leetcode-problemset/leetcode-cn/originData/find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k.json

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            "title": "Find the Minimum Number of Fibonacci Numbers Whose Sum Is K",
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            "content": "<p>Given an integer&nbsp;<code>k</code>, <em>return the minimum number of Fibonacci numbers whose sum is equal to </em><code>k</code>. The same Fibonacci number can be used multiple times.</p>\n\n<p>The Fibonacci numbers are defined as:</p>\n\n<ul>\n\t<li><code>F<sub>1</sub> = 1</code></li>\n\t<li><code>F<sub>2</sub> = 1</code></li>\n\t<li><code>F<sub>n</sub> = F<sub>n-1</sub> + F<sub>n-2</sub></code> for <code>n &gt; 2.</code></li>\n</ul>\nIt is guaranteed that for the given constraints we can always find such Fibonacci numbers that sum up to <code>k</code>.\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> k = 7\n<strong>Output:</strong> 2 \n<strong>Explanation:</strong> The Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, ... \nFor k = 7 we can use 2 + 5 = 7.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> k = 10\n<strong>Output:</strong> 2 \n<strong>Explanation:</strong> For k = 10 we can use 2 + 8 = 10.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> k = 19\n<strong>Output:</strong> 3 \n<strong>Explanation:</strong> For k = 19 we can use 1 + 5 + 13 = 19.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "和为 K 的最少斐波那契数字数目",
            "translatedContent": "<p>给你数字 <code>k</code>&nbsp;，请你返回和为&nbsp;<code>k</code>&nbsp;的斐波那契数字的最少数目，其中，每个斐波那契数字都可以被使用多次。</p>\n\n<p>斐波那契数字定义为：</p>\n\n<ul>\n\t<li>F<sub>1</sub> = 1</li>\n\t<li>F<sub>2</sub> = 1</li>\n\t<li>F<sub>n</sub> = F<sub>n-1</sub> + F<sub>n-2</sub>&nbsp;， 其中 n &gt; 2 。</li>\n</ul>\n\n<p>数据保证对于给定的 <code>k</code>&nbsp;，一定能找到可行解。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>k = 7\n<strong>输出：</strong>2 \n<strong>解释：</strong>斐波那契数字为：1，1，2，3，5，8，13，&hellip;&hellip;\n对于 k = 7 ，我们可以得到 2 + 5 = 7 。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>k = 10\n<strong>输出：</strong>2 \n<strong>解释：</strong>对于 k = 10 ，我们可以得到 2 + 8 = 10 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>k = 19\n<strong>输出：</strong>3 \n<strong>解释：</strong>对于 k = 19 ，我们可以得到 1 + 5 + 13 = 19 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= 10^9</code></li>\n</ul>\n",
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