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docs: update readme documents
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all.json

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lcci/08.11.Coin/README.md

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<p>你可以假设:</p>
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<ul>
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<li>0 &lt;= n (总金额) &lt;= 1000000</li>
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<li>0 &lt;= n (总金额) &lt;= 1000000</li>
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</ul>
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## 解法

lcci/16.02.Words Frequency/README_EN.md

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<p><strong>Note: </strong></p>
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<ul>
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<li><code>There are only lowercase letters in book[i].</code></li>
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<li><code>1 &lt;= book.length &lt;= 100000</code></li>
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<li><code>1 &lt;= book[i].length &lt;= 10</code></li>
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<li><code>get</code>&nbsp;function will not be called more than&nbsp;100000 times.</li>
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</ul>
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## Solutions

lcp/README.md

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# LCP

solution/0100-0199/0174.Dungeon Game/README.md

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<p><strong>说明:</strong></p>
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<ul>
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<li>
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<p>骑士的健康点数没有上限。</p>
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</li>
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<li>任何房间都可能对骑士的健康点数造成威胁,也可能增加骑士的健康点数,包括骑士进入的左上角房间以及公主被监禁的右下角房间。</li>
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</ul>
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## 解法

solution/0200-0299/0289.Game of Life/README_EN.md

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<p><b>Follow up</b>:</p>
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<ol>
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<li>Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.</li>
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<li>In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches the border of the array. How would you address these problems?</li>
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</ol>
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## Solutions

solution/0500-0599/0529.Minesweeper/README.md

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<p>现在给出在所有<strong>未挖出的</strong>方块中(&#39;M&#39;或者&#39;E&#39;)的下一个点击位置(行和列索引),根据以下规则,返回相应位置被点击后对应的面板:</p>
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<ol>
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<li>如果一个地雷(&#39;M&#39;)被挖出,游戏就结束了- 把它改为&nbsp;<strong>&#39;X&#39;</strong>。</li>
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<li>如果一个<strong>没有相邻地雷</strong>的空方块(&#39;E&#39;)被挖出,修改它为(&#39;B&#39;),并且所有和其相邻的方块都应该被递归地揭露。</li>
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<li>如果一个<strong>至少与一个地雷相邻</strong>的空方块(&#39;E&#39;)被挖出,修改它为数字(&#39;1&#39;到&#39;8&#39;),表示相邻地雷的数量。</li>
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<li>如果在此次点击中,若无更多方块可被揭露,则返回面板。</li>
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</ol>
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<p>&nbsp;</p>

solution/0800-0899/0825.Friends Of Appropriate Ages/README_EN.md

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<p>Person A will NOT friend request person B (B != A) if any of the following conditions are true:</p>
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<ul>
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<li><code>age[B]&nbsp;&lt;= 0.5 * age[A]&nbsp;+ 7</code></li>
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<li><code>age[B]&nbsp;&gt; age[A]</code></li>
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<li><code>age[B]&nbsp;&gt; 100 &amp;&amp;&nbsp;age[A]&nbsp;&lt; 100</code></li>
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</ul>
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<p>Otherwise, A will friend request B.</p>
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<p>Notes:</p>
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<ul>
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<li><code>1 &lt;= ages.length&nbsp;&lt;= 20000</code>.</li>
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<li><code>1 &lt;= ages[i] &lt;= 120</code>.</li>
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</ul>
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## Solutions

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