shadow the dog and

The JSON value of length n is too large and not supported

https://github.com/dotnet/runtime/issues/39953 I'm referring to this issue #30746 that was closed with limit of 125MB staying fixed opposed to being c ......
supported length value large JSON

Docker启动失败,报错Cannot connect to the Docker daemon at unix:///var/run/docker 完美解决方案

问题描述: Cannot connect to the Docker daemon at unix:///var/run/docker.sock. Is the docker daemon running? 首次安装docker后,服务没有自启动 解决方案: 查看docker状态 1 service ......
Docker 解决方案 connect 方案 Cannot

Docker启动报错:Cannot connect to the Docker daemon at unix:///var/run/docker.sock. Is the docker daemon running?

问题描述: Docker启动报错:Cannot connect to the Docker daemon at unix:///var/run/docker.sock. Is the docker daemon running? Cannot connect to the Docker daemon ......
Docker daemon docker the connect

Diagnostic Port on Electronic Engine and Transmission

Diagnostic ConnectorTest the Connection to the ECM using cat et diagnostic kit ToolkitDeutsch Connectors (6/9-Pin)NOTE: On the Deutsch 9-pin SAE Stand ......

sonarqube启动报错:You must address the points described in the following [2] lines before starting Elasticsearch.bootstrap check XXXmax numberXXXfor user[sonar] is too low .XXX check the logs at XXX/.log

You must address the points described in the following [2] lines before starting Elasticsearch.bootstrap check failure [1] of [2]: max number of threa ......

npm ERROR. node-sass and python

当前 install package 出现以下错误时 node-sass check python checking for Python executable "python2" in the PATH 建议安装 python@2 和 node@14 后再 install package. 下载 ......
node-sass python ERROR node sass

卷影副本(Volume Shadow Copy)是Windows操作系统提供的一项备份和恢复功能。它允许在文件被修改或删除之前,创建文件或文件夹的副本,以便在需要时进行数据的还原和恢复。卷影副本主要有以下作用和优势

卷影副本(Volume Shadow Copy)是Windows操作系统提供的一项备份和恢复功能。它允许在文件被修改或删除之前,创建文件或文件夹的副本,以便在需要时进行数据的还原和恢复。 卷影副本主要有以下作用和优势: 数据保护和恢复能力: 卷影副本可以保护用户的数据免受意外的文件修改、删除和损坏。 ......
副本 文件 文件夹 备份 作用

B. Chips on the Board

B. Chips on the Board 题意:保证在n*n的棋盘上对于每一个点的列或者行都有一个筹码,也可以两个都有,问最小筹码:对于每一个筹码计算:a[i]+b[i]; 要使每个点都满足,最少的筹码的个数应该为n 1.对于行来看,如果每一行都有一个筹码,那么列就可以随便填:直接全选择最小的列 ......
Chips Board the on

Vivado生成bitstream时报错[Opt 31-67] Problem: A LUT3 cell in the design is missing a connection on input pin I1, which is used by the LUT equation

这个原因主要是因为有一个引脚没有用到,解决方法。 1、打开Schematic。 2、根据提示的模块去找,比如说我的报错。 [Opt 31-67] Problem: A LUT3 cell in the design is missing a connection on input pin I1, w ......
connection LUT bitstream the equation

CF1100E Andrew and Taxi

套路题又来咯,最大值最小先直接上个二分答案\(lim\) 对于图中的边,若它的权值\(>lim\)的话这条边的方向就确定了,那么直接把这些边连出来跑个拓扑排序看看有没有环即可 如果有环则当前答案一定不合法,否则我们总存在如下的构造方法: 先把权值\(>lim\)的边得到的图的拓扑序搞出来,对于所有权 ......
Andrew 1100E 1100 Taxi and

[918] Copy the formatting from one cell in a table of a Word document to another cell in Python

To copy the formatting from one cell in a table of a Word document to another cell, you can use the python-docx library in Python. Here's a step-by-st ......
cell formatting document another Python

[919] Change the horizontal alignment of a cell to center within a table of a Word document using Python

To change the horizontal alignment of a cell to center within a table of a Word document using Python and the python-docx library, you can set the ali ......
horizontal alignment document Change Python

[920] Copy the font style from one cell in a table of a Word document to another cell using Python

To copy the font style from one cell in a table of a Word document to another cell using Python and the python-docx library, you can access the font p ......
cell document another Python style

[915] Implementation of zooming to layer and exporting to PDF in arcpy

ref: Camera - ArcGIS Pro ref: Introduction to arcpy.mp # Set the path to your project file (.aprx) project_file = r"Map 1.3 Heritage.aprx" # Reference ......
Implementation exporting zooming layer arcpy

Paper Reading: Sample and feature selecting based ensemble learning for imbalanced problems

为了克服现有集成方法的缺点,本文提出一种新的混合集成策略——样本和特征选择混合集成学习 SFSHEL。SFSHEL 考虑基于聚类的分层对大多数样本进行欠采样,并采用滑动窗口机制同时生成多样性的特征子集。然后将经过验证训练的权重分配给不同的基学习器,最后 SFSHEL 通过加权投票进行预测。SFSHE... ......

Codeforces Round 872 (Div. 2) B. LuoTianyi and the Table

给一个 \(n \times m\) 的矩阵和 \(n \times m\) 个数,你需要把这些数填入矩阵。保证 \[\sum_{i=1}^n \sum_{j=1}^m \left ( \mathop{max}\limits_{1 \leq x \leq i, 1 \leq y \leq j} a_ ......
Codeforces LuoTianyi Round Table 872

CF367C Sereja and the Arrangement of Numbers

这题首先上来会发现题目中的很多信息都是假的,核心就是问要构造一个\(x\)个点的完全图至少要多长的序列 我们把序列中相邻的两个元素看作图上的一条边,则可以把问题转化为:给一个\(x\)个点的完全图,问至少要走多长的路径才可以遍历图中的所有边至少一次 简单讨论下会发现当\(x\)为奇数时,此时图中每个 ......
Arrangement Numbers Sereja 367C 367

CF260D Black and White Tree

刚开始想复杂了,后面再细想了下发现是个傻逼题 考虑一下构造策略,每次从两种颜色集合中分别取出一个数\(u,v\),考虑连边\(u\leftrightarrow v\),边权为\(\min(s_u,s_v)\) 并在每次操作后将\(s_u,s_v\)中较小的那个直接删掉,并把较大的那个值减去\(\mi ......
Black White 260D Tree 260

CF821D Okabe and City

也是一个很经典的优化最短路的题,感觉在暑假前集训做过类似思想的题来着 首先发现我们可以把所有有路灯的点以及终点看作关键点,很显然我们只关心关键点之间的边权以及最短路 不难发现对于两个关键点\(i,j\),如果\(i,j\)相邻,则它们之间有边权为\(0\)的边;否则若\(|x_i-x_j|\le 2 ......
Okabe 821D City 821 and

struct and class

struct and class struct struct 和 class 都是由各种数据组成的集合(也叫做类),这些数据可以是整数,浮点数,字符,也可以是函数。在代码中,我们首先定义集合的名字,包含的数据类别。之后可以命名需用的集合,在主函数或者一些函数中对这些集合调用。 先对 struct 做 ......
struct class and

课程二第三周:Hyperparameter tuning, Batch Normalization and Programming Frameworks

Hyperparameter tuning Tuning process 到目前为止,接触到的超参数有: 学习效率learning-rate:\(\alpha\) Momentum算法的参数:\(\beta\) 加权平均的参数 Adam算法的参数:\(\beta_1、\beta_2、\epsilon ......

flex and bison usage in mysql

query parsing in mysql mysql source code version: 8.0.34 (from MYSQL_VERSION file) This an article from questions to understandings. which file does m ......
bison usage mysql flex and

python websocket server and client 用户认证

WebSocketServer.py pip install websockets #!/usr/bin/env python3 # -*- coding: utf-8 -*- # @mail : lshan523@163.com # @Time : 2023/10/18 9:58 # @Autho ......
websocket 用户 python client server

Signals and systems(1)

LEC 1 Introduction Signals Continuous signals EX1.Sound signals \(y = x(t)\) Continuous Time signal() One dimension signal(only have one variable time ......
Signals systems and

LuoguCF362B Petya and Staircases 题解

分析 简单排序题。 首先 Petya 可以通过跨过一个台阶和两个台阶保证不经过脏台阶,但是不可以通过跨过三个台阶来保证不经过脏台阶,所以只要看有没有连续的三个脏台阶即可。 同时,如果第一个台阶和最后一个台阶至少一个是脏台阶那么就不可以达成。 Accepted Code /*Code By Manip ......
题解 Staircases LuoguCF Petya 362B

[题解]CF1881G Anya and the Mysterious String

思路 发现如果一个字符串中有长度大于等于 \(2\) 回文子串,必定有长度为 \(2\) 的回文子串或长度为 \(3\) 的回文子串,并且形如:aa 和 aba。 所以考虑用线段树这两种情况。维护一段区间的最左、次左、最右、次右的元素,同时用两个标记变量 \(f_1,f_2\) 分别表示这个区间中是 ......
题解 Mysterious String 1881G 1881

PAT_A 1038 Recover the Smallest Number

Given a collection of number segments, you are supposed to recover the smallest number from them. For example, given { 32, 321, 3214, 0229, 87 }, we c ......
Smallest Recover Number PAT_A 1038

How to get macOS CPU details information in the command line All In One

How to get macOS CPU details information in the command line All In One 如何通过命令行获取 macOS CPU 的详细信息 ......
information details command macOS line

[QOJ6555] The 2nd Universal Cup. Stage 5. J : Sets May Be Good

先给 EI 磕三个 首先考虑用 \(n\) 个变量 \(x_1,x_2,\cdots,x_n\in\{0,1\}\) 表示第 \(i\) 个点选不选,那么导出子图的边数的奇偶性就是 \[f(x_1,x_2,\cdots,x_n)=\left(\sum_{(i,j)\in E}x_ix_j\right ......
Universal Stage 6555 Good Sets

asp.net core signalr 客户端调用服务端方法报:Error:Failed to invoke 'adduserToConnection' due to an error on the server

TS端调用方法为: connection.start() .then(() => { connection.invoke("adduserToConnection",account,connection.connectionId); }) .catch((err) => { console.erro ......