leetcode-problemset/leetcode-cn/originData/find-number-of-ways-to-reach-the-k-th-stair.json

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            "questionId": "3414",
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            "categoryTitle": "Algorithms",
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            "title": "Find Number of Ways to Reach the K-th Stair",
            "titleSlug": "find-number-of-ways-to-reach-the-k-th-stair",
            "content": "<p>You are given a <strong>non-negative</strong> integer <code>k</code>. There exists a staircase with an infinite number of stairs, with the <strong>lowest</strong> stair numbered 0.</p>\n\n<p>Alice has an integer <code>jump</code>, with an initial value of 0. She starts on stair 1 and wants to reach stair <code>k</code> using <strong>any</strong> number of <strong>operations</strong>. If she is on stair <code>i</code>, in one <strong>operation</strong> she can:</p>\n\n<ul>\n\t<li>Go down to stair <code>i - 1</code>. This operation <strong>cannot</strong> be used consecutively or on stair 0.</li>\n\t<li>Go up to stair <code>i + 2<sup>jump</sup></code>. And then, <code>jump</code> becomes <code>jump + 1</code>.</li>\n</ul>\n\n<p>Return the <em>total</em> number of ways Alice can reach stair <code>k</code>.</p>\n\n<p><strong>Note</strong> that it is possible that Alice reaches the stair <code>k</code>, and performs some operations to reach the stair <code>k</code> again.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">k = 0</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">2</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The 2 possible ways of reaching stair 0 are:</p>\n\n<ul>\n\t<li>Alice starts at stair 1.\n\t<ul>\n\t\t<li>Using an operation of the first type, she goes down 1 stair to reach stair 0.</li>\n\t</ul>\n\t</li>\n\t<li>Alice starts at stair 1.\n\t<ul>\n\t\t<li>Using an operation of the first type, she goes down 1 stair to reach stair 0.</li>\n\t\t<li>Using an operation of the second type, she goes up 2<sup>0</sup> stairs to reach stair 1.</li>\n\t\t<li>Using an operation of the first type, she goes down 1 stair to reach stair 0.</li>\n\t</ul>\n\t</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">k = 1</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">4</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The 4 possible ways of reaching stair 1 are:</p>\n\n<ul>\n\t<li>Alice starts at stair 1. Alice is at stair 1.</li>\n\t<li>Alice starts at stair 1.\n\t<ul>\n\t\t<li>Using an operation of the first type, she goes down 1 stair to reach stair 0.</li>\n\t\t<li>Using an operation of the second type, she goes up 2<sup>0</sup> stairs to reach stair 1.</li>\n\t</ul>\n\t</li>\n\t<li>Alice starts at stair 1.\n\t<ul>\n\t\t<li>Using an operation of the second type, she goes up 2<sup>0</sup> stairs to reach stair 2.</li>\n\t\t<li>Using an operation of the first type, she goes down 1 stair to reach stair 1.</li>\n\t</ul>\n\t</li>\n\t<li>Alice starts at stair 1.\n\t<ul>\n\t\t<li>Using an operation of the first type, she goes down 1 stair to reach stair 0.</li>\n\t\t<li>Using an operation of the second type, she goes up 2<sup>0</sup> stairs to reach stair 1.</li>\n\t\t<li>Using an operation of the first type, she goes down 1 stair to reach stair 0.</li>\n\t\t<li>Using an operation of the second type, she goes up 2<sup>1</sup> stairs to reach stair 2.</li>\n\t\t<li>Using an operation of the first type, she goes down 1 stair to reach stair 1.</li>\n\t</ul>\n\t</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= k &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "到达第 K 级台阶的方案数",
            "translatedContent": "<p>给你有一个 <strong>非负</strong>&nbsp;整数&nbsp;<code>k</code>&nbsp;。有一个无限长度的台阶，<strong>最低</strong>&nbsp;一层编号为 0 。</p>\n\n<p>虎老师有一个整数&nbsp;<code>jump</code>&nbsp;，一开始值为 0 。虎老师从台阶 1 开始，虎老师可以使用 <strong>任意</strong>&nbsp;次操作，目标是到达第&nbsp;<code>k</code> 级台阶。假设虎老师位于台阶 <code>i</code> ，一次 <strong>操作</strong> 中，虎老师可以：</p>\n\n<ul>\n\t<li>向下走一级到&nbsp;<code>i - 1</code>&nbsp;，但该操作&nbsp;<strong>不能</strong>&nbsp;连续使用，如果在台阶第 0 级也不能使用。</li>\n\t<li>向上走到台阶&nbsp;<code>i + 2<sup>jump</sup></code>&nbsp;处，然后&nbsp;<code>jump</code>&nbsp;变为&nbsp;<code>jump + 1</code>&nbsp;。</li>\n</ul>\n\n<p>请你返回虎老师到达台阶 <code>k</code>&nbsp;处的总方案数。</p>\n\n<p><b>注意</b>&nbsp;，虎老师可能到达台阶 <code>k</code>&nbsp;处后，通过一些操作重新回到台阶 <code>k</code>&nbsp;处，这视为不同的方案。</p>\n\n<p>&nbsp;</p>\n\n<p><b>示例 1：</b></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>k = 0</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>2</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>2 种到达台阶 0 的方案为：</p>\n\n<ul>\n\t<li>虎老师从台阶&nbsp;1 开始。\n\t<ul>\n\t\t<li>执行第一种操作，从台阶 1 向下走到台阶 0 。</li>\n\t</ul>\n\t</li>\n\t<li>虎老师从台阶 1 开始。\n\t<ul>\n\t\t<li>执行第一种操作，从台阶 1 向下走到台阶 0 。</li>\n\t\t<li>执行第二种操作，向上走 2<sup>0</sup>&nbsp;级台阶到台阶 1 。</li>\n\t\t<li>执行第一种操作，从台阶 1 向下走到台阶 0 。</li>\n\t</ul>\n\t</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>k = 1</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>4</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>4 种到达台阶 1 的方案为：</p>\n\n<ul>\n\t<li>虎老师从台阶 1 开始，已经到达台阶 1 。</li>\n\t<li>虎老师从台阶 1 开始。\n\t<ul>\n\t\t<li>执行第一种操作，从台阶 1 向下走到台阶 0 。</li>\n\t\t<li>执行第二种操作，向上走 2<sup>0</sup>&nbsp;级台阶到台阶 1 。</li>\n\t</ul>\n\t</li>\n\t<li>虎老师从台阶 1 开始。\n\t<ul>\n\t\t<li>执行第二种操作，向上走 2<sup>0</sup>&nbsp;级台阶到台阶 2 。</li>\n\t\t<li>执行第一种操作，向下走 1 级台阶到台阶 1 。</li>\n\t</ul>\n\t</li>\n\t<li>虎老师从台阶 1 开始。\n\t<ul>\n\t\t<li>执行第一种操作，从台阶 1 向下走到台阶 0 。</li>\n\t\t<li>执行第二种操作，向上走&nbsp;2<sup>0</sup>&nbsp;级台阶到台阶 1 。</li>\n\t\t<li>执行第一种操作，向下走 1 级台阶到台阶 0 。</li>\n\t\t<li>执行第二种操作，向上走 2<sup>1</sup>&nbsp;级台阶到台阶 2 。</li>\n\t\t<li>执行第一种操作，向下走&nbsp;1 级台阶到台阶 1 。</li>\n\t</ul>\n\t</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= k &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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                "On using <code>x</code> operations of the second type and <code>y</code> operations of the first type, the stair <code>2<sup>x + 1</sup> - y</code> is reached.",
                "Since first operations cannot be consecutive, there are exactly <code>x + 1</code> positions (before and after each power of 2) to perform the second operation.",
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