leetcode-problemset/leetcode-cn/originData/egg-drop-with-2-eggs-and-n-floors.json

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            "title": "Egg Drop With 2 Eggs and N Floors",
            "titleSlug": "egg-drop-with-2-eggs-and-n-floors",
            "content": "<p>You are given <strong>two identical</strong> eggs and you have access to a building with <code>n</code> floors labeled from <code>1</code> to <code>n</code>.</p>\n\n<p>You know that there exists a floor <code>f</code> where <code>0 &lt;= f &lt;= n</code> such that any egg dropped at a floor <strong>higher</strong> than <code>f</code> will <strong>break</strong>, and any egg dropped <strong>at or below</strong> floor <code>f</code> will <strong>not break</strong>.</p>\n\n<p>In each move, you may take an <strong>unbroken</strong> egg and drop it from any floor <code>x</code> (where <code>1 &lt;= x &lt;= n</code>). If the egg breaks, you can no longer use it. However, if the egg does not break, you may <strong>reuse</strong> it in future moves.</p>\n\n<p>Return <em>the <strong>minimum number of moves</strong> that you need to determine <strong>with certainty</strong> what the value of </em><code>f</code> is.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 2\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> We can drop the first egg from floor 1 and the second egg from floor 2.\nIf the first egg breaks, we know that f = 0.\nIf the second egg breaks but the first egg didn&#39;t, we know that f = 1.\nOtherwise, if both eggs survive, we know that f = 2.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 100\n<strong>Output:</strong> 14\n<strong>Explanation:</strong> One optimal strategy is:\n- Drop the 1st egg at floor 9. If it breaks, we know f is between 0 and 8. Drop the 2nd egg starting from floor 1 and going up one at a time to find f within 8 more drops. Total drops is 1 + 8 = 9.\n- If the 1st egg does not break, drop the 1st egg again at floor 22. If it breaks, we know f is between 9 and 21. Drop the 2nd egg starting from floor 10 and going up one at a time to find f within 12 more drops. Total drops is 2 + 12 = 14.\n- If the 1st egg does not break again, follow a similar process dropping the 1st egg from floors 34, 45, 55, 64, 72, 79, 85, 90, 94, 97, 99, and 100.\nRegardless of the outcome, it takes at most 14 drops to determine f.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 1000</code></li>\n</ul>\n",
            "translatedTitle": "鸡蛋掉落-两枚鸡蛋",
            "translatedContent": "<p>给你 <strong>2&nbsp;枚相同 </strong>的鸡蛋，和一栋从第 <code>1</code>&nbsp;层到第 <code>n</code> 层共有 <code>n</code> 层楼的建筑。</p>\n\n<p>已知存在楼层 <code>f</code> ，满足&nbsp;<code>0 &lt;= f &lt;= n</code> ，任何从 <strong>高于 </strong><code>f</code> 的楼层落下的鸡蛋都<strong> 会碎 </strong>，从 <strong><code>f</code> 楼层或比它低 </strong>的楼层落下的鸡蛋都 <strong>不会碎 </strong>。</p>\n\n<p>每次操作，你可以取一枚<strong> 没有碎</strong> 的鸡蛋并把它从任一楼层 <code>x</code> 扔下（满足&nbsp;<code>1 &lt;= x &lt;= 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>\n\n<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>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 2\n<strong>输出：</strong>2\n<strong>解释：</strong>我们可以将第一枚鸡蛋从 1 楼扔下，然后将第二枚从 2 楼扔下。\n如果第一枚鸡蛋碎了，可知 f = 0；\n如果第二枚鸡蛋碎了，但第一枚没碎，可知 f = 1；\n否则，当两个鸡蛋都没碎时，可知 f = 2。\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>14\n<strong>解释：\n</strong>一种最优的策略是：\n- 将第一枚鸡蛋从 9 楼扔下。如果碎了，那么 f 在 0 和 8 之间。将第二枚从 1 楼扔下，然后每扔一次上一层楼，在 8 次内找到 f 。总操作次数 = 1 + 8 = 9 。\n- 如果第一枚鸡蛋没有碎，那么再把第一枚鸡蛋从 22 层扔下。如果碎了，那么 f 在 9 和 21 之间。将第二枚鸡蛋从 10 楼扔下，然后每扔一次上一层楼，在 12 次内找到 f 。总操作次数 = 2 + 12 = 14 。\n- 如果第一枚鸡蛋没有再次碎掉，则按照类似的方法从 34, 45, 55, 64, 72, 79, 85, 90, 94, 97, 99 和 100 楼分别扔下第一枚鸡蛋。\n不管结果如何，最多需要扔 14 次来确定 f 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 1000</code></li>\n</ul>\n",
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