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CyC2018
2019-12-06 10:11:23 +08:00
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@ -12,7 +12,7 @@
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?x^n=\left\{\begin{array}{rcl}(x*x)^{n/2}&&{n\%2=0}\\x*(x*x)^{n/2}&&{n\%2=1}\end{array}\right." class="mathjax-pic"/></div> <br>-->
<div align="center"> <img src="pics/48b1d459-8832-4e92-938a-728aae730739.jpg" width="330px"> </div><br>
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/48b1d459-8832-4e92-938a-728aae730739.jpg" width="330px"> </div><br>
因为 (x\*x)<sup>n/2</sup> 可以通过递归求解并且每次递归 n 都减小一半因此整个算法的时间复杂度为 O(logN)