<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Popandsoss2022的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/Popandsoss2022</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Popandsoss2022]]></copyright><item><title><![CDATA[题解：CF2127D Root was Built by Love, Broken by Destiny]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/150431063</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/150431063</guid><author>Popandsoss2022</author><pubDate>Fri, 15 Aug 2025 21:31:24 +0800</pubDate><description><![CDATA[赛时吃了四发罚（我还是太菜了），但这题真的好玩。]]></description><category></category></item><item><title><![CDATA[Graham 算法求二维凸包]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/150156103</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/150156103</guid><author>Popandsoss2022</author><pubDate>Sun, 10 Aug 2025 20:21:50 +0800</pubDate><description><![CDATA[Graham 求解凸包]]></description><category></category></item><item><title><![CDATA[题解：AT_abc393_c [ABC393C] Make it Simple]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145789745</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145789745</guid><author>Popandsoss2022</author><pubDate>Sat, 22 Feb 2025 08:18:31 +0800</pubDate><description><![CDATA[给定一个图，问要删除多少条边才能使其变成一个简单图。简单图的定义是没有自环也没有重边的图。]]></description><category></category></item><item><title><![CDATA[题解：AT_abc393_d [ABC393D] Swap to Gather]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145789735</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145789735</guid><author>Popandsoss2022</author><pubDate>Sat, 22 Feb 2025 08:16:06 +0800</pubDate><description><![CDATA[给定一串只有0和1的字符串，可以将相邻的两个字符交换，问至少需要多少步才能让所有的1相邻。]]></description><category></category></item><item><title><![CDATA[【数论】—— 快速幂与扩展欧拉定理]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145680549</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145680549</guid><author>Popandsoss2022</author><pubDate>Mon, 17 Feb 2025 13:13:09 +0800</pubDate><description><![CDATA[快速幂，即很快的做幂运算，并且在求幂的同时支持取模操作。但这个算法究竟是怎么实现的，我们来探究一下。对于aaammm两数，若gcd⁡am1gcdam1aφm≡1modmaφm≡1modm。其实费马小定理就是欧拉定理的特殊情况，即当mmm为素数时。对于abmodma^b\bmod mabmodm这样一个式子，无论aaa与mmmabmodm≡b≥φmabmodφmφm。]]></description><category></category></item><item><title><![CDATA[【数论】 —— 最大公因数与最小公倍数]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145553309</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145553309</guid><author>Popandsoss2022</author><pubDate>Mon, 10 Feb 2025 16:39:32 +0800</pubDate><description><![CDATA[在两个数的公因数中最大的那个叫做这两个数的最大公因数。我们一般用gcd⁡ab\gcd(a,b)gcdab来表示aaabbb两数的最大公因数。接下来，我给大家介绍444种求最大公因数的方法。辗转相除法，又称欧几里得算法，是一种用来求最大公因数的算法。当amodb0a\bmod b=0amodb0时，便可以得到两数的最大公因数，即gcd⁡b。]]></description><category></category></item><item><title><![CDATA[【数论】—— 欧拉函数]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145532589</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145532589</guid><author>Popandsoss2022</author><pubDate>Sun, 09 Feb 2025 15:10:05 +0800</pubDate><description><![CDATA[对于任意正整数nnn，欧拉函数φn\varphi(n)φn表示在不大于nnn的正整数中与nnn互素的数的个数。φ11φ11，即111。φ62φ62，即111555。φ124φ124，即111555777111111。怎样求出一个数的欧拉函数呢？公式呈上：对于任意正整数nnn，若nP1k1×P2k2×⋯×PmkmnP1k1​​×P2k2​​×⋯×Pmkm​。]]></description><category></category></item><item><title><![CDATA[【数论】—— 素数]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145530871</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145530871</guid><author>Popandsoss2022</author><pubDate>Sun, 09 Feb 2025 12:54:54 +0800</pubDate><description><![CDATA[因数只有1和这个数本身的数被称作素数。1既不是素数也不是合数，2是最小的素数。]]></description><category></category></item><item><title><![CDATA[题解：P1006 [NOIP 2008 提高组] 传纸条]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145492921</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145492921</guid><author>Popandsoss2022</author><pubDate>Fri, 07 Feb 2025 13:10:43 +0800</pubDate><description><![CDATA[本人习惯先n后m，所以下面的n和m均与题目相反。]]></description><category></category></item><item><title><![CDATA[题解：P1005 [NOIP 2007 提高组] 矩阵取数游戏]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145471215</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145471215</guid><author>Popandsoss2022</author><pubDate>Thu, 06 Feb 2025 12:49:28 +0800</pubDate><description><![CDATA[给定一个n×m的矩阵，可以进行m次取数，每次只能取走每一行的第一个数或最后一个数。设第i次取出的数是b1​b2​⋯bn​，则第i次取数的得分是∑j1n​bj​×2i。问取完m次后的每次取数得分之和最大是多少？]]></description><category></category></item><item><title><![CDATA[题解：CF120F Spiders]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145439949</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145439949</guid><author>Popandsoss2022</author><pubDate>Tue, 04 Feb 2025 13:51:42 +0800</pubDate><description><![CDATA[给定n棵树，要求把n棵树拼在一起，使拼完后的树的直径最长，]]></description><category></category></item><item><title><![CDATA[题解：P5582 「SWTR-1」Escape]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145298749</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145298749</guid><author>Popandsoss2022</author><pubDate>Wed, 22 Jan 2025 10:37:03 +0800</pubDate><description><![CDATA[设所有可行的步数为序列。的情况，其它就没啥了。]]></description><category></category></item><item><title><![CDATA[题解：P3399 丝绸之路]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145270519</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145270519</guid><author>Popandsoss2022</author><pubDate>Mon, 20 Jan 2025 22:13:24 +0800</pubDate><description><![CDATA[给大家提供一种不同的 dp 方法。]]></description><category></category></item><item><title><![CDATA[题解：P1073 [NOIP2009 提高组] 最优贸易]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145265093</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145265093</guid><author>Popandsoss2022</author><pubDate>Mon, 20 Jan 2025 16:21:27 +0800</pubDate><description><![CDATA[各位大佬都用了分层图，但却只用了一种写法，蒟蒻过来补充另外一种写法。]]></description><category></category></item><item><title><![CDATA[题解：P1122 最大子树和]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145226105</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145226105</guid><author>Popandsoss2022</author><pubDate>Sat, 18 Jan 2025 13:21:28 +0800</pubDate><description><![CDATA[给大家提供一种不同的dp方法。]]></description><category></category></item><item><title><![CDATA[题解：P1063 [NOIP2006 提高组] 能量项链]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145202611</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145202611</guid><author>Popandsoss2022</author><pubDate>Fri, 17 Jan 2025 11:06:57 +0800</pubDate><description><![CDATA[一道区间 dp 的板子题。]]></description><category></category></item><item><title><![CDATA[题解：P8612 [蓝桥杯 2014 省 AB] 地宫取宝]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145136212</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145136212</guid><author>Popandsoss2022</author><pubDate>Tue, 14 Jan 2025 12:39:49 +0800</pubDate><description><![CDATA[顾名思义，记忆化搜索的意思便是把搜索到的结果记录下来，下次搜索到时便可以直接将结果返回，可以大大降低时间复杂度。比如说递归求斐波那契，如果不用记忆化，必超时。]]></description><category></category></item><item><title><![CDATA[题解：P6770 [USACO05MAR] Checking an Alibi 不在场的证明]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145136197</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145136197</guid><author>Popandsoss2022</author><pubDate>Tue, 14 Jan 2025 12:38:38 +0800</pubDate><description><![CDATA[看了一下，怎么都是最短路啊？这题 bfs 可以轻松水过，蒟蒻就来发一篇 bfs 的题解吧。]]></description><category></category></item><item><title><![CDATA[题解：P8035 [COCI2015-2016#7] Otpor]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145136182</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145136182</guid><author>Popandsoss2022</author><pubDate>Tue, 14 Jan 2025 12:37:31 +0800</pubDate><description><![CDATA[本题解主要为还没学物理的 OIer 们提供。做前须知：介绍一下公式：I=URI=\frac{U}{R}I=RU​。即电流等于电压与电阻之比。所以，可以得到推导式：R=UIR=\frac{U}{I}R=IU​。定义：几个连接起来的电阻所起的作用，可以用一个电阻来代替，这个电阻就是那些电阻的等效电阻。由串联电路中的电流处处相等，各用电器电压之和等于电源电压（等流分压）得：R总=U总I总=U1+U2+⋯+UnI=U1I+U2I+⋯+UnI=R1+R2+⋯+RnR_总=\frac{U_总}{I_总}=\frac{U]]></description><category></category></item><item><title><![CDATA[题解：P3753 国事访问]]></title><link>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145136164</link><guid>https://blog.youkuaiyun.com/Popandsoss2022/article/details/145136164</guid><author>Popandsoss2022</author><pubDate>Tue, 14 Jan 2025 12:35:31 +0800</pubDate><description><![CDATA[对于非最短路上的边，我们可以暴力枚举任意两点间的边，如果它们两个都在从。，分别代表能不能走。你需要求出最短路上的不能走的边以及非最短路上能走的边的和。读完题目后，思路大体就有了：跑一遍最短路，然后统计答案。对于最短路上的边，我们可以在 bfs 时用一个。在 bfs 后，我们便可以用循环来找。否则判断这条边是否需要销毁，若需要，答案加一。，对这种图，我们可以使用 bfs 来求最短路。给你一个无向图，图上的边权变成了。，所以我们便可以使用邻接矩阵存图。]]></description><category></category></item></channel></rss>