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feat: add biweekly contest 146 (doocs#3876)
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solution/0300-0399/0357.Count Numbers with Unique Digits/README.md

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<pre>
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<strong>输入:</strong>n = 2
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<strong>输出:</strong>91
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<strong>解释:</strong>答案应为除去 <code>11、22、33、44、55、66、77、88、99 </code>外,在 0 ≤ x &lt; 100 范围内的所有数字。
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<strong>解释:</strong>答案应为除去 <code>11、22、33、44、55、66、77、88、99 </code>外,在 0 ≤ x &lt; 100 范围内的所有数字。
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</pre>
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<p><strong>示例 2:</strong></p>

solution/0500-0599/0552.Student Attendance Record II/README.md

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<strong>输出:</strong>8
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<strong>解释:
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</strong>有 8 种长度为 2 的记录将被视为可奖励:
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"PP" , "AP", "PA", "LP", "PL", "AL", "LA", "LL"
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"PP" , "AP", "PA", "LP", "PL", "AL", "LA", "LL"
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只有"AA"不会被视为可奖励,因为缺勤次数为 2 次(需要少于 2 次)。
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</pre>
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solution/0600-0699/0600.Non-negative Integers without Consecutive Ones/README.md

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<pre>
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<strong>输入:</strong> n = 5
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<strong>输出:</strong> 5
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<strong>解释:</strong>
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<strong>解释:</strong>
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下面列出范围在 [0, 5] 的非负整数与其对应的二进制表示:
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0 : 0
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1 : 1

solution/0600-0699/0600.Non-negative Integers without Consecutive Ones/README_EN.md

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3 : 11
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4 : 100
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5 : 101
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Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule.
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Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule.
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</pre>
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<p><strong class="example">Example 2:</strong></p>

solution/0700-0799/0788.Rotated Digits/README.md

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<pre><strong>输入:</strong> 10
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<strong>输出:</strong> 4
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<strong>解释:</strong>
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<strong>解释:</strong>
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在[1, 10]中有四个好数: 2, 5, 6, 9。
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注意 1 和 10 不是好数, 因为他们在旋转之后不变。
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</pre>

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/0800-0899/0877.Stone Game/README_EN.md

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<pre>
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<strong>Input:</strong> piles = [5,3,4,5]
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<strong>Output:</strong> true
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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Alice starts first, and can only take the first 5 or the last 5.
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Say she takes the first 5, so that the row becomes [3, 4, 5].
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If Bob takes 3, then the board is [4, 5], and Alice takes 5 to win with 10 points.

solution/0800-0899/0887.Super Egg Drop/README.md

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<p>每次操作,你可以取一枚没有碎的鸡蛋并把它从任一楼层 <code>x</code> 扔下(满足 <code>1 <= x <= n</code>)。如果鸡蛋碎了,你就不能再次使用它。如果某枚鸡蛋扔下后没有摔碎,则可以在之后的操作中 <strong>重复使用</strong> 这枚鸡蛋。</p>
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<p>请你计算并返回要确定 <code>f</code> <strong>确切的值</strong> 的 <strong>最小操作次数</strong> 是多少?</p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>k = 1, n = 2
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<strong>输出:</strong>2
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<strong>解释:</strong>
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鸡蛋从 1 楼掉落。如果它碎了,肯定能得出 f = 0 。
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否则,鸡蛋从 2 楼掉落。如果它碎了,肯定能得出 f = 1 。
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如果它没碎,那么肯定能得出 f = 2 。
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因此,在最坏的情况下我们需要移动 2 次以确定 f 是多少。
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鸡蛋从 1 楼掉落。如果它碎了,肯定能得出 f = 0 。
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否则,鸡蛋从 2 楼掉落。如果它碎了,肯定能得出 f = 1 。
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如果它没碎,那么肯定能得出 f = 2 。
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因此,在最坏的情况下我们需要移动 2 次以确定 f 是多少。
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</pre>
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<p><strong>示例 2:</strong></p>

solution/0900-0999/0977.Squares of a Sorted Array/README.md

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<ul>
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</ul>
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<p> </p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<strong>输出:</strong>[4,9,9,49,121]
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</pre>
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<p> </p>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code><span>1 <= nums.length <= </span>10<sup>4</sup></code></li>
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<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>
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<li><code><span>1 &lt;= nums.length &lt;= </span>10<sup>4</sup></code></li>
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<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
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<li><code>nums</code> 已按 <strong>非递减顺序</strong> 排序</li>
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</ul>
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<p> </p>
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<p>&nbsp;</p>
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<p><strong>进阶:</strong></p>
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<ul>
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<li>请你<span style="color: rgb(36, 41, 46); font-family: -apple-system, BlinkMacSystemFont, &quot;Segoe UI&quot;, Helvetica, Arial, sans-serif, &quot;Apple Color Emoji&quot;, &quot;Segoe UI Emoji&quot;; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: rgb(255, 255, 255); text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">设计时间复杂度为 <code>O(n)</code> 的算法解决本问题</span></li>
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<li>请你设计时间复杂度为 <code>O(n)</code> 的算法解决本问题</li>
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</ul>
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