<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[IOLOand32sn的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/IOLOand32sn</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; IOLOand32sn]]></copyright><item><title><![CDATA[洛谷 P8918 『MdOI R5』Jump 题解]]></title><link>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692665</link><guid>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692665</guid><author>IOLOand32sn</author><pubDate>Mon, 15 May 2023 20:57:50 +0800</pubDate><description><![CDATA[规律很难找，但是我们发现不管怎么样，都有 $n\le\mid\star2。因此，在走出第一步后无论怎么走也到达不了奇数的距离，即无法走到。，这其实很好解释，因为如果不满足这个关系式，那么就算。看上去很离奇，实际上略加思索就会发现，如果输入。\star$"处全部使用正号，也无法达到。为偶数，那么就铁定无解。所以我们便可以利用这个特性，让一个变量。，则当走出第一步（步长为。，此时累加的次数就是答案。]]></description><category></category></item><item><title><![CDATA[Mondriaan‘s Dream 【POJ2411】 题解]]></title><link>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692682</link><guid>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692682</guid><author>IOLOand32sn</author><pubDate>Mon, 15 May 2023 20:57:37 +0800</pubDate><description><![CDATA[行号（枚举顺序）压缩后的状态方案数于是我们可以定义fijxfij​x表示第iii行的状态为jjj（假设前i−1i - 1i−1行都已确定）时的方案数。由于一共有mmm行，我们可以将数组略定大一点，最终求到fm10fm10​的值，便是包括到mmm行的方案总数。截止到上一行的方案数即为上一行所有合法的情况数总和。]]></description><category></category></item><item><title><![CDATA[CF594A Warrior and Archer 题解]]></title><link>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692643</link><guid>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692643</guid><author>IOLOand32sn</author><pubDate>Mon, 15 May 2023 20:55:29 +0800</pubDate><description><![CDATA[博弈论]]></description><category></category></item><item><title><![CDATA[洛谷 P8762 [蓝桥杯 2021 国 ABC] 123 题解]]></title><link>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692613</link><guid>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692613</guid><author>IOLOand32sn</author><pubDate>Mon, 15 May 2023 20:54:55 +0800</pubDate><description><![CDATA[于是本题变得好思考了些：我们只要记录每一行的总和，再记录到数列第。初读题目，我们发现这个数列很迷惑，似乎不能使用数学方法来解。所在的行最后一个数位置的前缀和，减去数。的方式来估计行，最后再进行前缀和相减。两数所在行之间所有行的和，再减去数。项的前缀和，然后在每次输入。后面的数的总和，减去。]]></description><category></category></item><item><title><![CDATA[洛谷 P9009 [入门赛 #9] 牵连的世界 (Hard Version) 题解]]></title><link>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692596</link><guid>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692596</guid><author>IOLOand32sn</author><pubDate>Mon, 15 May 2023 20:54:06 +0800</pubDate><description><![CDATA[如果实在不会做 hack 题，怎么办？直接无脑输出极限数据，很大概率可以过。这里的极限指数据规模及数据大小都达到极限。]]></description><category></category></item><item><title><![CDATA[洛谷 P8742 [蓝桥杯 2021 省 AB] 砝码称重]]></title><link>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692563</link><guid>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692563</guid><author>IOLOand32sn</author><pubDate>Mon, 15 May 2023 20:53:16 +0800</pubDate><description><![CDATA[01背包模板题之一]]></description><category></category></item><item><title><![CDATA[二次根式进阶：]]></title><link>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692547</link><guid>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692547</guid><author>IOLOand32sn</author><pubDate>Mon, 15 May 2023 20:50:41 +0800</pubDate><description><![CDATA[二体根式进阶：计算：(1465​​−14−65​​415​​4−15​​​421​2​1​32​23​1​⋯10099​99100​1​​2005×2006×2007×20081​−20062​。]]></description><category></category></item><item><title><![CDATA[OI常见缩写]]></title><link>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692518</link><guid>https://blog.youkuaiyun.com/IOLOand32sn/article/details/130692518</guid><author>IOLOand32sn</author><pubDate>Mon, 15 May 2023 20:49:44 +0800</pubDate><description><![CDATA[搞笑]]></description><category></category></item></channel></rss>