<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[最后一次修改 ]]></title><description><![CDATA[于 **/**/****]]></description><link>https://blog.youkuaiyun.com/Estia_</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Estia_]]></copyright><item><title><![CDATA[Riesz表示定理]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/133531877</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/133531877</guid><author>Estia_</author><pubDate>Wed, 04 Oct 2023 03:59:40 +0800</pubDate><description><![CDATA[……Hilbert空间的基本定理之一]]></description><category></category></item><item><title><![CDATA[简单无向图计数]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/124331152</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/124331152</guid><author>Estia_</author><pubDate>Thu, 21 Apr 2022 23:18:21 +0800</pubDate><description><![CDATA[图的同构计数]]></description><category></category></item><item><title><![CDATA[在Qt6.2下使用QCustomPlot报错的解决方法]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/122160339</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/122160339</guid><author>Estia_</author><pubDate>Sun, 26 Dec 2021 20:40:55 +0800</pubDate><description><![CDATA[分别报了好几个错
error: template argument 1 is invalid
Call to non-static member function without an object argument
error: undefined reference to `__imp__ZN8QPrinterC1ENS_11PrinterModeE’
第一个报错只需要把 namespace QCP 删掉，换上下面的 class QCP就行了
第二个只需要把 class QCP 的 getMarg]]></description><category></category></item><item><title><![CDATA[CF-GYM102900L. Traveling in the Grid World]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/116176878</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/116176878</guid><author>Estia_</author><pubDate>Mon, 26 Apr 2021 23:56:38 +0800</pubDate><description><![CDATA[Pick's Theorem, EXGCD]]></description><category></category></item><item><title><![CDATA[堆]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/115978320</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/115978320</guid><author>Estia_</author><pubDate>Thu, 22 Apr 2021 00:31:24 +0800</pubDate><description><![CDATA[左偏树 优先队列 对顶堆]]></description><category></category></item><item><title><![CDATA[CF-GYM102222F Moving On]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/115843152</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/115843152</guid><author>Estia_</author><pubDate>Mon, 19 Apr 2021 00:22:38 +0800</pubDate><description><![CDATA[Floyd]]></description><category></category></item><item><title><![CDATA[CF-GYM102222J. Nested Triangles]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/115842732</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/115842732</guid><author>Estia_</author><pubDate>Mon, 19 Apr 2021 00:01:05 +0800</pubDate><description><![CDATA[计算几何+LIS]]></description><category></category></item><item><title><![CDATA[【Contest】Codeforces Round #672 (Div. 2)]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/108901342</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/108901342</guid><author>Estia_</author><pubDate>Mon, 05 Oct 2020 22:03:56 +0800</pubDate><description><![CDATA[Problems: http://codeforces.com/contest/1420


A. Cubes Sorting
When the array is strictly descending we need n(n−1)2\frac{n(n-1)}{2}2n(n−1)​ operations.
The number of operations we need is also the number of inversions in the array.

Submitted wrong file.]]></description><category></category></item><item><title><![CDATA[【Luogu4735】最大异或和【可持久化01trie】]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/108858431</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/108858431</guid><author>Estia_</author><pubDate>Wed, 30 Sep 2020 17:29:57 +0800</pubDate><description><![CDATA[可持久化 01trie 的模板题]]></description><category></category></item><item><title><![CDATA[【Luogu2824】【HEOI2016/TJOI2016】排序 【二分答案+线段树 / 权值线段树分裂&合并】]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/108814198</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/108814198</guid><author>Estia_</author><pubDate>Mon, 28 Sep 2020 21:35:58 +0800</pubDate><description><![CDATA[线段树……还可以的题目？]]></description><category></category></item><item><title><![CDATA[【Luogu5494】【模板】线段树分裂]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/108675009</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/108675009</guid><author>Estia_</author><pubDate>Sat, 26 Sep 2020 16:37:44 +0800</pubDate><description><![CDATA[[Link\frak{Link}Link]


不开 long long 见祖宗！！

不难发现应该是开多颗权值线段树
感觉像是把线段树拆开？有点类似 Lazytag 那样就不需要处理中间部分
然后外面部分染上你的颜色
其实就是类似主席树那个感觉嘛

...]]></description><category></category></item><item><title><![CDATA[【Graphics】直线裁剪 Cohen-Sutherland]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/108806026</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/108806026</guid><author>Estia_</author><pubDate>Sat, 26 Sep 2020 11:47:36 +0800</pubDate><description><![CDATA[直线裁剪 Cohen-Sutherland 算法
Computer Graphics #001

编码

把平面按照裁剪窗口分为九块，用四位二进制数表示每一块区域。

在 A 、C 、E 、G 四个区域分别填入 2i,  i∈{0,1,2,3}2^i,\;i\in\{0,1,2,3\}2i,i∈{0,1,2,3}  （不要求位置对应）
然后每个角落填入两个邻接块的值的或
比如假设 A=(0001)2,  C=(0100)2A=(0001)_2,\;C=(0100)_2A=(0001)2​,C=(0100)2]]></description><category></category></item><item><title><![CDATA[「Note」Math not for OI]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/84945764</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/84945764</guid><author>Estia_</author><pubDate>Fri, 25 Sep 2020 23:08:13 +0800</pubDate><description><![CDATA[[Link\frak{Link}Link]


欧拉公式：eiθ=cosθ+isinθ\mathcal {e^{i\theta}=cos\theta+isin\theta}eiθ=cosθ+isinθ
单位根：wnk=cos(2πkn)+isin(2πkn)\mathcal {w_n^k=cos(\frac{2\pi k}{n})+isin(\frac{2\pi k}{n})}wnk​=cos(...]]></description><category></category></item><item><title><![CDATA[【Luogu3834】【模板】可持久化线段树 2（主席树）]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/108627721</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/108627721</guid><author>Estia_</author><pubDate>Fri, 18 Sep 2020 20:32:23 +0800</pubDate><description><![CDATA[[Link\frak{Link}Link]


求区间 kth 显然可以离散化权值，对每个位置用树维护此前每个权值的出现次数，
于是树上可以查询得到一定范围内所有权值的出现次数和
把这些树合并一下就可以了。。得到一个朝下散开的分叉拖把头
查询的时候在 L-1 和 R 两个版本上同时跑，根据相减结果导航到区间 kth 的位置上
O(mlog⁡n)O(m\log n)O(mlogn)



...]]></description><category></category></item><item><title><![CDATA[「Note」倍增LCA]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/108522847</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/108522847</guid><author>Estia_</author><pubDate>Mon, 14 Sep 2020 21:24:15 +0800</pubDate><description><![CDATA[[Link\frak{Link}Link]

没想到吧我已经连LCA都不能一次过了
不过我应该一直就是这个水平

倍增。Θ(nlog⁡n)\Theta(n\log n)Θ(nlogn)
不能折叠有点可惜，
倍增可以根据两个点的深度差直接找到起始跳跃层数
一定要注意不要把 while 跟 for + if 弄混， while 停下来就走不了了。。
倍增 70pts 常数巨大代码巨丑版
#include&lt;cstdio&gt;
#include&lt;cstdlib&gt;
#include&lt;cma.]]></description><category></category></item><item><title><![CDATA[【BZOJ3750】【POI2015】Pieczęć]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/108478215</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/108478215</guid><author>Estia_</author><pubDate>Wed, 09 Sep 2020 00:23:42 +0800</pubDate><description><![CDATA[BZOJ挂了
数据下载

如果用 AC自动机/KMP 可以得到 Θ(n3)\Theta(n^3)Θ(n3) 的做法
这是万万不行的
而 bitset 并不支持相关操作
我们先不要考虑算法，，
考虑操作的时候会出现什么情况

显然
要染黑所有点就要让所有点被染黑
所以挑出所有左上角的点就可以了
Θ(n2)\Theta(n^2)Θ(n2)
，，

很遗憾我下不下来测试数据 不过起码上面的思路应该是没有问题的
细节还挺多。。
#include&lt;cstdio&gt;
#include&lt;cstdlib&.]]></description><category></category></item><item><title><![CDATA[[Luogu4239] 【模板】多项式求逆（加强版） [多项式求逆][MTT][三模数NTT]]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/89186270</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/89186270</guid><author>Estia_</author><pubDate>Tue, 25 Aug 2020 23:53:35 +0800</pubDate><description><![CDATA[#include&lt;cstdio&gt;
#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;cstdlib&gt;
#include&lt;cstring&gt;
#include&lt;cmath&gt;
#include&lt;ctime&gt;
#include&lt;cctype&gt;
using names...]]></description><category></category></item><item><title><![CDATA[[POJ2154] Color [Polya定理][莫比乌斯反演]]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/90261621</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/90261621</guid><author>Estia_</author><pubDate>Thu, 16 May 2019 13:18:56 +0800</pubDate><description><![CDATA[题意:
从N种颜色的珠子中选出N个串成一圈，问有多少种方案。（不考虑翻转）

读入X(X &lt;= 3500)
对于X组数据，每一组给出N(1 &lt;= N &lt;= 1000000000)和P(1 &lt;= P &lt;= 30000)，要求输出Ans%P


环是可以旋转的（本题不考虑翻转）
显然是置换。
有N种置换（包括不转动）；
第i种置换（也就是旋转了i格）有r=gcd(N,i)...]]></description><category></category></item><item><title><![CDATA[[Luogu4339] [ZJOI2018] 迷宫 [有限状态自动机]]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/90261460</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/90261460</guid><author>Estia_</author><pubDate>Thu, 16 May 2019 13:03:38 +0800</pubDate><description><![CDATA[参考：
对DFA和NFA的简单理解：http://blog.163.com/ppt_compiler/blog/static/202813007201222873720918/
NFA确定化和DFA最小化：https://blog.youkuaiyun.com/u014541281/article/details/52423191
题解：https://blog.youkuaiyun.com/qq_16267919/ar...]]></description><category></category></item><item><title><![CDATA[[Note] Burnsde 引理 & Polya 定理]]></title><link>https://blog.youkuaiyun.com/Estia_/article/details/90261412</link><guid>https://blog.youkuaiyun.com/Estia_/article/details/90261412</guid><author>Estia_</author><pubDate>Thu, 16 May 2019 12:57:14 +0800</pubDate><description><![CDATA[是很好的总结了。因为总是会找不到所以贴一下（


polya定理小结

　　polya定理是组合数学中比较难的一部分。首先需要对置换群、集合论有一定的了解，这样有助于理解burnside引理的证明。其次，polya定理只是对于在环上存在旋转、反射等等价的变换的一种计数方法，实际的题目中很多需要其他的知识来进行辅助。
　　环上的计数主要就是处理置换 -&gt; 着色这种情况。很关键的一点是同一循环内...]]></description><category></category></item></channel></rss>