leetcode-problemset/leetcode-cn/originData/stone-game-ix.json

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            "title": "Stone Game IX",
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            "content": "<p>Alice and Bob continue their games with stones. There is a row of n stones, and each stone has an associated value. You are given an integer array <code>stones</code>, where <code>stones[i]</code> is the <strong>value</strong> of the <code>i<sup>th</sup></code> stone.</p>\n\n<p>Alice and Bob take turns, with <strong>Alice</strong> starting first. On each turn, the player may remove any stone from <code>stones</code>. The player who removes a stone <strong>loses</strong> if the <strong>sum</strong> of the values of <strong>all removed stones</strong> is divisible by <code>3</code>. Bob will win automatically if there are no remaining stones (even if it is Alice&#39;s turn).</p>\n\n<p>Assuming both players play <strong>optimally</strong>, return <code>true</code> <em>if Alice wins and</em> <code>false</code> <em>if Bob wins</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> stones = [2,1]\n<strong>Output:</strong> true\n<strong>Explanation:</strong>&nbsp;The game will be played as follows:\n- Turn 1: Alice can remove either stone.\n- Turn 2: Bob removes the remaining stone. \nThe sum of the removed stones is 1 + 2 = 3 and is divisible by 3. Therefore, Bob loses and Alice wins the game.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> stones = [2]\n<strong>Output:</strong> false\n<strong>Explanation:</strong>&nbsp;Alice will remove the only stone, and the sum of the values on the removed stones is 2. \nSince all the stones are removed and the sum of values is not divisible by 3, Bob wins the game.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> stones = [5,1,2,4,3]\n<strong>Output:</strong> false\n<strong>Explanation:</strong> Bob will always win. One possible way for Bob to win is shown below:\n- Turn 1: Alice can remove the second stone with value 1. Sum of removed stones = 1.\n- Turn 2: Bob removes the fifth stone with value 3. Sum of removed stones = 1 + 3 = 4.\n- Turn 3: Alices removes the fourth stone with value 4. Sum of removed stones = 1 + 3 + 4 = 8.\n- Turn 4: Bob removes the third stone with value 2. Sum of removed stones = 1 + 3 + 4 + 2 = 10.\n- Turn 5: Alice removes the first stone with value 5. Sum of removed stones = 1 + 3 + 4 + 2 + 5 = 15.\nAlice loses the game because the sum of the removed stones (15) is divisible by 3. Bob wins the game.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= stones.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= stones[i] &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "石子游戏 IX",
            "translatedContent": "<p>Alice 和 Bob 再次设计了一款新的石子游戏。现有一行 n 个石子，每个石子都有一个关联的数字表示它的价值。给你一个整数数组 <code>stones</code> ，其中 <code>stones[i]</code> 是第 <code>i</code> 个石子的价值。</p>\n\n<p>Alice 和 Bob 轮流进行自己的回合，<strong>Alice</strong> 先手。每一回合，玩家需要从 <code>stones</code>&nbsp;中移除任一石子。</p>\n\n<ul>\n\t<li>如果玩家移除石子后，导致 <strong>所有已移除石子</strong> 的价值&nbsp;<strong>总和</strong> 可以被 3 整除，那么该玩家就 <strong>输掉游戏</strong> 。</li>\n\t<li>如果不满足上一条，且移除后没有任何剩余的石子，那么 Bob 将会直接获胜（即便是在 Alice 的回合）。</li>\n</ul>\n\n<p>假设两位玩家均采用&nbsp;<strong>最佳</strong> 决策。如果 Alice 获胜，返回 <code>true</code> ；如果 Bob 获胜，返回 <code>false</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>stones = [2,1]\n<strong>输出：</strong>true\n<strong>解释：</strong>游戏进行如下：\n- 回合 1：Alice 可以移除任意一个石子。\n- 回合 2：Bob 移除剩下的石子。 \n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。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>stones = [2]\n<strong>输出：</strong>false\n<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 。 \n由于所有石子都已移除，且值总和无法被 3 整除，Bob 获胜。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>stones = [5,1,2,4,3]\n<strong>输出：</strong>false\n<strong>解释：</strong>Bob 总会获胜。其中一种可能的游戏进行方式如下：\n- 回合 1：Alice 可以移除值为 1 的第 2 个石子。已移除石子值总和为 1 。\n- 回合 2：Bob 可以移除值为 3 的第 5 个石子。已移除石子值总和为 = 1 + 3 = 4 。\n- 回合 3：Alices 可以移除值为 4 的第 4 个石子。已移除石子值总和为 = 1 + 3 + 4 = 8 。\n- 回合 4：Bob 可以移除值为 2 的第 3 个石子。已移除石子值总和为 = 1 + 3 + 4 + 2 = 10.\n- 回合 5：Alice 可以移除值为 5 的第 1 个石子。已移除石子值总和为 = 1 + 3 + 4 + 2 + 5 = 15.\nAlice 输掉游戏，因为已移除石子值总和（15）可以被 3 整除，Bob 获胜。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= stones.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= stones[i] &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
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