feat: add biweekly contest 154

solution/3500-3599/3512.Minimum Operations to Make Array Sum Divisible by K/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3512.Minimum%20Operations%20to%20Make%20Array%20Sum%20Divisible%20by%20K/README.md
+---
+
+<!-- problem:start -->
+
+# [3512. 使数组和能被 K 整除的最少操作次数](https://leetcode.cn/problems/minimum-operations-to-make-array-sum-divisible-by-k)
+
+[English Version](/solution/3500-3599/3512.Minimum%20Operations%20to%20Make%20Array%20Sum%20Divisible%20by%20K/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个整数数组 <code>nums</code> 和一个整数 <code>k</code>。你可以执行以下操作任意次：</p>
+
+<ul>
+	<li>选择一个下标&nbsp;<code>i</code>，并将 <code>nums[i]</code> 替换为 <code>nums[i] - 1</code>。</li>
+</ul>
+
+<p>返回使数组元素之和能被 <code>k</code> 整除所需的<strong>最小</strong>操作次数。</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">nums = [3,9,7], k = 5</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">4</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>对 <code>nums[1] = 9</code> 执行 4 次操作。现在 <code>nums = [3, 5, 7]</code>。</li>
+	<li>数组之和为 15，可以被 5 整除。</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">nums = [4,1,3], k = 4</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">0</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>数组之和为 8，已经可以被 4 整除。因此不需要操作。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">nums = [3,2], k = 6</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">5</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>对 <code>nums[0] = 3</code> 执行 3 次操作，对 <code>nums[1] = 2</code> 执行 2 次操作。现在 <code>nums = [0, 0]</code>。</li>
+	<li>数组之和为 0，可以被 6 整除。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
+	<li><code>1 &lt;= k &lt;= 100</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/3500-3599/3512.Minimum Operations to Make Array Sum Divisible by K/README_EN.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3512.Minimum%20Operations%20to%20Make%20Array%20Sum%20Divisible%20by%20K/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3512. Minimum Operations to Make Array Sum Divisible by K](https://leetcode.com/problems/minimum-operations-to-make-array-sum-divisible-by-k)
+
+[中文文档](/solution/3500-3599/3512.Minimum%20Operations%20to%20Make%20Array%20Sum%20Divisible%20by%20K/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. You can perform the following operation any number of times:</p>
+
+<ul>
+	<li>Select an index <code>i</code> and replace <code>nums[i]</code> with <code>nums[i] - 1</code>.</li>
+</ul>
+
+<p>Return the <strong>minimum</strong> number of operations required to make the sum of the array divisible by <code>k</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [3,9,7], k = 5</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>Perform 4 operations on <code>nums[1] = 9</code>. Now, <code>nums = [3, 5, 7]</code>.</li>
+	<li>The sum is 15, which is divisible by 5.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [4,1,3], k = 4</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>The sum is 8, which is already divisible by 4. Hence, no operations are needed.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [3,2], k = 6</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">5</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>Perform 3 operations on <code>nums[0] = 3</code> and 2 operations on <code>nums[1] = 2</code>. Now, <code>nums = [0, 0]</code>.</li>
+	<li>The sum is 0, which is divisible by 6.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
+	<li><code>1 &lt;= k &lt;= 100</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/3500-3599/3513.Number of Unique XOR Triplets I/README.md

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+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3513.Number%20of%20Unique%20XOR%20Triplets%20I/README.md
+---
+
+<!-- problem:start -->
+
+# [3513. 不同 XOR 三元组的数目 I](https://leetcode.cn/problems/number-of-unique-xor-triplets-i)
+
+[English Version](/solution/3500-3599/3513.Number%20of%20Unique%20XOR%20Triplets%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个长度为 <code>n</code> 的整数数组 <code>nums</code>，其中 <code>nums</code> 是范围 <code>[1, n]</code> 内所有数的&nbsp;<strong>排列&nbsp;</strong>。</p>
+
+<p><strong>XOR 三元组</strong> 定义为三个元素的异或值 <code>nums[i] XOR nums[j] XOR nums[k]</code>，其中 <code>i &lt;= j &lt;= k</code>。</p>
+
+<p>返回所有可能三元组 <code>(i, j, k)</code> 中&nbsp;<strong>不同&nbsp;</strong>的 XOR 值的数量。</p>
+
+<p><strong>排列</strong> 是一个集合中所有元素的重新排列。</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">nums = [1,2]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">2</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>所有可能的 XOR 三元组值为：</p>
+
+<ul>
+	<li><code>(0, 0, 0) → 1 XOR 1 XOR 1 = 1</code></li>
+	<li><code>(0, 0, 1) → 1 XOR 1 XOR 2 = 2</code></li>
+	<li><code>(0, 1, 1) → 1 XOR 2 XOR 2 = 1</code></li>
+	<li><code>(1, 1, 1) → 2 XOR 2 XOR 2 = 2</code></li>
+</ul>
+
+<p>不同的 XOR 值为 <code>{1, 2}</code>，因此输出为 2。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">nums = [3,1,2]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">4</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>可能的 XOR 三元组值包括：</p>
+
+<ul>
+	<li><code>(0, 0, 0) → 3 XOR 3 XOR 3 = 3</code></li>
+	<li><code>(0, 0, 1) → 3 XOR 3 XOR 1 = 1</code></li>
+	<li><code>(0, 0, 2) → 3 XOR 3 XOR 2 = 2</code></li>
+	<li><code>(0, 1, 2) → 3 XOR 1 XOR 2 = 0</code></li>
+</ul>
+
+<p>不同的 XOR 值为 <code>{0, 1, 2, 3}</code>，因此输出为 4。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= n</code></li>
+	<li><code>nums</code> 是从 <code>1</code> 到 <code>n</code> 的整数的一个排列。</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 -->