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Copy file name to clipboardExpand all lines: solution/0800-0899/0829.Consecutive Numbers Sum/README_EN.md
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Mathematical Derivation
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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.
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Given that $a \geq 1$, it follows that $n \times 2 = (a \times 2 + k - 1) \times k \geq k \times (k + 1)$.
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In summary, we can conclude:
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1. $k$ must divide $n \times 2$ evenly;
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2. $k \times (k + 1) \leq n \times 2$;
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3. $(n \times 2) / k - k + 1$ must be an even number.
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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.
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After finishing the enumeration, we return the answer.
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The time complexity is $O(\sqrt{n})$, where $n$ is the given positive integer. The space complexity is $O(1)$.
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<!-- tabs:start -->
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@@ -69,7 +85,7 @@ class Solution:
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n <<=1
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ans, k =0, 1
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while k * (k +1) <= n:
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if n % k ==0and (n // k +1- k) %2==0:
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if n % k ==0and (n // k - k+1) %2==0:
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ans +=1
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k +=1
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return ans
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<p>Write a solution to find the students who meet the following criteria:</p>
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<p>编写一个解决方案来查找满足下述标准的学生：</p>
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<ul>
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<li>Have<strong> taken all mandatory courses</strong> and <strong>at least two</strong> elective courses offered in <strong>their major.</strong></li>
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<li>Achieved a grade of <strong>A</strong> in <strong>all mandatory courses</strong> and at least <strong>B</strong> in<strong> elective courses</strong>.</li>
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<li>Maintained an average <code>GPA</code> of at least <code>2.5</code> across all their courses (including those outside their major).</li>
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+------------+
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</pre>
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<p><strong>Explanation:</strong></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>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.</li>
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<li>Bob (student_id 2) is a Computer Science major but did not receive an A in all required courses.</li>
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<li>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.</li>
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<li>David (student_id 4) is a Mathematics major but did not receive an A in all required courses.</li>
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<li>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。</li>
First, we calculate the average GPA of each student and store it in a temporary table `T`.
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First, we filter out students with an average GPA greater than or equal to 2.5 and record them in table `T`.
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Next, we join the `students` table with the `courses` table based on `major`, and then join with the `T` table based on `student_id`, followed by a left join with the `enrollments` table based on `student_id` and `course_id`.
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Next, we join the `T` table with the `students` table based on `student_id`, then join with the `courses` table based on `major`, and finally perform a left join with the `enrollments` table based on `student_id` and `course_id`.
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After that, we filter out students with an average GPA greater than or equal to 2.5, group by student ID, use the `HAVING` clause to filter students who meet the criteria, and finally sort by student ID.
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After that, we group by student ID, use the `HAVING` clause to filter out students who meet the conditions, and finally sort by student ID.
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<!-- tabs:start -->
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT student_id, AVG(GPA) AS avg_gpa
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SELECT student_id
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FROM enrollments
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GROUP BY1
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HAVINGAVG(GPA) >=2.5
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)
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SELECT student_id
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FROM
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students
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T
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JOIN students USING (student_id)
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JOIN courses USING (major)
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JOIN T USING (student_id)
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LEFT JOIN enrollments USING (student_id, course_id)
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