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leetcode-problemset/算法题(国内版)/problem (Chinese)/数值的整数次方 [shu-zhi-de-zheng-shu-ci-fang-lcof].html

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<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>
<p> </p>
<p><strong>示例 1</strong></p>
<pre>
<strong>输入:</strong>x = 2.00000, n = 10
<strong>输出:</strong>1024.00000
</pre>
<p><strong>示例 2</strong></p>
<pre>
<strong>输入:</strong>x = 2.10000, n = 3
<strong>输出:</strong>9.26100</pre>
<p><strong>示例 3</strong></p>
<pre>
<strong>输入:</strong>x = 2.00000, n = -2
<strong>输出:</strong>0.25000
<strong>解释:</strong>2<sup>-2</sup> = 1/2<sup>2</sup> = 1/4 = 0.25</pre>
<p> </p>
<p><strong>提示:</strong></p>
<ul>
<li><code>-100.0 < x < 100.0</code></li>
<li><code>-2<sup>31</sup> <= n <= 2<sup>31</sup>-1</code></li>
<li><code>-10<sup>4</sup> <= x<sup>n</sup> <= 10<sup>4</sup></code></li>
</ul>
<p> </p>
<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>