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Commit d8fd6a9

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feat: update lc problems (doocs#3414)
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‎solution/0800-0899/0813.Largest Sum of Averages/README.md‎

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<pre>
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<strong>输入:</strong> nums = [9,1,2,3,9], k = 3
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<strong>输出:</strong> 20.00000
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<strong>解释:</strong>
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nums 的最优分组是[9], [1, 2, 3], [9]. 得到的分数是 9 + (1 + 2 + 3) / 3 + 9 = 20.
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我们也可以把 nums 分成[9, 1], [2], [3, 9].
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<strong>解释:</strong>
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nums 的最优分组是[9], [1, 2, 3], [9]. 得到的分数是 9 + (1 + 2 + 3) / 3 + 9 = 20.
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我们也可以把 nums 分成[9, 1], [2], [3, 9].
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这样的分组得到的分数为 5 + 2 + 6 = 13, 但不是最大值.
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</pre>
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‎solution/0800-0899/0813.Largest Sum of Averages/README_EN.md‎

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<pre>
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<strong>Input:</strong> nums = [9,1,2,3,9], k = 3
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<strong>Output:</strong> 20.00000
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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The best choice is to partition nums into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
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We could have also partitioned nums into [9, 1], [2], [3, 9], for example.
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That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.

‎solution/0800-0899/0814.Binary Tree Pruning/README_EN.md‎

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<pre>
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<strong>Input:</strong> root = [1,null,0,0,1]
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<strong>Output:</strong> [1,null,0,null,1]
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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Only the red nodes satisfy the property &quot;every subtree not containing a 1&quot;.
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The diagram on the right represents the answer.
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</pre>

‎solution/0800-0899/0849.Maximize Distance to Closest Person/README.md‎

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<strong>解释:
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</strong>如果亚历克斯坐在第二个空位(seats[2])上,他到离他最近的人的距离为 2 。
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如果亚历克斯坐在其它任何一个空位上,他到离他最近的人的距离为 1 。
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因此,他到离他最近的人的最大距离是 2 。
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因此,他到离他最近的人的最大距离是 2 。
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</pre>
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<p><strong>示例 2:</strong></p>

‎solution/1300-1399/1310.XOR Queries of a Subarray/README.md‎

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<pre>
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<strong>输入:</strong>arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
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<strong>输出:</strong>[2,7,14,8]
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<strong>输出:</strong>[2,7,14,8]
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<strong>解释:</strong>
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数组中元素的二进制表示形式是:
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1 = 0001
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3 = 0011
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4 = 0100
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8 = 1000
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1 = 0001
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3 = 0011
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4 = 0100
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8 = 1000
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查询的 XOR 值为:
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[0,1] = 1 xor 3 = 2
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[1,2] = 3 xor 4 = 7
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[0,3] = 1 xor 3 xor 4 xor 8 = 14
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[0,1] = 1 xor 3 = 2
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[1,2] = 3 xor 4 = 7
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[0,3] = 1 xor 3 xor 4 xor 8 = 14
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[3,3] = 8
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</pre>
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‎solution/1300-1399/1310.XOR Queries of a Subarray/README_EN.md‎

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<pre>
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<strong>Input:</strong> arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
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<strong>Output:</strong> [2,7,14,8]
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<strong>Explanation:</strong>
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<strong>Output:</strong> [2,7,14,8]
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<strong>Explanation:</strong>
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The binary representation of the elements in the array are:
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1 = 0001
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3 = 0011
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4 = 0100
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8 = 1000
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1 = 0001
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3 = 0011
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4 = 0100
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8 = 1000
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The XOR values for queries are:
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[0,1] = 1 xor 3 = 2
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[1,2] = 3 xor 4 = 7
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[0,3] = 1 xor 3 xor 4 xor 8 = 14
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[0,1] = 1 xor 3 = 2
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[1,2] = 3 xor 4 = 7
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[0,3] = 1 xor 3 xor 4 xor 8 = 14
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[3,3] = 8
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</pre>
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‎solution/1300-1399/1311.Get Watched Videos by Your Friends/README.md‎

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<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1311.Get%20Watched%20Videos%20by%20Your%20Friends/images/leetcode_friends_1.png" style="height: 179px; width: 129px;"></strong></p>
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<pre><strong>输入:</strong>watchedVideos = [[&quot;A&quot;,&quot;B&quot;],[&quot;C&quot;],[&quot;B&quot;,&quot;C&quot;],[&quot;D&quot;]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 1
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<strong>输出:</strong>[&quot;B&quot;,&quot;C&quot;]
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<strong>输出:</strong>[&quot;B&quot;,&quot;C&quot;]
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<strong>解释:</strong>
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你的 id 为 0(绿色),你的朋友包括(黄色):
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id 为 1 -&gt; watchedVideos = [&quot;C&quot;]&nbsp;

‎solution/1300-1399/1311.Get Watched Videos by Your Friends/README_EN.md‎

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<pre>
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<strong>Input:</strong> watchedVideos = [[&quot;A&quot;,&quot;B&quot;],[&quot;C&quot;],[&quot;B&quot;,&quot;C&quot;],[&quot;D&quot;]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 1
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<strong>Output:</strong> [&quot;B&quot;,&quot;C&quot;]
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<strong>Explanation:</strong>
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<strong>Output:</strong> [&quot;B&quot;,&quot;C&quot;]
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<strong>Explanation:</strong>
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You have id = 0 (green color in the figure) and your friends are (yellow color in the figure):
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Person with id = 1 -&gt; watchedVideos = [&quot;C&quot;]&nbsp;
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Person with id = 2 -&gt; watchedVideos = [&quot;B&quot;,&quot;C&quot;]&nbsp;
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<pre>
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<strong>Input:</strong> watchedVideos = [[&quot;A&quot;,&quot;B&quot;],[&quot;C&quot;],[&quot;B&quot;,&quot;C&quot;],[&quot;D&quot;]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 2
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<strong>Output:</strong> [&quot;D&quot;]
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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You have id = 0 (green color in the figure) and the only friend of your friends is the person with id = 3 (yellow color in the figure).
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</pre>
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‎solution/2000-2099/2029.Stone Game IX/README.md‎

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<strong>输出:</strong>true
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<strong>解释:</strong>游戏进行如下:
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- 回合 1:Alice 可以移除任意一个石子。
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- 回合 2:Bob 移除剩下的石子。
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- 回合 2:Bob 移除剩下的石子。
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已移除的石子的值总和为 1 + 2 = 3 且可以被 3 整除。因此,Bob 输,Alice 获胜。
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</pre>
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<pre>
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<strong>输入:</strong>stones = [2]
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<strong>输出:</strong>false
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<strong>解释:</strong>Alice 会移除唯一一个石子,已移除石子的值总和为 2 。
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<strong>解释:</strong>Alice 会移除唯一一个石子,已移除石子的值总和为 2 。
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由于所有石子都已移除,且值总和无法被 3 整除,Bob 获胜。
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</pre>
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‎solution/2000-2099/2029.Stone Game IX/README_EN.md‎

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<strong>Output:</strong> true
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<strong>Explanation:</strong>&nbsp;The game will be played as follows:
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- Turn 1: Alice can remove either stone.
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- Turn 2: Bob removes the remaining stone.
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- Turn 2: Bob removes the remaining stone.
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The sum of the removed stones is 1 + 2 = 3 and is divisible by 3. Therefore, Bob loses and Alice wins the game.
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</pre>
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<pre>
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<strong>Input:</strong> stones = [2]
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<strong>Output:</strong> false
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<strong>Explanation:</strong>&nbsp;Alice will remove the only stone, and the sum of the values on the removed stones is 2.
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<strong>Explanation:</strong>&nbsp;Alice will remove the only stone, and the sum of the values on the removed stones is 2.
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Since all the stones are removed and the sum of values is not divisible by 3, Bob wins the game.
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</pre>
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