feat: add biweekly contest 126 (doocs#2450)

Author: yanglbme

solution/3000-3099/3079.Find the Sum of Encrypted Integers/README.md

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+# [3079. 求出加密整数的和](https://leetcode.cn/problems/find-the-sum-of-encrypted-integers)
+
+[English Version](/solution/3000-3099/3079.Find%20the%20Sum%20of%20Encrypted%20Integers/README_EN.md)
+
+<!-- tags: -->
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个整数数组&nbsp;<code>nums</code>&nbsp;，数组中的元素都是&nbsp;<strong>正</strong>&nbsp;整数。定义一个加密函数&nbsp;<code>encrypt</code>&nbsp;，<code>encrypt(x)</code>&nbsp;将一个整数 <code>x</code>&nbsp;中 <strong>每一个</strong>&nbsp;数位都用 <code>x</code>&nbsp;中的&nbsp;<strong>最大</strong>&nbsp;数位替换。比方说&nbsp;<code>encrypt(523) = 555</code> 且&nbsp;<code>encrypt(213) = 333</code>&nbsp;。</p>
+
+<p>请你返回数组中所有元素加密后的 <strong>和</strong>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>输入：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">nums = [1,2,3]</span></p>
+
+<p><strong>输出：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">6</span></p>
+
+<p><b>解释：</b>加密后的元素位&nbsp;<code>[1,2,3]</code>&nbsp;。加密元素的和为&nbsp;<code>1 + 2 + 3 == 6</code>&nbsp;。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>输入：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">nums = [10,21,31]</span></p>
+
+<p><strong>输出：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">66</span></p>
+
+<p><b>解释：</b>加密后的元素为&nbsp;<code>[11,22,33]</code>&nbsp;。加密元素的和为&nbsp;<code>11 + 22 + 33 == 66</code> 。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
+</ul>
+
+## 解法
+
+### 方法一
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/3000-3099/3079.Find the Sum of Encrypted Integers/README_EN.md

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+# [3079. Find the Sum of Encrypted Integers](https://leetcode.com/problems/find-the-sum-of-encrypted-integers)
+
+[中文文档](/solution/3000-3099/3079.Find%20the%20Sum%20of%20Encrypted%20Integers/README.md)
+
+<!-- tags: -->
+
+## Description
+
+<p>You are given an integer array <code>nums</code> containing <strong>positive</strong> integers. We define a function <code>encrypt</code> such that <code>encrypt(x)</code> replaces <strong>every</strong> digit in <code>x</code> with the <strong>largest</strong> digit in <code>x</code>. For example, <code>encrypt(523) = 555</code> and <code>encrypt(213) = 333</code>.</p>
+
+<p>Return <em>the <strong>sum </strong>of encrypted elements</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">nums = [1,2,3]</span></p>
+
+<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">6</span></p>
+
+<p><strong>Explanation:</strong> The encrypted elements are&nbsp;<code>[1,2,3]</code>. The sum of encrypted elements is <code>1 + 2 + 3 == 6</code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">nums = [10,21,31]</span></p>
+
+<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">66</span></p>
+
+<p><strong>Explanation:</strong> The encrypted elements are <code>[11,22,33]</code>. The sum of encrypted elements is <code>11 + 22 + 33 == 66</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
+</ul>
+
+## Solutions
+
+### Solution 1
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/3000-3099/3080.Mark Elements on Array by Performing Queries/README.md

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+# [3080. 执行操作标记数组中的元素](https://leetcode.cn/problems/mark-elements-on-array-by-performing-queries)
+
+[English Version](/solution/3000-3099/3080.Mark%20Elements%20on%20Array%20by%20Performing%20Queries/README_EN.md)
+
+<!-- tags: -->
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个长度为 <code>n</code>&nbsp;下标从 <strong>0</strong>&nbsp;开始的正整数数组&nbsp;<code>nums</code>&nbsp;。</p>
+
+<p>同时给你一个长度为 <code>m</code>&nbsp;的二维操作数组&nbsp;<code>queries</code>&nbsp;，其中&nbsp;<code>queries[i] = [index<sub>i</sub>, k<sub>i</sub>]</code>&nbsp;。</p>
+
+<p>一开始，数组中的所有元素都 <strong>未标记</strong>&nbsp;。</p>
+
+<p>你需要依次对数组执行 <code>m</code>&nbsp;次操作，第 <code>i</code>&nbsp;次操作中，你需要执行：</p>
+
+<ul>
+	<li>如果下标&nbsp;<code>index<sub>i</sub></code>&nbsp;对应的元素还没标记，那么标记这个元素。</li>
+	<li>然后标记&nbsp;<code>k<sub>i</sub></code>&nbsp;个数组中还没有标记的&nbsp;<strong>最小</strong>&nbsp;元素。如果有元素的值相等，那么优先标记它们中下标较小的。如果少于&nbsp;<code>k<sub>i</sub></code>&nbsp;个未标记元素存在，那么将它们全部标记。</li>
+</ul>
+
+<p>请你返回一个长度为 <code>m</code>&nbsp;的数组 <code>answer</code>&nbsp;，其中<em>&nbsp;</em><code>answer[i]</code>是第&nbsp;<code>i</code>&nbsp;次操作后数组中还没标记元素的&nbsp;<strong>和</strong>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>输入：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">nums = [1,2,2,1,2,3,1], queries = [[1,2],[3,3],[4,2]]</span></p>
+
+<p><strong>输出：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">[8,3,0]</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>我们依次对数组做以下操作：</p>
+
+<ul>
+	<li>标记下标为&nbsp;<code>1</code>&nbsp;的元素，同时标记&nbsp;<code>2</code>&nbsp;个未标记的最小元素。标记完后数组为&nbsp;<code>nums = [<em><strong>1</strong></em>,<em><strong>2</strong></em>,2,<em><strong>1</strong></em>,2,3,1]</code>&nbsp;。未标记元素的和为&nbsp;<code>2 + 2 + 3 + 1 = 8</code>&nbsp;。</li>
+	<li>标记下标为&nbsp;<code>3</code>&nbsp;的元素，由于它已经被标记过了，所以我们忽略这次标记，同时标记最靠前的&nbsp;<code>3</code>&nbsp;个未标记的最小元素。标记完后数组为&nbsp;<code>nums = [<em><strong>1</strong></em>,<em><strong>2</strong></em>,<em><strong>2</strong></em>,<em><strong>1</strong></em>,<em><strong>2</strong></em>,3,<em><strong>1</strong></em>]</code>&nbsp;。未标记元素的和为&nbsp;<code>3</code>&nbsp;。</li>
+	<li>标记下标为 <code>4</code>&nbsp;的元素，由于它已经被标记过了，所以我们忽略这次标记，同时标记最靠前的 <code>2</code>&nbsp;个未标记的最小元素。标记完后数组为&nbsp;<code>nums = [<em><strong>1</strong></em>,<em><strong>2</strong></em>,<em><strong>2</strong></em>,<em><strong>1</strong></em>,<em><strong>2</strong></em>,<em><strong>3</strong></em>,<em><strong>1</strong></em>]</code>&nbsp;。未标记元素的和为&nbsp;<code>0</code>&nbsp;。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>输入：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">nums = [1,4,2,3], queries = [[0,1]]</span></p>
+
+<p><strong>输出：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">[7]</span></p>
+
+<p><strong>解释：</strong>我们执行一次操作，将下标为&nbsp;<code>0</code>&nbsp;处的元素标记，并且标记最靠前的&nbsp;<code>1</code>&nbsp;个未标记的最小元素。标记完后数组为&nbsp;<code>nums = [<em><strong>1</strong></em>,4,<em><strong>2</strong></em>,3]</code>&nbsp;。未标记元素的和为&nbsp;<code>4 + 3 = 7</code>&nbsp;。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>n == nums.length</code></li>
+	<li><code>m == queries.length</code></li>
+	<li><code>1 &lt;= m &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>queries[i].length == 2</code></li>
+	<li><code>0 &lt;= index<sub>i</sub>, k<sub>i</sub> &lt;= n - 1</code></li>
+</ul>
+
+## 解法
+
+### 方法一
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/3000-3099/3080.Mark Elements on Array by Performing Queries/README_EN.md

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+# [3080. Mark Elements on Array by Performing Queries](https://leetcode.com/problems/mark-elements-on-array-by-performing-queries)
+
+[中文文档](/solution/3000-3099/3080.Mark%20Elements%20on%20Array%20by%20Performing%20Queries/README.md)
+
+<!-- tags: -->
+
+## Description
+
+<p>You are given a <strong>0-indexed</strong> array <code>nums</code> of size <code>n</code> consisting of positive integers.</p>
+
+<p>You are also given a 2D array <code>queries</code> of size <code>m</code> where <code>queries[i] = [index<sub>i</sub>, k<sub>i</sub>]</code>.</p>
+
+<p>Initially all elements of the array are <strong>unmarked</strong>.</p>
+
+<p>You need to apply <code>m</code> queries on the array in order, where on the <code>i<sup>th</sup></code> query you do the following:</p>
+
+<ul>
+	<li>Mark the element at index <code>index<sub>i</sub></code> if it is not already marked.</li>
+	<li>Then mark <code>k<sub>i</sub></code> unmarked elements in the array with the <strong>smallest</strong> values. If multiple such elements exist, mark the ones with the smallest indices. And if less than <code>k<sub>i</sub></code> unmarked elements exist, then mark all of them.</li>
+</ul>
+
+<p>Return <em>an array answer of size </em><code>m</code><em> where </em><code>answer[i]</code><em> is the <strong>sum</strong> of unmarked elements in the array after the </em><code>i<sup>th</sup></code><em> query</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">nums = [1,2,2,1,2,3,1], queries = [[1,2],[3,3],[4,2]]</span></p>
+
+<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">[8,3,0]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We do the following queries on the array:</p>
+
+<ul>
+	<li>Mark the element at index <code>1</code>, and <code>2</code> of the smallest unmarked elements with the smallest indices if they exist, the marked elements now are <code>nums = [<strong><u>1</u></strong>,<u><strong>2</strong></u>,2,<u><strong>1</strong></u>,2,3,1]</code>. The sum of unmarked elements is <code>2 + 2 + 3 + 1 = 8</code>.</li>
+	<li>Mark the element at index <code>3</code>, since it is already marked we skip it. Then we mark <code>3</code> of the smallest unmarked elements with the smallest indices, the marked elements now are <code>nums = [<strong><u>1</u></strong>,<u><strong>2</strong></u>,<u><strong>2</strong></u>,<u><strong>1</strong></u>,<u><strong>2</strong></u>,3,<strong><u>1</u></strong>]</code>. The sum of unmarked elements is <code>3</code>.</li>
+	<li>Mark the element at index <code>4</code>, since it is already marked we skip it. Then we mark <code>2</code> of the smallest unmarked elements with the smallest indices if they exist, the marked elements now are <code>nums = [<strong><u>1</u></strong>,<u><strong>2</strong></u>,<u><strong>2</strong></u>,<u><strong>1</strong></u>,<u><strong>2</strong></u>,<strong><u>3</u></strong>,<u><strong>1</strong></u>]</code>. The sum of unmarked elements is <code>0</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">nums = [1,4,2,3], queries = [[0,1]]</span></p>
+
+<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">[7]</span></p>
+
+<p><strong>Explanation: </strong> We do one query which is mark the element at index <code>0</code> and mark the smallest element among unmarked elements. The marked elements will be <code>nums = [<strong><u>1</u></strong>,4,<u><strong>2</strong></u>,3]</code>, and the sum of unmarked elements is <code>4 + 3 = 7</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == nums.length</code></li>
+	<li><code>m == queries.length</code></li>
+	<li><code>1 &lt;= m &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>queries[i].length == 2</code></li>
+	<li><code>0 &lt;= index<sub>i</sub>, k<sub>i</sub> &lt;= n - 1</code></li>
+</ul>
+
+## Solutions
+
+### Solution 1
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->