diff --git a/solution/0800-0899/0829.Consecutive Numbers Sum/README.md b/solution/0800-0899/0829.Consecutive Numbers Sum/README.md
index 6a41f2529e51a..5f674cfe5eaa2 100644
--- a/solution/0800-0899/0829.Consecutive Numbers Sum/README.md
+++ b/solution/0800-0899/0829.Consecutive Numbers Sum/README.md
@@ -58,15 +58,21 @@ tags:
### 方法一：数学推导
-连续正整数构成一个等差数列($d=1$)。我们假设等差数列的第一项为 $a$，项数为 $k$，则 $n=(a+a+k-1)*k/2$，即 $n*2=(a*2+k-1)*k$ ①。
+连续正整数构成一个公差 $d = 1$ 的等差数列。我们假设等差数列的第一项为 $a$，项数为 $k$，那么 $n = (a + a + k - 1) \times k / 2$，即 $n \times 2 = (a \times 2 + k - 1) \times k$。这里我们可以得出 $k$ 一定能整除 $n \times 2$，并且 $(n \times 2) / k - k + 1$ 一定是偶数。
-由于是连续正整数， $a>=1$，所以 ① 可以化为 $n*2>=(k+1)*k$，即 $k*(k+1)<=n*2$ ②。
+由于 $a \geq 1$，所以 $n \times 2 = (a \times 2 + k - 1) \times k \geq k \times (k + 1)$。
-因此，$k$ 的范围需要满足 $k>=1$ 并且 $k*(k+1)<=n*2$。另外，我们从 ① 式可以发现，$k$ 必须能整除 $n*2$。
+综上，我们可以得出：
-综上，我们枚举 $k$，累加满足条件的 $k$ 的个数即可。
+1. $k$ 一定能整除 $n \times 2$；
+2. $k \times (k + 1) \leq n \times 2$；
+3. $(n \times 2) / k - k + 1$ 一定是偶数。
-时间复杂度 $O(\sqrt{n})$。
+我们从 $k = 1$ 开始枚举，当 $k \times (k + 1) > n \times 2$ 时，我们可以结束枚举。在枚举的过程中，我们判断 $k$ 是否能整除 $n \times 2$，并且 $(n \times 2) / k - k + 1$ 是否是偶数，如果是则满足条件，答案加一。
+
+枚举结束后，返回答案即可。
+
+时间复杂度 $O(\sqrt{n})$，其中 $n$ 为给定的正整数。空间复杂度 $O(1)$。
@@ -135,6 +141,21 @@ func consecutiveNumbersSum(n int) int {
}
```
+#### TypeScript
+
+```ts
+function consecutiveNumbersSum(n: number): number {
+ let ans = 0;
+ n <<= 1;
+ for (let k = 1; k * (k + 1) <= n; ++k) {
+ if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
+ ++ans;
+ }
+ }
+ return ans;
+}
+```
+
diff --git a/solution/0800-0899/0829.Consecutive Numbers Sum/README_EN.md b/solution/0800-0899/0829.Consecutive Numbers Sum/README_EN.md
index 7531ec06aeb2c..44a89815dac70 100644
--- a/solution/0800-0899/0829.Consecutive Numbers Sum/README_EN.md
+++ b/solution/0800-0899/0829.Consecutive Numbers Sum/README_EN.md
@@ -57,7 +57,23 @@ tags:
-### Solution 1
+### Solution 1: Mathematical Derivation
+
+Consecutive positive integers form an arithmetic sequence with a common difference $d = 1$. Let's assume the first term of the sequence is $a$, and the number of terms is $k$. Then, $n = (a + a + k - 1) \times k / 2$, which simplifies to $n \times 2 = (a \times 2 + k - 1) \times k$. From this, we can deduce that $k$ must divide $n \times 2$ evenly, and $(n \times 2) / k - k + 1$ must be an even number.
+
+Given that $a \geq 1$, it follows that $n \times 2 = (a \times 2 + k - 1) \times k \geq k \times (k + 1)$.
+
+In summary, we can conclude:
+
+1. $k$ must divide $n \times 2$ evenly;
+2. $k \times (k + 1) \leq n \times 2$;
+3. $(n \times 2) / k - k + 1$ must be an even number.
+
+We start enumerating from $k = 1$, and we can stop when $k \times (k + 1) > n \times 2$. During the enumeration, we check if $k$ divides $n \times 2$ evenly, and if $(n \times 2) / k - k + 1$ is an even number. If both conditions are met, it satisfies the criteria, and we increment the answer by one.
+
+After finishing the enumeration, we return the answer.
+
+The time complexity is $O(\sqrt{n})$, where $n$ is the given positive integer. The space complexity is $O(1)$.
@@ -69,7 +85,7 @@ class Solution:
n <<= 1
ans, k = 0, 1
while k * (k + 1) <= n:
- if n % k == 0 and (n // k + 1 - k) % 2 == 0:
+ if n % k == 0 and (n // k - k + 1) % 2 == 0:
ans += 1
k += 1
return ans
@@ -126,6 +142,21 @@ func consecutiveNumbersSum(n int) int {
}
```
+#### TypeScript
+
+```ts
+function consecutiveNumbersSum(n: number): number {
+ let ans = 0;
+ n <<= 1;
+ for (let k = 1; k * (k + 1) <= n; ++k) {
+ if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
+ ++ans;
+ }
+ }
+ return ans;
+}
+```
+
diff --git a/solution/0800-0899/0829.Consecutive Numbers Sum/Solution.py b/solution/0800-0899/0829.Consecutive Numbers Sum/Solution.py
index b0dbe1960676a..fe343173266d7 100644
--- a/solution/0800-0899/0829.Consecutive Numbers Sum/Solution.py
+++ b/solution/0800-0899/0829.Consecutive Numbers Sum/Solution.py
@@ -3,7 +3,7 @@ def consecutiveNumbersSum(self, n: int) -> int:
n <<= 1
ans, k = 0, 1
while k * (k + 1) <= n:
- if n % k == 0 and (n // k + 1 - k) % 2 == 0:
+ if n % k == 0 and (n // k - k + 1) % 2 == 0:
ans += 1
k += 1
return ans
diff --git a/solution/0800-0899/0829.Consecutive Numbers Sum/Solution.ts b/solution/0800-0899/0829.Consecutive Numbers Sum/Solution.ts
new file mode 100644
index 0000000000000..053f93127b0c4
--- /dev/null
+++ b/solution/0800-0899/0829.Consecutive Numbers Sum/Solution.ts
@@ -0,0 +1,10 @@
+function consecutiveNumbersSum(n: number): number {
+ let ans = 0;
+ n <<= 1;
+ for (let k = 1; k * (k + 1) <= n; ++k) {
+ if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
+ ++ans;
+ }
+ }
+ return ans;
+}
diff --git a/solution/3100-3199/3188.Find Top Scoring Students II/README.md b/solution/3100-3199/3188.Find Top Scoring Students II/README.md
index f044524f1257b..750b6d123dce2 100644
--- a/solution/3100-3199/3188.Find Top Scoring Students II/README.md
+++ b/solution/3100-3199/3188.Find Top Scoring Students II/README.md
@@ -6,7 +6,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3188.Fi
-# [3188. Find Top Scoring Students II 🔒](https://leetcode.cn/problems/find-top-scoring-students-ii)
+# [3188. 查找得分最高的学生 II 🔒](https://leetcode.cn/problems/find-top-scoring-students-ii)
[English Version](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README_EN.md)
@@ -14,7 +14,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3188.Fi
-Table: students
+表：students
+-------------+----------+
@@ -24,11 +24,11 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3188.Fi
| name | varchar |
| major | varchar |
+-------------+----------+
-student_id is the primary key for this table.
-Each row contains the student ID, student name, and their major.
+student_id 是这张表的主键（有不同值的列的组合）。
+这张表的每一行包含学生 ID，学生姓名和他们的专业。
-Table: courses
+表：courses
+-------------+-------------------+
@@ -40,12 +40,12 @@ Each row contains the student ID, student name, and their major.
| major | varchar |
| mandatory | enum |
+-------------+-------------------+
-course_id is the primary key for this table.
-mandatory is an enum type of ('Yes', 'No').
-Each row contains the course ID, course name, credits, major it belongs to, and whether the course is mandatory.
+course_id 是这张表的主键。
+mandatory 是 ('Yes', 'No') 的枚举类型。
+每一行包含课程 ID，课程名，学分，所属专业，以及该课程是否必修。
-Table: enrollments
+表：enrollments
+-------------+----------+
@@ -57,27 +57,28 @@ Each row contains the course ID, course name, credits, major it belongs to, and
| grade | varchar |
| GPA | decimal |
+-------------+----------+
-(student_id, course_id, semester) is the primary key (combination of columns with unique values) for this table.
-Each row contains the student ID, course ID, semester, and grade received.
+(student_id, course_id, semester) 是这张表的主键（有不同值的列的组合）。
+这张表的每一行包含学生 ID，课程 ID，学期和获得的学分。
-Write a solution to find the students who meet the following criteria:
+编写一个解决方案来查找满足下述标准的学生：
- - Have taken all mandatory courses and at least two elective courses offered in their major.
- - Achieved a grade of A in all mandatory courses and at least B in elective courses.
- - Maintained an average
GPA of at least 2.5 across all their courses (including those outside their major).
+ - 已经 修完他们专业中所有的必修课程 和 至少两个 选修课程。
+ - 在 所有必修课程 中取得等级 A 并且 选修课程 至少取得 B。
+ - 保持他们所有课程（包括不属于他们专业的）的平均
GPA 至少在 2.5 以上。
-Return the result table ordered by student_id in ascending order.
+返回结果表以 student_id 升序 排序。
-Example:
+
+示例：
-
Input:
+
输入：
-
students table:
+
students 表：
+------------+------------------+------------------+
@@ -90,7 +91,7 @@ Each row contains the student ID, course ID, semester, and grade received.
+------------+------------------+------------------+
-
courses table:
+
courses 表：
+-----------+-------------------+---------+------------------+----------+
@@ -107,7 +108,7 @@ Each row contains the student ID, course ID, semester, and grade received.
+-----------+-------------------+---------+------------------+----------+
-
enrollments table:
+
enrollments 表：
+------------+-----------+-------------+-------+-----+
@@ -128,7 +129,7 @@ Each row contains the student ID, course ID, semester, and grade received.
+------------+-----------+-------------+-------+-----+
-
Output:
+
输出：
+------------+
@@ -139,16 +140,16 @@ Each row contains the student ID, course ID, semester, and grade received.
+------------+
-
Explanation:
+
解释：
- - Alice (student_id 1) is a Computer Science major and has taken both Algorithms and Data Structures, receiving an A in both. She has also taken Machine Learning and Operating Systems as electives, receiving an A and B respectively.
- - Bob (student_id 2) is a Computer Science major but did not receive an A in all required courses.
- - Charlie (student_id 3) is a Mathematics major and has taken both Calculus and Linear Algebra, receiving an A in both. He has also taken Probability and Statistics as electives, receiving an A and B respectively.
- - David (student_id 4) is a Mathematics major but did not receive an A in all required courses.
+ - Alice (student_id 1) 是计算机科学专业并且修了 Algorithms 和 Data Structures，都取得了 A。她同时选修了 Machine Learning 和 Operating Systems，分别取得了 A 和 B。
+ - Bob (student_id 2) 是计算机科学专业但没有在所有需求的课程中取得 A。
+ - Charlie (student_id 3) 是数学专业并且修了 Calculus 和 Linear Algebra，都取得了 A。他同时选修了 Probability 和 Statistics，分别取得了 A 和 B。
+ - David (student_id 4) 是数学专业但没有在所有需要的课程中取得 A。
-
Note: Output table is ordered by student_id in ascending order.
+
注意：输出表以 student_id 升序排序。