leetcode-problemset/leetcode-cn/originData/soup-servings.json

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            "questionId": "826",
            "questionFrontendId": "808",
            "categoryTitle": "Algorithms",
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            "title": "Soup Servings",
            "titleSlug": "soup-servings",
            "content": "<p>You have two soups, <strong>A</strong> and <strong>B</strong>, each starting with <code>n</code> mL. On every turn, one of the following four serving operations is chosen <em>at random</em>, each with probability <code>0.25</code> <strong>independent</strong> of all previous turns:</p>\n\n<ul>\n\t<li>pour 100 mL from type A and 0 mL from type B</li>\n\t<li>pour 75 mL from type A and 25 mL from type B</li>\n\t<li>pour 50 mL from type A and 50 mL from type B</li>\n\t<li>pour 25 mL from type A and 75 mL from type B</li>\n</ul>\n\n<p><strong>Note:</strong></p>\n\n<ul>\n\t<li>There is no operation that pours 0 mL from A and 100 mL from B.</li>\n\t<li>The amounts from A and B are poured <em>simultaneously</em> during the turn.</li>\n\t<li>If an operation asks you to pour <strong>more than</strong> you have left of a soup, pour all that remains of that soup.</li>\n</ul>\n\n<p>The process stops immediately after any turn in which <em>one of the soups</em> is used up.</p>\n\n<p>Return the probability that A is used up <em>before</em> B, plus half the probability that both soups are used up in the<strong> same turn</strong>. Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 50\n<strong>Output:</strong> 0.62500\n<strong>Explanation:</strong> \nIf we perform either of the first two serving operations, soup A will become empty first.\nIf we perform the third operation, A and B will become empty at the same time.\nIf we perform the fourth operation, B will become empty first.\nSo the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 100\n<strong>Output:</strong> 0.71875\n<strong>Explanation:</strong> \nIf we perform the first serving operation, soup A will become empty first.\nIf we perform the second serving operations, A will become empty on performing operation [1, 2, 3], and both A and B become empty on performing operation 4.\nIf we perform the third operation, A will become empty on performing operation [1, 2], and both A and B become empty on performing operation 3.\nIf we perform the fourth operation, A will become empty on performing operation 1, and both A and B become empty on performing operation 2.\nSo the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.71875.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "分汤",
            "translatedContent": "<p>你有两种汤，<strong>A</strong> 和 <strong>B</strong>，每种初始为 <code>n</code>&nbsp;毫升。在每一轮中，会随机选择以下四种操作中的一种，每种操作的概率为 <code>0.25</code>，且与之前的所有轮次 <strong>无关</strong>：</p>\n\n<ol>\n\t<li>从汤 A 取 100 毫升，从汤 B 取 0 毫升</li>\n\t<li>从汤 A 取 75 毫升，从汤 B 取 25 毫升</li>\n\t<li>从汤 A 取 50 毫升，从汤 B 取 50 毫升</li>\n\t<li>从汤 A 取 25 毫升，从汤 B 取 75 毫升</li>\n</ol>\n\n<p><strong>注意：</strong></p>\n\n<ul>\n\t<li>不存在从汤 A 取&nbsp;<code>0</code>&nbsp;ml 和从汤 B 取&nbsp;<code>100</code> ml 的操作。</li>\n\t<li>汤 A 和 B 在每次操作中同时被取出。</li>\n\t<li>如果一次操作要求你取出比剩余的汤更多的量，请取出该汤剩余的所有部分。</li>\n</ul>\n\n<p>操作过程在任何回合中任一汤被取完后立即停止。</p>\n\n<p>返回汤 A 在 B 前取完的概率，加上两种汤在 <strong>同一回合&nbsp;</strong>取完概率的一半。返回值在正确答案&nbsp;<code>10<sup>-5</sup></code>&nbsp;的范围内将被认为是正确的。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 50\n<strong>输出：</strong>0.62500\n<strong>解释：\n</strong>如果我们选择前两个操作<strong>，</strong>A 首先将变为空。\n对于第三个操作，A 和 B 会同时变为空。\n对于第四个操作，B 首先将变为空。<strong>\n</strong>所以 A 变为空的总概率加上 A 和 B 同时变为空的概率的一半是 0.25 *(1 + 1 + 0.5 + 0)= 0.625。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 100\n<strong>输出：</strong>0.71875\n<strong>解释：</strong>\n如果我们选择第一个操作，A 首先将变为空。\n如果我们选择第二个操作，A 将在执行操作 [1, 2, 3] 时变为空，然后 A 和 B 在执行操作 4 时同时变空。\n如果我们选择第三个操作，A 将在执行操作 [1, 2] 时变为空，然后 A 和 B 在执行操作 3 时同时变空。\n如果我们选择第四个操作，A 将在执行操作 1 时变为空，然后 A 和 B 在执行操作 2 时同时变空。\n所以 A 变为空的总概率加上 A 和 B 同时变为空的概率的一半是 0.71875。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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