feat: add new lc problem

README.md

@@ -52,7 +52,7 @@
 - [乘积小于 K 的子数组](/solution/0700-0799/0713.Subarray%20Product%20Less%20Than%20K/README.md) - `双指针`
 - [位 1 的个数](/solution/0100-0199/0191.Number%20of%201%20Bits/README.md) - `位运算`、`lowbit`
 - [合并区间](/solution/0000-0099/0056.Merge%20Intervals/README.md) - `区间合并`
-  <!-- 排序算法、待补充 -->
+      <!-- 排序算法、待补充 -->
 
 ### 2. 数据结构
 

solution/3800-3899/3802.Number of Ways to Paint Sheets/README.md

@@ -0,0 +1,147 @@
+---
+comments: true
+difficulty: 困难
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3802.Number%20of%20Ways%20to%20Paint%20Sheets/README.md
+---
+
+<!-- problem:start -->
+
+# [3802. 给纸张涂色的方式数量 🔒](https://leetcode.cn/problems/number-of-ways-to-paint-sheets)
+
+[English Version](/solution/3800-3899/3802.Number%20of%20Ways%20to%20Paint%20Sheets/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给定一个整数&nbsp;<code>n</code>&nbsp;表示纸张的数量。</p>
+
+<p>同时给定一个长度为&nbsp;<code>m</code>&nbsp;的整数数组&nbsp;<code>limit</code>，其中&nbsp;<code>limit[i]</code> 是使用颜色 <code>i</code>&nbsp;能够涂色的最大纸张数。</p>
+
+<p>你必须在下列条件下给 <strong>所有</strong>&nbsp;<code>n</code>&nbsp;张纸涂色：</p>
+
+<ul>
+	<li><strong>恰好使用两种不同</strong>&nbsp;颜色。</li>
+	<li>每种颜色必须覆盖 <strong>连续的一段</strong> 纸张。</li>
+	<li>用颜色 <code>i</code> 涂的纸张数量不能超过 <code>limit[i]</code>。</li>
+</ul>
+
+<p>返回一个整数表示给所有纸张染色的 <strong>不同</strong>&nbsp;方式数量。由于答案可能很大，返回答案对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;取模的结果。</p>
+
+<p><strong>注意：</strong>如果 <strong>至少</strong> 有一张纸涂上了不同的颜色，就是不同的两种方式。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>n = 4, limit = [3,1,2]</span></p>
+
+<p><span class="example-io"><b>输出：</b>6</span></p>
+
+<p><b>解释：</b></p>
+对于每个有序数对&nbsp;<code>(i, j)</code>，其中颜色&nbsp;<code>i</code>&nbsp;被用于给第一段涂色，颜色&nbsp;<code>j</code>&nbsp;被用于给第二段涂色（<code>i != j</code>），<code>x</code> 和&nbsp;<code>4 - x</code>&nbsp;的分割是有效的，当且仅当&nbsp;<code>1 &lt;= x &lt;= limit[i]</code> 且&nbsp;<code>1 &lt;= 4 - x &lt;= limit[j]</code>。
+
+<p>合法的数对以及数量是：</p>
+
+<ul>
+	<li><code>(0, 1): x = 3</code></li>
+	<li><code>(0, 2): x = 2, 3</code></li>
+	<li><code>(1, 0): x = 1</code></li>
+	<li><code>(2, 0): x = 1, 2</code></li>
+</ul>
+
+<p>因此，总共有 6 种有效的方式。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>n = 3, limit = [1,2]</span></p>
+
+<p><span class="example-io"><b>输出：</b>2</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>对于每个有序数对&nbsp;<code>(i, j)</code>，其中颜色&nbsp;<code>i</code>&nbsp;被用于给第一段涂色，颜色&nbsp;<code>j</code>&nbsp;被用于给第二段涂色（<code>i != j</code>），<code>x</code> 和 <code>3 - x</code>&nbsp;的分割是有效的，当且仅当&nbsp;<code>1 &lt;= x &lt;= limit[i]</code> 且&nbsp;<code>1 &lt;= 3 - x &lt;= limit[j]</code>。</p>
+
+<p>合法的数对和数量是：</p>
+
+<ul>
+	<li><code>(0, 1): x = 1</code></li>
+	<li><code>(1, 0): x = 2</code></li>
+</ul>
+
+<p>因此，总共有 2 种合法的方式。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>n = 3, limit = [2,2]</span></p>
+
+<p><span class="example-io"><b>输出：</b>4</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>对于每个有序数对&nbsp;<code>(i, j)</code>，其中颜色&nbsp;<code>i</code>&nbsp;被用于给第一段涂色，颜色&nbsp;<code>j</code>&nbsp;被用于给第二段涂色（<code>i != j</code>），<code>x</code> 和 <code>3 - x</code>&nbsp;的分割是有效的，当且仅当&nbsp;<code>1 &lt;= x &lt;= limit[i]</code> 且&nbsp;<code>1 &lt;= 3 - x &lt;= limit[j]</code>。</p>
+
+<p>合法的数对和数量是：</p>
+
+<ul>
+	<li><code>(0, 1): x = 1, 2</code></li>
+	<li><code>(1, 0): x = 1, 2</code></li>
+</ul>
+
+<p>因此，总共有 4 种合法的方式。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &lt;= 10<sup>9</sup></code></li>
+	<li><code>2 &lt;= m == limit.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= limit[i] &lt;= 10<sup>9</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 -->

solution/3800-3899/3802.Number of Ways to Paint Sheets/README_EN.md

@@ -0,0 +1,145 @@
+---
+comments: true
+difficulty: Hard
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3802.Number%20of%20Ways%20to%20Paint%20Sheets/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3802. Number of Ways to Paint Sheets 🔒](https://leetcode.com/problems/number-of-ways-to-paint-sheets)
+
+[中文文档](/solution/3800-3899/3802.Number%20of%20Ways%20to%20Paint%20Sheets/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an integer <code>n</code> representing the number of sheets.</p>
+
+<p>You are also given an integer array <code>limit</code> of size <code>m</code>, where <code>limit[i]</code> is the <strong>maximum</strong> number of sheets that can be painted using color <code>i</code>.</p>
+
+<p>You must paint <strong>all</strong> <code>n</code> sheets under the following conditions:</p>
+
+<ul>
+	<li><strong>Exactly two distinct</strong> colors are used.</li>
+	<li>Each color must cover a <strong>single contiguous</strong> segment of sheets.</li>
+	<li>The number of sheets painted with color <code>i</code> cannot exceed <code>limit[i]</code>.</li>
+</ul>
+
+<p>Return an integer denoting the number of <strong>distinct</strong> ways to paint all sheets. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p><strong>Note:</strong> Two ways differ if <strong>at least</strong> one sheet is painted with a different color.</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 = 4, limit = [3,1,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">6</span></p>
+
+<p><strong>Explanation:</strong>​​​​​​​</p>
+For each ordered pair <code>(i, j)</code>, where color <code>i</code> is used for the first segment and color <code>j</code> for the second segment (<code>i != j</code>), a split of <code>x</code> and <code>4 - x</code> is valid if <code>1 &lt;= x &lt;= limit[i]</code> and <code>1 &lt;= 4 - x &lt;= limit[j]</code>.
+
+<p>Valid pairs and counts are:</p>
+
+<ul>
+	<li><code>(0, 1): x = 3</code></li>
+	<li><code>(0, 2): x = 2, 3</code></li>
+	<li><code>(1, 0): x = 1</code></li>
+	<li><code>(2, 0): x = 1, 2</code></li>
+</ul>
+
+<p>Therefore, there are 6 valid ways in total.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">n = 3, limit = [1,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">2</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>For each ordered pair <code>(i, j)</code>, where color <code>i</code> is used for the first segment and color <code>j</code> for the second segment (<code>i != j</code>), a split of <code>x</code> and <code>3 - x</code> is valid if <code>1 &lt;= x &lt;= limit[i]</code> and <code>1 &lt;= 3 - x &lt;= limit[j]</code>.</p>
+
+<p>Valid pairs and counts are:</p>
+
+<ul>
+	<li><code>(0, 1): x = 1</code></li>
+	<li><code>(1, 0): x = 2</code></li>
+</ul>
+
+<p>Hence, there are 2 valid ways in total.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">n = 3, limit = [2,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>For each ordered pair <code>(i, j)</code>, where color <code>i</code> is used for the first segment and color <code>j</code> for the second segment (<code>i != j</code>), a split of <code>x</code> and <code>3 - x</code> is valid if <code>1 &lt;= x &lt;= limit[i]</code> and <code>1 &lt;= 3 - x &lt;= limit[j]</code>.</p>
+
+<p>Valid pairs and counts are:</p>
+
+<ul>
+	<li><code>(0, 1): x = 1, 2</code></li>
+	<li><code>(1, 0): x = 1, 2</code></li>
+</ul>
+
+<p>Therefore, there are 4 valid ways in total.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &lt;= 10<sup>9</sup></code></li>
+	<li><code>2 &lt;= m == limit.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= limit[i] &lt;= 10<sup>9</sup></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/README.md

@@ -3812,6 +3812,7 @@
 |  3799  |  [单词方块 II](/solution/3700-3799/3799.Word%20Squares%20II/README.md)  |    |  中等  |  第 483 场周赛  |
 |  3800  |  [使二进制字符串相等的最小成本](/solution/3800-3899/3800.Minimum%20Cost%20to%20Make%20Two%20Binary%20Strings%20Equal/README.md)  |    |  中等  |  第 483 场周赛  |
 |  3801  |  [合并有序列表的最小成本](/solution/3800-3899/3801.Minimum%20Cost%20to%20Merge%20Sorted%20Lists/README.md)  |    |  困难  |  第 483 场周赛  |
+|  3802  |  [给纸张涂色的方式数量](/solution/3800-3899/3802.Number%20of%20Ways%20to%20Paint%20Sheets/README.md)  |    |  困难  |  🔒  |
 
 ## 版权
 

solution/README_EN.md

@@ -3810,6 +3810,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
 |  3799  |  [Word Squares II](/solution/3700-3799/3799.Word%20Squares%20II/README_EN.md)  |    |  Medium  |  Weekly Contest 483  |
 |  3800  |  [Minimum Cost to Make Two Binary Strings Equal](/solution/3800-3899/3800.Minimum%20Cost%20to%20Make%20Two%20Binary%20Strings%20Equal/README_EN.md)  |    |  Medium  |  Weekly Contest 483  |
 |  3801  |  [Minimum Cost to Merge Sorted Lists](/solution/3800-3899/3801.Minimum%20Cost%20to%20Merge%20Sorted%20Lists/README_EN.md)  |    |  Hard  |  Weekly Contest 483  |
+|  3802  |  [Number of Ways to Paint Sheets](/solution/3800-3899/3802.Number%20of%20Ways%20to%20Paint%20Sheets/README_EN.md)  |    |  Hard  |  🔒  |
 
 ## Copyright