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# [50. Pow(x, n)](https://leetcode.com/problems/powx-n/)
## 题目
Implement [pow(*x*, *n*)](http://www.cplusplus.com/reference/valarray/pow/), which calculates *x* raised to the power *n* (xn).
**Example 1:**
Input: 2.00000, 10
Output: 1024.00000
**Example 2:**
Input: 2.10000, 3
Output: 9.26100
**Example 3:**
Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
**Note:**
- -100.0 < *x* < 100.0
- *n* is a 32-bit signed integer, within the range [2^31, 2^31 1]
## 题目大意
实现 pow(x, n) 即计算 x n 次幂函数
## 解题思路
- 要求计算 Pow(x, n)
- 这一题用递归的方式不断的将 n 2 分下去注意 n 的正负数n 的奇偶性