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leetcode-cn/problem (Chinese)/Pow(x, n) [powx-n].html Normal file
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<p>实现&nbsp;<a href="https://www.cplusplus.com/reference/valarray/pow/" target="_blank">pow(<em>x</em>, <em>n</em>)</a>&nbsp;，即计算 <code>x</code> 的 <code>n</code> 次幂函数（即，<code>x<sup>n</sup></code><sup><span style="font-size:10.8333px"> </span></sup>）。</p>
<p>&nbsp;</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>&nbsp;</p>
<p><strong>提示：</strong></p>
<ul>
<li><code>-100.0 &lt; x &lt; 100.0</code></li>
<li><code>-2<sup>31</sup> &lt;= n &lt;= 2<sup>31</sup>-1</code></li>
<li><code>-10<sup>4</sup> &lt;= x<sup>n</sup> &lt;= 10<sup>4</sup></code></li>
</ul>