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leetcode-problemset/leetcode-cn/originData/shu-zhi-de-zheng-shu-ci-fang-lcof.json
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"questionFrontendId": "剑指 Offer 16",
"categoryTitle": "LCOF",
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"title": "数值的整数次方 LCOF",
"titleSlug": "shu-zhi-de-zheng-shu-ci-fang-lcof",
"content": "English description is not available for the problem. Please switch to Chinese.",
"translatedTitle": "数值的整数次方",
"translatedContent": "<p>实现 <a href=\"https://www.cplusplus.com/reference/valarray/pow/\">pow(<em>x</em>, <em>n</em>)</a> ,即计算 x 的 n 次幂函数x<sup>n</sup>)。不得使用库函数,同时不需要考虑大数问题。</p>\n\n<p> </p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>x = 2.00000, n = 10\n<strong>输出:</strong>1024.00000\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>x = 2.10000, n = 3\n<strong>输出:</strong>9.26100</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<strong>输入:</strong>x = 2.00000, n = -2\n<strong>输出:</strong>0.25000\n<strong>解释:</strong>2<sup>-2</sup> = 1/2<sup>2</sup> = 1/4 = 0.25</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>-100.0 < x < 100.0</code></li>\n\t<li><code>-2<sup>31</sup> <= n <= 2<sup>31</sup>-1</code></li>\n\t<li><code>-10<sup>4</sup> <= x<sup>n</sup> <= 10<sup>4</sup></code></li>\n</ul>\n\n<p> </p>\n\n<p>注意:本题与主站 50 题相同:<a href=\"https://leetcode-cn.com/problems/powx-n/\">https://leetcode-cn.com/problems/powx-n/</a></p>\n",
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