increasing decreasing 1864a and

D. Trees and Segments

D. Trees and Segments The teachers of the Summer Informatics School decided to plant $n$ trees in a row, and it was decided to plant only oaks and fir ......
Segments Trees and

C++ Constructor And Destructor

if you have problems with constructors and destructors, you can insert such print statements in constructors for your real classes to see that they wo ......
Constructor Destructor And

C++ Cast And Go Cast

## C++ * A static_cast can be used to explicitly convert between related pointer types, such as void* and double* * reinterpret_cast can cast between ......
Cast And Go

springmvc 开启异步请求报错 Java code using the Servlet API or by adding "true" to servlet and filter declarations in web.xml.

报错内容: java.lang.IllegalStateException: Async support must be enabled on a servlet and for all filters involved in async request processing. This is do ......
quot declarations springmvc Servlet servlet

CF437C The Child and Toy

### 题目大意 $n$ 个带权点,$m$ 条无向边,删除一个点就要付出所有与之连接且没有被删除的点的点权之和的代价。 求删除所有点的最小代价。 ### 思路 考虑点的贡献异常麻烦,我们可以把点的贡献转化为边的贡献。 对于一条边,我们有如下几点: 1. 伴随着所有的点被删掉,所有的边也会被删掉; 2 ......
Child 437C 437 The and

CF1060E Sergey and Subway 题解

[题面](https://codeforces.com/problemset/problem/1060/E) 由题意可知,在原图中经过边数为 $2$ 的一对点,在新图中经过边数为 $1$。所以每对点在新图中的距离为: $$ \begin{aligned} \lceil \frac{dis(i,j)} ......
题解 Sergey Subway 1060E 1060

【转载】The Dog and the Shadow

## Article Format Source ChatGPT ## Main Content Source Title: The City Mouse and the Country Mouse Source: https://www.zhihu.com/question/263840407/a ......
Shadow The Dog and the

AGC064C Erase and Divide Game

[题面传送门](https://atcoder.jp/contests/agc064/tasks/agc064_c) 首先考虑你只插入若干个数怎么做:按位从低到高插入一棵 Trie,问题就变成:在 Trie 上每次可以往左儿子走或者往右儿子走,如果当某个人操作的时候为空节点那么这个人就输了。 如果我 ......
Divide Erase 064C Game AGC

NetSuite: Get specific Custom Record Types and related sub Custom Fields

背景 以前当使用search.create({})来获取数据时,我们需要制定特定的数据返回列;例如:search.createColumn(options) 而query可以使用 SELECT * FROM 来动态返回所有的数据列(这在有的时候是一个优点),那么如何让search也动态返回所有的数据 ......
Custom NetSuite specific related Fields

CodeForces 1858D Trees and Segments

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CodeForces Segments 1858D Trees 1858

[Typescript] Don't compare generic function, instead compare function arguments and return type

Typescript has its problem that when you try to compare generic function to a function signature, you will run into issue. Because for one function, i ......

Unveiling the Power of Diagnostic Tools: Cat Caterpillar ET, Volvo Vocom II Interface, JLR DoIP VCI, WABCO Diagnostic Kit, and GM Tech 2

In the realm of automotive diagnostics, the right tools can make all the difference. Whether you're a professional mechanic or an avid car enthusiast, ......

card,swallow-and-electron-cloud

贺卡,燕子和电子云 Created: 2023-07-16T17:52+08:00 Published: 2023-08-15T17:36+08:00 Categories: Fragment [toc] # 贺卡 翻日记,看到高三时候班级举行过随机互赠贺卡的活动,对于这件事我只有一点点模糊的印象, ......

CF1513D GCD and MST 题解

## 题面 对于一个序列,若有 $(i,j)(i typedef long long valueType; typedef std::vector ValueVector; typedef std::pair ValuePair; typedef std::vector PairVector; ty ......
题解 1513D 1513 GCD and

CF1859D Andrey and Escape from Capygrad 题解

## 思路 思考贪心,容易得出我们只有不断往右跳跃才能走得更远。 所以,对于一个线段 $[l, r]$ 可以轻易到达 $[a, b]$,那么只对 $[l, b]$ 有用,这些点都可以跳到 $b$,$[b + 1, r]$ 这一部分不能往回跳,所以不用考虑。 那么我们就可以把这些线段都当成 $[l, ......
题解 Capygrad Andrey Escape 1859D

Go - Keywords, Operators and punctuation

Keywords The following keywords are reserved and may not be used as identifiers. break default func interface select case defer go map struct chan els ......
punctuation Operators Keywords and Go

"Don't be shy. Speak English loudly and crazily!"

"Don't be shy. Speak English loudly and crazily!" 俞敏洪犀利点评马云、王石、刘强东、雷军英语水平 俞敏洪:马云8岁学英语,考上杭师范读专科,而我在北大读本科! 李阳、马云、俞敏洪,这三个中国最著名的英语老师,只有李阳仍坚守在一线。 是啊,当马云再次登 ......
quot English crazily loudly Speak

Codechef - Longest AND Subarray(位运算)

题目大意 给定一个正整数N,其序列为[1, 2, 3, ..., N],找到一个长度最大的连续子列,使得其所有元素取与运算的结果为正(最终输出只需要输出最大长度即可)。 思路 刚开始可能并不好注意到,可以举一些小的样例来找规律。比如2:1 & 2 = 0,不满足条件,所以只能取长度为1的数组[1]或 ......
Codechef Subarray Longest AND

PnP and Perspective Projection and Pose Computation

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Perspective Computation Projection and Pose

Mongodb or and 查询

查询两个条件符合一个即可 { $or : [ { $and : [ { "Type" : "Unorganized" }, { "Tel" : "Procedure" }, { "Factor" : { "$ne" : { } } } ] }, { $and : [ { "Type" : "Orga ......
Mongodb and or

Git:Vscode提交报错Make sure you configure your "user.name" and "user.email" in git

使用VScode编辑代码后,Push到云端报错:Make sure you configure your "user.name" and "user.email" in git 解决步骤: 1.进入本地端的文件夹,右键Git Bash; 2.输入命令: $ git config --global u ......
quot user configure Vscode email

CaltechCS122 笔记:Assignment 1: NanoDB Set-Up and Storage Layer

# Assignment 1: NanoDB Set-Up and Storage Layer NanoDB 是加州理工大学 Caltech CS122 课程使用的教学数据库系统 ## buffer pool manager lab1 的第二部分是实现充分利用空间的 bpm,当前所给出的 bpm 代 ......
Assignment CaltechCS Storage 笔记 NanoDB

「解题报告」AGC064C Erase and Divide Game

第二次打 AGC,场上过了 C,还是很开心的。 而且是在一整天没碰 OI 的情况下。所以长时间 AFO 再回来打比赛会不会对 OI 有一定的 buff?有时候思维过度疲惫的情况下貌似打比赛脑子是真的转不动。 不过为啥 D 过的比 C 多啊,我觉得 C 很简单啊,基本没有任何转化,可能 AT 的这类计 ......
报告 Divide Erase 064C Game

UOI 2023 An Array And Addition Again

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Addition Again Array 2023 UOI

D. Andrey and Escape from Capygrad

D. Andrey and Escape from Capygrad An incident occurred in Capygrad, the capital of Tyagoland, where all the capybaras in the city went crazy and star ......
Capygrad Andrey Escape from and

Data structure and algorithm-One

右边界左边界 函数式编程 传统的实现方式通常是将具体的功能代码直接写在方法内部。这样的实现方式对于固定的功能来说可能是足够的,但它的灵活性和复用性较低。每当需要不同的功能时,您需要编写新的方法或修改原有方法,这可能会导致代码的冗余和不易维护。 而函数式编程的优点在于它将函数本身作为参数传递,使得调用 ......
algorithm-One structure algorithm Data and

【翻译】为什么我们需要极限和无穷小?(Why Do We Need Limits and Infinitesimals?)

那么多数学课,没有任何上下文,就跳到极限,无穷小,非常小的数(T)。但是我们为什么要在乎呢?数学帮助我们模拟世界。我们可以把一个复杂的想法(一条蜿蜒的曲线)分解成更简单的部分(矩形): ![image](https://img2023.cnblogs.com/blog/2469253/202308/ ......
无穷小 Infinitesimals 极限 Limits Need

CF643F Bears and Juice

[题面传送门](https://www.luogu.com.cn/problem/CF643F) 感觉这个从信息的角度去考虑比较高妙。 首先取 $p=\min(p,n-1)$。 我们来考虑每个桶对应喝的熊,这样会有一个长度为 $n$ 的数组,假设有 $t$ 天,那么每个位置会写一个 $[1,t+1] ......
Bears Juice 643F 643 and

【Oracle】SEGMENT SHRINK and Details. (Doc ID 242090.1)

``` PURPOSE In Oracle10g and onwards, we have the option to shrink a segment, which will help DBAs to manage the space in better way. This feature als ......
242090.1 Details SEGMENT Oracle 242090

[图论记录] CF1082G Petya and Graph

([题目传送门](https://www.luogu.com.cn/problem/CF1082G)) 一道最大权闭合子图的模板题 把所有点当作负权点,所有边当作正权点。 考虑一条边 $e=(x,y,z)$ 能选择的条件,当 $x,y$ 均被选择时就可以选,那对应到最大权闭合子图的模型中,就可以将 ......
1082G Petya Graph 1082 and