<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[JUST___enjoy的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/JUST___enjoy</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; JUST___enjoy]]></copyright><item><title><![CDATA[Python爬虫爬取肯德基餐厅信息，基于‘路飞学城IT’UP主的视频。]]></title><link>https://blog.youkuaiyun.com/JUST___enjoy/article/details/132150447</link><guid>https://blog.youkuaiyun.com/JUST___enjoy/article/details/132150447</guid><author>JUST___enjoy</author><pubDate>Mon, 07 Aug 2023 17:17:45 +0800</pubDate><description><![CDATA[使用Python爬虫获取KFC官网门店信息的案例实现，代码比较少，希望对别人有所帮助]]></description><category></category></item><item><title><![CDATA[vscode在调试c时，报错Segmentation fault。]]></title><link>https://blog.youkuaiyun.com/JUST___enjoy/article/details/131699332</link><guid>https://blog.youkuaiyun.com/JUST___enjoy/article/details/131699332</guid><author>JUST___enjoy</author><pubDate>Thu, 13 Jul 2023 12:02:55 +0800</pubDate><description><![CDATA[vscode在调试c时，报错Segmentation fault的解决方法。]]></description><category></category></item><item><title><![CDATA[二分字符串：给定正整数N，计算所有长度为N但没有连续1的二分字符。]]></title><link>https://blog.youkuaiyun.com/JUST___enjoy/article/details/120976423</link><guid>https://blog.youkuaiyun.com/JUST___enjoy/article/details/120976423</guid><author>JUST___enjoy</author><pubDate>Tue, 26 Oct 2021 17:32:42 +0800</pubDate><description><![CDATA[思路:

1. 先找出所有长度为N的二进制字符串，也就是2的N次方个数

    2. 判断这个数是否符合题目的条件，符合就加到返回列表中，不符合就舍弃

思路很简单，怎么实现呢？

    1. 找出长度为N的二进制数，用一个for循环就可以，代码中的i就是对应的各个数了


for i in rang(2**N)

    2. 再将他们转变为2进制数就可以了(具体的bin用法可以自行查看，比较简单)


bin_2 = bin(i)[2:].zfill(N)

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