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feat: add lc problems and solutions
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‎solution/0600-0699/0633.Sum of Square Numbers/README.md

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<li><code>0 &lt;= c &lt;= 2<sup>31</sup> - 1</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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![](./images/table.png)
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![](https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0633.Sum%20of%20Square%20Numbers/images/table.png)
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上图为 a，b，c 之间的关系，这题其实就是在这张“表”里查找 c
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool judgeSquareSum(int c) {
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long i = 0, j = (long) sqrt(c);
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while (i <= j) {
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long s = i * i + j * j;
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if (s < c) ++i;
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else if (s > c) --j;
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else return true;
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}
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return false;
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}
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};
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```
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### **Go**
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```go
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func judgeSquareSum(c int) bool {
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i, j := 0, int(math.Sqrt(float64(c)))
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for i <= j {
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s := i*i + j*j
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if s < c {
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i++
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} else if s > c {
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j--
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} else {
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return true
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}
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}
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return false
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}
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```
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### **...**
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```

‎solution/0600-0699/0633.Sum of Square Numbers/README_EN.md

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<li><code>0 &lt;= c &lt;= 2<sup>31</sup> - 1</code></li>
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</ul>
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## Solutions
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![](./images/table.png)
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![](https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0633.Sum%20of%20Square%20Numbers/images/table.png)
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The picture above shows the relationship between `a`, `b`, and `c`. This question is actually looking up `c` in this table
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool judgeSquareSum(int c) {
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long i = 0, j = (long) sqrt(c);
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while (i <= j) {
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long s = i * i + j * j;
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if (s < c) ++i;
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else if (s > c) --j;
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else return true;
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}
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return false;
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}
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};
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```
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### **Go**
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```go
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func judgeSquareSum(c int) bool {
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i, j := 0, int(math.Sqrt(float64(c)))
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for i <= j {
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s := i*i + j*j
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if s < c {
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i++
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} else if s > c {
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j--
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} else {
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return true
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}
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}
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return false
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}
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```
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### **...**
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```

‎solution/0600-0699/0633.Sum of Square Numbers/Solution.cpp

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class Solution {
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public:
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bool judgeSquareSum(int c) {
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long i = 0, j = (long) sqrt(c);
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while (i <= j) {
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long s = i * i + j * j;
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if (s < c) ++i;
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else if (s > c) --j;
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else return true;
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}
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return false;
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}
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};

‎solution/0600-0699/0633.Sum of Square Numbers/Solution.go

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func judgeSquareSum(c int) bool {
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i, j := 0, int(math.Sqrt(float64(c)))
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for i <= j {
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s := i*i + j*j
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if s < c {
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i++
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} else if s > c {
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j--
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} else {
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return true
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}
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}
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return false
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}

‎solution/1200-1299/1204.Last Person to Fit in the Bus/README.md

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# [1204. 最后一个能进入电梯的人](https://leetcode-cn.com/problems/last-person-to-fit-in-the-bus)
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[English Version](/solution/1200-1299/1204.Last%20Person%20to%20Fit%20in%20the%20Bus/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>表: <code>Queue</code></p>
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<pre>+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| person_id | int |
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| person_name | varchar |
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| weight | int |
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| turn | int |
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+-------------+---------+
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person_id 是这个表的主键。
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该表展示了所有等待电梯的人的信息。
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表中 person_id 和 turn 列将包含从 1 到 n 的所有数字，其中 n 是表中的行数。
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</pre>
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<p>&nbsp;</p>
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<p>电梯最大载重量为 <strong>1000</strong>。</p>
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<p>写一条 SQL 查询语句查找最后一个能进入电梯且不超过重量限制的 <code>person_name</code> 。题目确保队列中第一位的人可以进入电梯 。</p>
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<p>查询结果如下所示 :</p>
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<pre>Queue 表
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+-----------+-------------------+--------+------+
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| person_id | person_name | weight | turn |
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+-----------+-------------------+--------+------+
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| 5 | George Washington | 250 | 1 |
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| 3 | John Adams | 350 | 2 |
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| 6 | Thomas Jefferson | 400 | 3 |
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| 2 | Will Johnliams | 200 | 4 |
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| 4 | Thomas Jefferson | 175 | 5 |
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| 1 | James Elephant | 500 | 6 |
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+-----------+-------------------+--------+------+
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Result 表
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+-------------------+
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| person_name |
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+-------------------+
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| Thomas Jefferson |
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+-------------------+
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为了简化，Queue 表按 turn 列由小到大排序。
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上例中 George Washington(id 5), John Adams(id 3) 和 Thomas Jefferson(id 6) 将可以进入电梯,因为他们的体重和为 250 + 350 + 400 = 1000。
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Thomas Jefferson(id 6) 是最后一个体重合适并进入电梯的人。
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</pre>
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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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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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```
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<!-- tabs:end -->

‎solution/1200-1299/1204.Last Person to Fit in the Bus/README_EN.md

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# [1204. Last Person to Fit in the Bus](https://leetcode.com/problems/last-person-to-fit-in-the-bus)
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[中文文档](/solution/1200-1299/1204.Last%20Person%20to%20Fit%20in%20the%20Bus/README.md)
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## Description
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<p>Table: <code>Queue</code></p>
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<pre>
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| person_id | int |
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| person_name | varchar |
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| weight | int |
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| turn | int |
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+-------------+---------+
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person_id is the primary key column for this table.
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This table has the information about all people waiting for a bus.
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The person_id and turn columns will contain all numbers from 1 to n, where n is the number of rows in the table.
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turn determines the order of which the people will board the bus, where turn=1 denotes the first person to board and turn=n denotes the last person to board.
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weight is the weight of the person in kilograms.
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</pre>
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<p>&nbsp;</p>
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<p>There is a queue of people waiting to board a bus. However, the bus has a weight limit of <strong>1000 kilograms</strong>, so there may be some people who cannot board.</p>
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<p>Write an SQL query to find the <code>person_name</code> of the <strong>last person</strong> that can fit on the bus without exceeding the weight limit. The testcases are generated such that the first person does not exceed the weight limit.</p>
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<p>The query result format is in the following example:</p>
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<p>&nbsp;</p>
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<pre>
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Queue table:
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+-----------+-------------+--------+------+
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| person_id | person_name | weight | turn |
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+-----------+-------------+--------+------+
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| 5 | Alice | 250 | 1 |
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| 4 | Bob | 175 | 5 |
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| 3 | Alex | 350 | 2 |
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| 6 | John Cena | 400 | 3 |
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| 1 | Winston | 500 | 6 |
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| 2 | Marie | 200 | 4 |
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+-----------+-------------+--------+------+
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Result table
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+-------------+
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| person_name |
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+-------------+
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| John Cena |
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+-------------+
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The table is ordered by the turn for simplicity.
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+------+----+-----------+--------+--------------+
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| Turn | ID | Name | Weight | Total Weight |
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+------+----+-----------+--------+--------------+
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| 1 | 5 | Alice | 250 | 250 |
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| 2 | 3 | Alex | 350 | 600 |
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| 3 | 6 | John Cena | 400 | 1000 | (last person to board)
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| 4 | 2 | Marie | 200 | 1200 | (cannot board)
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| 5 | 4 | Bob | 175 | ___ |
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| 6 | 1 | Winston | 500 | ___ |
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+------+----+-----------+--------+--------------+
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</pre>
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## Solutions
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<!-- tabs:start -->
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### **SQL**
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```sql
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```
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<!-- tabs:end -->

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