<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Whyckck的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/Whyckck</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Whyckck]]></copyright><item><title><![CDATA[Verilog相关基础笔记]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/118363527</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/118363527</guid><author>Whyckck</author><pubDate>Wed, 30 Jun 2021 16:10:06 +0800</pubDate><description><![CDATA[]]></description><category></category></item><item><title><![CDATA[带小数的二进制与十六进制互相转换（超暴力版）]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/103229863</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/103229863</guid><author>Whyckck</author><pubDate>Sun, 24 Nov 2019 22:06:13 +0800</pubDate><description><![CDATA[啊，就写着玩了，很暴力

如果转换后的字符串是回文串，顺便输出个Beautiful


#include &lt;bits/stdc++.h&gt;
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
//typedef __int128 bll;
const int maxn ...]]></description><category></category></item><item><title><![CDATA[【思维】D. Sequence Sorting]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/102844946</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/102844946</guid><author>Whyckck</author><pubDate>Thu, 31 Oct 2019 20:23:29 +0800</pubDate><description><![CDATA[D. Sequence Sorting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a sequencea1,a2,…,ana1,a2,…,an, consisting of ...]]></description><category></category></item><item><title><![CDATA[【换根DP】E. Tree Painting]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/102825755</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/102825755</guid><author>Whyckck</author><pubDate>Wed, 30 Oct 2019 19:47:39 +0800</pubDate><description><![CDATA[E. Tree Painting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a tree (an undirected connected acyclic graph) con...]]></description><category></category></item><item><title><![CDATA[【回文自动机】 2018 南京网络赛 Skr]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/100782571</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/100782571</guid><author>Whyckck</author><pubDate>Thu, 12 Sep 2019 20:24:36 +0800</pubDate><description><![CDATA[题目链接:https://nanti.jisuanke.com/t/A1955

题意：给定一个数字串，求每个不同回文数字串的和

思路：由回文树可知，每个节点回文是由上一个节点转移而来，深搜奇回文树与偶回文树，过程中O1计算出每个节点的值，加起来即可

也可以通过每个回文的区间+Hash O1 获得该串数字 算出答案

代码为方深搜代码


#include &lt;bits/stdc++.h&...]]></description><category></category></item><item><title><![CDATA[洛谷P5496 【模板】回文自动机（PAM）]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/100748882</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/100748882</guid><author>Whyckck</author><pubDate>Wed, 11 Sep 2019 20:55:12 +0800</pubDate><description><![CDATA[题目链接：https://www.luogu.org/problem/P5496

题意：给定字符串，求每个位置为结尾的回文串个数

思路：裸题，不过注意，是last指针在fail树上跳，并不是直接输出num数组


#include &lt;bits/stdc++.h&gt;
#define endl '\n'
using namespace std;
typedef long long ll;...]]></description><category></category></item><item><title><![CDATA[【回文自动机】The Preliminary Contest for ICPC Asia Xuzhou 2019 G Colorful String]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/100748586</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/100748586</guid><author>Whyckck</author><pubDate>Wed, 11 Sep 2019 20:45:17 +0800</pubDate><description><![CDATA[题目链接：https://nanti.jisuanke.com/t/41389

题意：给定3e5的字符串，求每个不同的回文串中不同的字母总和

思路：预处出Pre保存前缀和以此来计算区间内不同字母的个数，构造回文树，那么每个回文串对答案的贡献就是串的长度乘上串出现的次数了


#include &lt;bits/stdc++.h&gt;
#define endl '\n'
using names...]]></description><category></category></item><item><title><![CDATA[【PAM】回文自动机 && 回文树]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/100748394</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/100748394</guid><author>Whyckck</author><pubDate>Wed, 11 Sep 2019 20:39:13 +0800</pubDate><description><![CDATA[学习博客：https://www.cnblogs.com/nbwzyzngyl/p/8260921.html

博客中讲的很清楚了，在此我就不过多赘述了

len[i], 以i结尾的最长回文子串的长度
cnt[i]：以i结尾的最长回文子串相同的子串的个数 在count后得到全部
num[i] 表示以i结尾的回文串的种类数
str[] 存放添加的字符
fail[] 失配后跳转到的不等于自身的最...]]></description><category></category></item><item><title><![CDATA[【模板】 扫描线]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/99862675</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/99862675</guid><author>Whyckck</author><pubDate>Tue, 20 Aug 2019 17:29:47 +0800</pubDate><description><![CDATA[模板题:https://www.luogu.org/problem/P5490与http://poj.org/problem?id=1151

扫描线从左向右扫,rwa[]数组表示整数值 i 对应的原始 y 坐标值。 c[i] 记录扫描线上第 i 段覆盖的次数，ans 每次累加上 len*(x2-x1)即可 ，朴素做法O(n^2) ，用线段树维护 c 数组后复杂度 O（n*longn）


...]]></description><category></category></item><item><title><![CDATA[【并查集+组合数】2019牛客暑期多校训练营（第九场） E	All men are brothers]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/99683844</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/99683844</guid><author>Whyckck</author><pubDate>Fri, 16 Aug 2019 19:09:50 +0800</pubDate><description><![CDATA[题目链接：https://ac.nowcoder.com/acm/contest/889/E

题意：给你n个人，m轮，每轮会让其中的两个人认识（人之间具有传递性）,问每次挑出4个互相不认识的有多少种

思路：首先很容易确定是第一轮是,（注意精度）,其次我们考虑每一轮的减少量.

这时候我们考虑每次合并会带来什么影响：

假设有集合X,Y,Z，我们合并X,Y集合

那么有如下影响

合并前

1....]]></description><category></category></item><item><title><![CDATA[【离散化+线段树维护区间最大子段和】2019 Multi-University Training Contest 6  Snowy Smile]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/99678560</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/99678560</guid><author>Whyckck</author><pubDate>Fri, 16 Aug 2019 15:13:13 +0800</pubDate><description><![CDATA[题目：http://acm.hdu.edu.cn/showproblem.php?pid=6638

题意：2000个点，让你用矩形框起来，使得矩阵和最大

思路：首先离散化使数据将到O（2000），按y升序排列（按x也可以）后枚举上下界，在上界浮动的时候顺便将在上界上的点扔进线段树维护。复杂度O（n^2long(n)）


#include &lt;bits/stdc++.h&gt;
#defi...]]></description><category></category></item><item><title><![CDATA[线段树维护区间最大子段和]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/99543092</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/99543092</guid><author>Whyckck</author><pubDate>Wed, 14 Aug 2019 09:59:32 +0800</pubDate><description><![CDATA[一段区间的最大子段和出现在以下三种情况：左区间/右区间/跨越中间


void push(node &amp;now,node &amp;lt,node &amp;rt)
{	
	now.sum = lt.sum + rt.sum;
	now.lgss = max(lt.lgss,lt.sum+rt.lgss);
	now.rgss = max(rt.rgss,rt.sum+lt.rgss);
...]]></description><category></category></item><item><title><![CDATA[【模板】KMP && EXKMP]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/99406508</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/99406508</guid><author>Whyckck</author><pubDate>Tue, 13 Aug 2019 10:45:05 +0800</pubDate><description><![CDATA[KMP博客：https://www.cnblogs.com/SYCstudio/p/7194315.html

注：pre数组中保存的是该最长前后缀的前缀结束下标（字符串从0开始）,如果要求最长相同前缀后缀的长度，要输出pre[i]+1。

获取next函数


void get_pre()
{
	pre[0] = -1;
	
	for(ll i = 1; i &lt; len2; i++)
	{...]]></description><category></category></item><item><title><![CDATA[【十进制矩阵快速幂】 2019牛客暑期多校训练营（第五场） B]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/98945353</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/98945353</guid><author>Whyckck</author><pubDate>Fri, 09 Aug 2019 11:12:43 +0800</pubDate><description><![CDATA[原文：https://blog.youkuaiyun.com/henucm/article/details/98492753

题意：给你x0、x1a、b、mod，根据求出



根据上述公式，再套用十进制矩阵快速幂即可（二进制好像会超时...）

用十进制

设为res 则可得

我们再设

就是分解n为每一位，再去相乘。

例如

讲计算出来的即可


#include &lt;bits/st...]]></description><category></category></item><item><title><![CDATA[【线段树+单调栈】2019牛客暑期多校训练营（第四场）C	sequence]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/98762557</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/98762557</guid><author>Whyckck</author><pubDate>Wed, 07 Aug 2019 17:09:22 +0800</pubDate><description><![CDATA[链接：https://ac.nowcoder.com/acm/contest/884/C
来源：牛客网


时间限制：C/C++ 3秒，其他语言6秒
空间限制：C/C++ 524288K，其他语言1048576K
64bit IO Format: %lld

题目描述

Your are given two sequences a1…na_{1 \dots n}a1…n​ and b1…nb_{...]]></description><category></category></item><item><title><![CDATA[【关于%3前缀和性质证明】]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/98724488</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/98724488</guid><author>Whyckck</author><pubDate>Wed, 07 Aug 2019 09:19:54 +0800</pubDate><description><![CDATA[命题：
有数字串str，str[i]的前缀和为sum[i]，若(sum[a] mod 3) == (sum[b] mod 3)，则子串(a, b]为3的倍数

证明：
当sum[i] mod 3 == 0时，sum[i]为3的倍数；若有sum[j] mod 3 == 0（j &gt;= i），则sum[j]也为3的倍数，那么sum[j] - sum[i]也为3的倍数，那么子串(i, j]是3的倍数...]]></description><category></category></item><item><title><![CDATA[CCPC-Wannafly Summer Camp 2019 Day1   (无 I 题)]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/98209147</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/98209147</guid><author>Whyckck</author><pubDate>Fri, 02 Aug 2019 16:38:57 +0800</pubDate><description><![CDATA[比赛链接：https://vjudge.net/contest/312902#overview

目录

A - Jzzhu and Cities

B - Phillip and Trains

C - A Mist of Florescence

D - Unbearable Controversy of Being

E - Igor In the Museum

F - The Chi...]]></description><category></category></item><item><title><![CDATA[【模板】 树形DP求树直径（可处理负边权直径）]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/90758286</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/90758286</guid><author>Whyckck</author><pubDate>Mon, 03 Jun 2019 20:20:40 +0800</pubDate><description><![CDATA[DP【x】表示从x出发走向以x为根的子树，能够到达的最远节点的距离


void get_dp(ll now)
{
	flag[now] = true; //标记是否处理过
	
	for(ll i = head[now]; i != 0; i = E[i].next)
	{
		ll to = E[i].to;
		
		if(!flag[to])
		{
			get_dp(to);
			
...]]></description><category></category></item><item><title><![CDATA[【模板】除法的大质数取模问题]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/90740657</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/90740657</guid><author>Whyckck</author><pubDate>Sun, 02 Jun 2019 16:45:38 +0800</pubDate><description><![CDATA[long long inv(ll x)
{
	long long b = mod - 2,ans = 1;
	while(b)
	{
		if(b&amp;1)
			ans = ans*x%mod;
		
		x = x*x%mod;
		b &gt;&gt;= 1;
	}
	
	return ans;
	
}



]]></description><category></category></item><item><title><![CDATA[【树直径】P3629 [APIO2010]巡逻]]></title><link>https://blog.youkuaiyun.com/Whyckck/article/details/90576061</link><guid>https://blog.youkuaiyun.com/Whyckck/article/details/90576061</guid><author>Whyckck</author><pubDate>Sun, 26 May 2019 14:58:41 +0800</pubDate><description><![CDATA[题目描述

在一个地区中有 n 个村庄，编号为 1, 2, ..., n。有 n – 1 条道路连接着这些村 庄，每条道路刚好连接两个村庄，从任何一个村庄，都可以通过这些道路到达其 他任一个村庄。每条道路的长度均为 1 个单位。 为保证该地区的安全，巡警车每天要到所有的道路上巡逻。警察局设在编号 为 1 的村庄里，每天巡警车总是从警察局出发，最终又回到警察局。 下图表示一个有 8 个村庄的地区，其...]]></description><category></category></item></channel></rss>