codeforces complete 1508c the
AtCoder World Tour 2022 B The Greatest Two
原题面：https://atcoder.jp/contests/wtf22-day2/tasks/wtf22_day2_b 题面翻译： 一个长度为 \(n\) 的排列 \(p\)，每次可以把一个长 \(k\) 区间的最大与次大值交换，问操作任意次数后可以得到的排列数量对 \(998244353\) ......
Introducing the incident management
Forward Today, we talk about the significance of incident management. Firstly, we start it by some simple concepts. What exactly the incident is? Or w ......
A connection was successfully established with the server but then an error
在使用EFCore生成数据库的时候，报上面的错误！ 解决方法： 加（Encrypt=True;TrustServerCertificate=True;）即可： "ConnectionStrings": { "DefaultConnection": "Data Source=LAPTOP;Initia ......
An improved LSTM-based model for identifying high working intensity load segments of the tractor load spectrum
一区top Computers and Electronics in Agriculture 题目： “基于改进 lstm 的拖拉机载荷谱高工作强度载荷段识别模型” (pdf) “An improved LSTM-based model for identifying high working in ......
web项目启动时dubbo报错： No provider available for the service com.davidhu.shopguide.api.service.UserEventService from the url zookeeper ://localhost:2181/org.apache.du
Caused by: java.lang.IllegalStateException: Failed to check the status of the service com.davidhu.shopguide.api.service.UserEventService. No provider ......
Hey left 1 Codeforces Round 918 (Div. 4)
题目链接 A. 3个数，其中2个数相同，输出不相同的那个 可以用if else判断，较为麻烦 用的map，输出出现一次的 #include <bits/stdc++.h> using namespace std; const int N=1e5+10; void solve(){ map<int,i ......
初中英语优秀范文100篇-056I have the courage to accept the challenge-我有勇气接受挑战
PDF格式公众号回复关键字:SHCZFW056 记忆树 1 Every year there is a singing competition in our school. 翻译 每一年，我们学校都会举行一场歌唱比赛。 简化记忆 比赛 句子结构 主语 ("Every year")：表示时间状语的短语 ......
org.springframework.kafka.listener.ListenerExecutionFailedException: Listener method could not be invoked with the incoming message
问题描述 kafka在yml文件中未开启批量消费时，程序正常运行；但一开启正常消费后，就直接报错；排查问题的过程中一直觉得是配置文件里的问题，最后发现是消费者接受的参数类型错误 问题本质 消费者开启批量消费数据后，不能用单个实体类接收参数，而应该用list 解决方法 修改消费者函数参数类型 ......
redis报错MISCONF Errors writing to the AOF file: No space left on device，磁盘满无法写入数据报错
查询报错远程调用接口异常，F12查看网络和标投和预览到是mgr的接口此错误#根据报错提示：Error in execution; nested excepti on is io.lettuce.core.RedisCommandExecutionException: MISCONF Errors w ......
[linux kernel] struct pid && the way the kernel used to cast uint32_t pid to struct task_struct
摘抄内容引用自linux kernel document description pre the initialization and the manage of the struct pid is use IDR.(implement by hash table->(why not array?- ......
CodeForces 1312G Autocompletion
洛谷传送门 CF 传送门 考虑直接在题目给的 Trie 上 dp，设 \(f_u\) 为打出 \(u\) 结点的串的最小代价。 首先我们有 \(f_u \gets f_{fa_u} + 1\)。 我们有 \(f_u \gets \min\limits_v f_v + t + 1\)，要求 \(u\) ......
Educational Codeforces Round 152 (Rated for Div. 2)
layout: ../../layouts/MarkdownPostLayout.astro title: 'Educational Codeforces Round 152 (Rated for Div. 2)' pubDate: 2024-01-11 description: '一些训练' au ......
Redis - (error) WRONGTYPE Operation against a key holding the wrong kind of value
zzh@ZZHPC:~$ docker exec -it redis redis-cli 127.0.0.1:6379> SET bike:1 "Process 134" OK 127.0.0.1:6379> GET bike:1 "Process 134" 127.0.0.1:6379> HSET ......
The Acdamic Guideline
How to write your papers？ Introduction Problem Methods Results Simulation How to start your research？ Observation Problem Reserach Hypothesis Experime ......
codeforces 918 div4
C 题意 : 给定一些数,问这些数的和是不是完全平方数(5*5这样) #include<cmath> #include<iostream> #define ll long long using namespace std; bool issq(ll n){ ll root = sqrt(n); re ......
解决前后端的跨域问题：Access to XMLHttpRequest at '**' from origin '**' has been blocked by CORS policy: No 'Access-Control-Allow-Origin' header is present on the requested resource.
报错信息： Access to XMLHttpRequest at 'http://localhost:8182/cooperationRequest/getList' from origin 'http://localhost:3004' has been blocked by CORS poli ......
Understanding the linux kernel Chapter2 Memory Addressing
Physical Memory Layout unavailable address for kernel either because they map hardware devices’ I/O shared memory or because the corresponding page fr ......
No 'Access-Control-Allow-Origin' header is present on the requested resource'， 跨域访问的解决方法
https://blog.csdn.net/dear_little_bear/article/details/83999391 1. 当请求不在同一域名下的资源文件（ip地址+端口号）时，会报如下错误：“No ‘Access-Control-Allow-Origin’ header is prese ......
Unlocking the Road to Success: The Benefits of Online Driver's Education
In the fast-paced world we live in, online education has become a staple for acquiring new skills and knowledge. This trend extends to driver's educat ......
The Evolution of Smart Car Technology: A Glimpse into the Future of Mobility
In the last decade, the automotive industry has witnessed a transformative shift towards smart car technology. Once a futuristic concept, smart cars a ......
vue报错：You may use special comments to disable some warnings. Use // eslint-disable-next-line to ignore the next line. Use /* eslint-disable */ to ignore all warnings in a file.
报错信息： 解决方法： ESLint 语法错误，那就取消ESLint验证规则 然后重启即可 ......
the ObjectARX system dynamic linker object
ObjectARX Reference Guide > Macros > AcRx Macros > acrxDynamicLinker Macro acrxDynamicLinkerC++ define acrxDynamicLinker \ AcRxDynamicLinker::cast(acr ......
MySQL Ignoring the redo log due to missing MLOG_CHECKPOINT between the checkpoint
错误信息： 2023-12-12T09:32:31.383149Z 0 [ERROR] InnoDB: Ignoring the redo log due to missing MLOG_CHECKPOINT between the checkpoint 5777611209 and the end ......
CodeForces 1329D Dreamoon Likes Strings
洛谷传送门 CF 传送门 考虑构造一个新串 \(t\)，只保留原串 \(s_{i - 1} = s_i\) 的字符 \(s_i\)。设 \(a_i\) 为 \(t_i\) 在原串的位置。 那么新串上我们有两种操作： \(\forall i\)，删除 \(t_i\)（相当于删除原串中的 \([a_i, ......
CodeForces 1237H Balanced Reversals
洛谷传送门 CF 传送门 容易想到把 \(s, t\) 分成长度为 \(2\) 的段考虑。容易发现 \(00, 11\) 的个数在操作过程中不会改变，所以若两串的 \(00\) 或 \(11\) 个数不相等则无解。 考虑依次对 \(i = 2, 4, \ldots, n\) 构造 \(s[1 : i ......
vp Codeforces Round 915 (Div. 2)
vp还是比正式打舒服一些。。AB很顺畅，A题。。我只能说玩MC的都一眼秒了好吧 C题，我卡住了，结论非常好推，我直接退出来了，但是，问题是我对特例的判断不是很熟悉，或者说不是很敏感。这是一个大问题，我在wa on test 2的时候，第一反应是去看看这个算法整个有没有什么问题，事实上是没有的那么问题 ......
Codeforces [Hello 2024]
Codeforces Hello 2024 主打一个昏了头 A. Wallet Exchange #include <bits/stdc++.h> #define endl '\n' //#define int long long using namespace std; const int N = ......
CodeForces 1379E Inverse Genealogy
洛谷传送门 CF 传送门 \(n\) 为偶数显然无解。 否则我们可以构造一棵 \(n\) 个点的完全二叉树，当 \(n + 1\) 是 \(2\) 的幂时满足 \(m = 1\)，否则 \(m = 0\)。 当 \(n \ge 5\) 时可以递归至 \((n - 2, m - 1)\)，再挂一个叶子 ......
CF1523H Hopping Around the Array
首先考虑 \(k = 0\) 的情况。 贪心，最后一步之前每个 \(i\) 只会跳到 \(j \in [i, i + a_i]\) 且 \(j + a_j\) 最大的点 \(j\)，这个信息或许可以线性处理？但是我没脑子，我用线段树维护，时间复杂度 \(\mathcal O(n \log n)\)。 ......
CodeForces 1919F2 Wine Factory (Hard Version)
洛谷传送门 CF 传送门 题目看着感觉很像最大流，不妨建模，\(S \to i\)，容量为 \(a_i\)；\(i \to T\)，容量为 \(b_i\)；\(i \to i + 1\)，容量为 \(c_i\)。答案是这个图的最大流。 考虑最大流转最小割。观察到 \(S \to i\) 和 \(i ......