Beautiful%20Substrings%20I/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个字符串 <code>s</code> 和一个正整数 <code>k</code> 。</p>
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<p>用 <code>vowels</code> 和 <code>consonants</code> 分别表示字符串中元音字母和辅音字母的数量。</p>
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<p>如果某个字符串满足以下条件,则称其为 <strong>美丽字符串</strong> :</p>
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<ul>
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<li><code>vowels == consonants</code>,即元音字母和辅音字母的数量相等。</li>
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<li><code>(vowels * consonants) % k == 0</code>,即元音字母和辅音字母的数量的乘积能被 <code>k</code> 整除。</li>
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</ul>
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<p>返回字符串 <code>s</code> 中 <strong>非空美丽子字符串</strong> 的数量。</p>
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<p>子字符串是字符串中的一个连续字符序列。</p>
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<p>英语中的<strong> 元音字母 </strong>为 <code>'a'</code>、<code>'e'</code>、<code>'i'</code>、<code>'o'</code> 和 <code>'u'</code> 。</p>
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<p>英语中的<strong> 辅音字母 </strong>为除了元音字母之外的所有字母。</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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<pre>
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<strong>输入:</strong>s = "baeyh", k = 2
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<strong>输出:</strong>2
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<strong>解释:</strong>字符串 s 中有 2 个美丽子字符串。
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- 子字符串 "b<em><strong>aeyh</strong></em>",vowels = 2(["a","e"]),consonants = 2(["y","h"])。
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可以看出字符串 "aeyh" 是美丽字符串,因为 vowels == consonants 且 vowels * consonants % k == 0 。
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- 子字符串 "<em><strong>baey</strong></em>h",vowels = 2(["a","e"]),consonants = 2(["b","y"])。
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可以看出字符串 "baey" 是美丽字符串,因为 vowels == consonants 且 vowels * consonants % k == 0 。
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可以证明字符串 s 中只有 2 个美丽子字符串。
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</pre>
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<p><strong class="example">示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>s = "abba", k = 1
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<strong>输出:</strong>3
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<strong>解释:</strong>字符串 s 中有 3 个美丽子字符串。
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- 子字符串 "<strong><em>ab</em></strong>ba",vowels = 1(["a"]),consonants = 1(["b"])。
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- 子字符串 "ab<strong><em>ba</em></strong>",vowels = 1(["a"]),consonants = 1(["b"])。
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- 子字符串 "<em><strong>abba</strong></em>",vowels = 2(["a","a"]),consonants = 2(["b","b"])。
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可以证明字符串 s 中只有 3 个美丽子字符串。
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</pre>
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<p><strong class="example">示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>s = "bcdf", k = 1
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<strong>输出:</strong>0
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<strong>解释:</strong>字符串 s 中没有美丽子字符串。
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</pre>
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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;= 1000</code></li>
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<li><code>1 &lt;= k &lt;= 1000</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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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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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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### **...**
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```
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```
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<!-- tabs:end -->

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