<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Ghostsman的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/Ghostsman</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Ghostsman]]></copyright><item><title><![CDATA[Educational Codeforces Round 159 (Rated for Div. 2) A - Binary Imbalance]]></title><link>https://blog.youkuaiyun.com/Ghostsman/article/details/134797820</link><guid>https://blog.youkuaiyun.com/Ghostsman/article/details/134797820</guid><author>Ghostsman</author><pubDate>Tue, 05 Dec 2023 08:46:31 +0800</pubDate><description><![CDATA[题意：给定一个由0，1字符构成的字符串s，你可以选择在s的第i到第(i+1)个字符之间插入一个字符（i，i+1的范围在0到字符串s的长度-1），当si=si+1，则可以在这两个字符中间插入字符‘1’，反之，则可以在这两个字符中间插入字符‘0’。问有没有可能使用任意数量的操作(可能是不进行任何操作)使字符串中的0的数量严格大于1的数量？思路：当字符串中没有‘1’的时候，一定满足题意。剩下的情况中，可以发现字符串中如果存在一组si≠si+1，就能在这之间无限产生出‘0’，也都能满足题意。]]></description><category></category></item><item><title><![CDATA[初学移动应用开发]]></title><link>https://blog.youkuaiyun.com/Ghostsman/article/details/124645733</link><guid>https://blog.youkuaiyun.com/Ghostsman/article/details/124645733</guid><author>Ghostsman</author><pubDate>Mon, 09 May 2022 10:05:25 +0800</pubDate><description><![CDATA[web服务器、Tomcat详解、HTTP、Maven、idea配置maven、idea配置tomcat]]></description><category></category></item><item><title><![CDATA[Codeforces Round #772 (Div. 2)C - Differential Sorting]]></title><link>https://blog.youkuaiyun.com/Ghostsman/article/details/123075022</link><guid>https://blog.youkuaiyun.com/Ghostsman/article/details/123075022</guid><author>Ghostsman</author><pubDate>Tue, 22 Feb 2022 19:12:42 +0800</pubDate><description><![CDATA[​C - Differential Sorting​]]></description><category></category></item><item><title><![CDATA[Codeforces Round #772 (Div. 2)A - Min Or Sum]]></title><link>https://blog.youkuaiyun.com/Ghostsman/article/details/123066093</link><guid>https://blog.youkuaiyun.com/Ghostsman/article/details/123066093</guid><author>Ghostsman</author><pubDate>Tue, 22 Feb 2022 13:38:22 +0800</pubDate><description><![CDATA[A - Min Or Sum]]></description><category></category></item><item><title><![CDATA[2022牛客寒假算法基础集训营1 H - 牛牛看云]]></title><link>https://blog.youkuaiyun.com/Ghostsman/article/details/122981753</link><guid>https://blog.youkuaiyun.com/Ghostsman/article/details/122981753</guid><author>Ghostsman</author><pubDate>Thu, 17 Feb 2022 13:54:07 +0800</pubDate><description><![CDATA[H - 牛牛看云]]></description><category></category></item><item><title><![CDATA[Codeforces Round #771 (Div. 2) A - Reverse]]></title><link>https://blog.youkuaiyun.com/Ghostsman/article/details/122974157</link><guid>https://blog.youkuaiyun.com/Ghostsman/article/details/122974157</guid><author>Ghostsman</author><pubDate>Wed, 16 Feb 2022 23:43:22 +0800</pubDate><description><![CDATA[A - Reverse]]></description><category></category></item><item><title><![CDATA[Codeforces Round #771 (Div. 2) B - Odd Swap Sort]]></title><link>https://blog.youkuaiyun.com/Ghostsman/article/details/122973783</link><guid>https://blog.youkuaiyun.com/Ghostsman/article/details/122973783</guid><author>Ghostsman</author><pubDate>Wed, 16 Feb 2022 23:27:26 +0800</pubDate><description><![CDATA[​B - Odd Swap Sort​]]></description><category></category></item><item><title><![CDATA[Codeforces Round #771 (Div. 2) C - Inversion Graph]]></title><link>https://blog.youkuaiyun.com/Ghostsman/article/details/122960504</link><guid>https://blog.youkuaiyun.com/Ghostsman/article/details/122960504</guid><author>Ghostsman</author><pubDate>Wed, 16 Feb 2022 12:19:55 +0800</pubDate><description><![CDATA[C - Inversion Graph]]></description><category></category></item></channel></rss>