feiyue 03

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Linux Beginner Notes 03

. Repo[Redhatsoftware] # # #仓库名称Name=localsoftware # # #仓库描述Baseurl=file:///hello # # #仓库地址Gpgcheck=0 # # #是否检测软件的gpgkey(5). Check if the local Yum warehouse is successful?Yum Clean All # #清空yum仓库的缓存Yum Repolist # #列出该yum仓库中可安装的软件包3. Build the network Yum repository: Build a yum repository, and the other hosts just need to modify the Yum configuration file to install the software(1). Install the HTTPD service (yum install httpd-y), turn off the firewall, start the httpd service(2). Clean up sett

Array-03. Bubble Method Sort

Array-03. Bubble sort (20) time limit MS Memory limit 65536 KB code length limit 8000 B procedure StandardAuthor Xu Jianchun (Zhejiang University)The bubbling sort, which sorts n integers from small to large, works like this: compare adjacent two elements from beginning to end, and swap them if the preceding elements are larger than the following elements immediately after them. By scanning again, the last element must be the largest element. The firs

Linux No. 03 Day

Linux No. 03 Day 1. Linux account and ACL Rights Management 1. Account number and user group 1.1 user identifier ———— UID (root 0 System user is 1~499 Normal user is 500~65535) 1.2 user group identifier ———— GID () 1.3 User account information file ————/etc/ Passwdhadoop:x:500:500:hadoop:/home/hadoop:/bin/bashhadoop is the user name X for the password cipher 500 for UID500 to Gidhadoop as the annotation information/home/ Hadoop for user home directory

Java Learning 03-in-Process learning

quotient and the remainder, so that until the quotient is 0 o'clock, and then the first obtained remainder as a binary number of the low-effective bit, the remainder as a binary number of the high-level effective bit, and then arranged together 6/2 =3---03/2 =1---01/2 =0---1 , decimal 6 calculation process:2. Conversion between octal and binaryOctal 145, Binary 1100101Calculation process:The binary three-bit group is grouped, and the values of each g

2018-03-11 20165235 Qi Ying "Java programming" the first week of study summary

2018-03-11 20165235 Qi Ying "java program design" the first week of learning summary textbook Learning content Summary Chapter II Highlights: In this chapter I have learned a lot of things: (1) Boolean type bool, the assignment of a Boolean type can only be true or false. (2) Unicode code (3) System.out.println () input + indicates end-to-end connection (4) Float:8 digit valid number, Double:8 bit valid digit.

2018-03-19 Linux Learning

派的资源占用CPU百分比,长时间大于50,说明系统资源紧张sy 系统本身占用CPU百分比id 空闲wa 等待CPU的百分比,数值大,CPU不够用st 被偷走了CPU百分比10.3 Topus 重要zombie 僵尸进程默认按CPU百分比排序按 M 按内存排序按 P 按CPU排序top -c 显示进程完整路径top -bn1 一次性把所有进程都列出来10.4 SAR监控系统状态 sar -n DEV 网卡流量 sar -q 系统负载 sar -b 磁盘读写 sar -f /var/log/sa/saxx 历史文件 nload yum install -y epel-release yum install -y nloadsar# yum install -y sysstatrx 接收tx 发送rxpck/s 几千数据包比较正常,上万时负载过重txpck/s [[emailprotected] ~]# sar -q Linux 3.10.0-693.17.1.el7

Linux Command 2018-03-01 Update

改变文件内容sed ‘/^test/‘d test.txt # 删除所有开头是test的行sed ‘/^$/d‘ test.txt # 删除空白行sed ‘2d‘ test.txt # 删除第2行sed ‘$d‘ test.txt # 删除最后1行sed ‘2d,$d‘ test.txt # 删除第2到最后1行之间所有行sed -n ‘20,30p‘ test.txt # 打印文件20到30行之间的内容# 把递归查找到的所有test.txt中所有的alex替换成lemased -i ‘s/alex/lema/g‘ `find . -name "test.txt"` awk -F ‘:‘ ‘{print $NF,$(NF-1)}‘ /etc/passwd # 打印最后两列awk ‘{if(NR>19NRFile Directory Management################### #增 ################### #touch把已存在文件的时间更新为当前系统时间或者创建新文件cp sshd_config sshd_config.lema.20180301 # Bac

Linux Fundamentals-Disk Management-03

Tags: Linux cloud computing Disk Management partition file management DahneIn the previous article, we talked about the basic composition, basic concepts, and basic principles of the disk.So today's article, let's look at how these devices are represented in the Linux system,such as how different types of hard disks are represented,For example, how different types of partitions are represented,Only by figuring out how these "devices" are represented in the system, in the latter study, canFigure

"Java Learning" 03.MyEclipse installation and configuration tomcat

FeedEclipse is commonly used for j2se,myeclipse for EE;Steps1, install MyEclipse, on-line download is good;2, crack myeclipse, to be cracked ...3, configure Tomcat, disable MyEclipse with Tomcat, because there is no Tomcat8 in MyEclipse, can only be configured to tomcat7.x, the top menu to find the server configuration icon,Configure servce--services--integrated sandbox--myeclipse Tomcat 6--disable--OK4. Configure Tomcat, configure the install-free Tomcat 8, enter configure Servce--services--tom

Python notes 03 List Dictionary

falseDictionary traversalFor key in XXXXFor value in XXXXFor item in XXXX elementFor Key,value in XXXX key value pairEnumerate () use If you want to traverse the index and iterate through the elements on a list, you can first write this: list1 = ["这", "是", "一个", "测试"]for i in range (len(list1)): print i ,list1[i] The above method is somewhat cumbersome, and the use of enumerate () will be more direct and graceful: list1 = ["这", "是", "一个", "测试"]for index, item in enum

Data structure and algorithm learning 1th season 03 Stack Queue tree Chart

2015 Study Plan Arrangement:Http://www.cnblogs.com/cyrus-ho/p/4182275.htmlStack: Linear table for LIFOQueue: Linear table of FIFOTree: (recursive definition) n nodes have a poor set, to a non-empty tree, there is only one node called the root, n>1, the remaining nodes are divided into M-disjoint finite sets, and each set itself is a tree, and is called the root subtreeFigure: An organizational structure in which relationships can exist between any two data elementsBinary tree storage: Data struc

Linux Common Commands-03

directory destination file or directoryOptions:-P maintains the specified property (default: Mode, ownership, timestamp)-R recursively replicates all contents in directory and directory-D If the source file is a linked file, copy the connection properties-A equals-pdr650) this.width=650; "src=" Http://s3.51cto.com/wyfs02/M02/6C/D7/wKiom1VTRxfSD-_rAARIfumramM540.jpg "title=" Cp1.jpg "style=" Float:none; "alt=" wkiom1vtrxfsd-_raarifumramm540.jpg "/>650" this.width=650; "src=" http:// S3.51cto.com

Linux Learning -03-First Login

Displays the languages currently supported by the system: Echo $LANGZh_cn. UTF-8If the system does not support Chinese and garbled, then you need to modify the language: Lang=en_usEcho $LANG again, show en_USLed that shows the date is not time: dateShow led order for calendar: CalEasy to use calculator: BC[Tab], command to be padded[Ctrl]-c, interrupt Program[Ctrl]-d,exit exitSimple text Editor: Nano + file nameOpen the old file if it exists, open the new file if it doesn't existShutdown:Who sho

Crazy C + + 's essential c++03

3. Compose an expressionA. Short-circuit evaluation methodA | | b If A is true, B is not evaluated and the expression is trueA b If A is false, B is not evaluated and the expression is falseB.if () statement and the expression in sizeof ()650) this.width=650; "title=" 1.jpg "src=" Http://s3.51cto.com/wyfs02/M01/6C/5F/wKioL1VIPdWRAmghAAJBBeNuqBQ706.jpg " alt= "Wkiol1vipdwramghaajbbenuqbq706.jpg"/>This article is from the "very blog" blog, make sure to keep this source http://10093949.blog.51cto.

03-First script and input/output _python programming path

)   >>> age = Input ("Please enter your Age:") Please enter your ages: + >>> type   >>> result = input ("Please enter an expression:") Enter an expression: 2*50 >>> Print (result) 2*50   Of course Python also provides a way to easily hide the entered passwordWe only need to import this module before useImport Getpass   Then use the Getpass.getpass () method>>> passwd = Getpass.getpass ("Please enter your password: ") Please enter your password: >>> print (passwd) 123456 Default

Effective C + + clause 03 Collation

indicates that the object is a constant object, and any of its members cannot be modified. The same is true for object pointers and object references. Const-Decorated object, any non-const member function of the object cannot be called because any non-const member function has an attempt to modify the member variable. Example: Class aaa{ (5) Some suggestions for using constThe bold use of const will bring you endless benefits, but only if you have to figure it out;To avoid the m

Fluent_python_section2 data types, 03-dict-set, dictionaries, and collections

Key must be hashable. The dictionary has a huge overhead in memory because hash table is a sparse array Key query is fast because hash table is the space change time of classic The order of keys depends on the order of additions Adding a new key to a dictionary may change the order of existing keys, and do not iterate and modify the dictionary at the same time. Because the Python interpreter might make a decision to expand the dictionary, a hash conflict may occur

Algorithm note generation arrangement of 03--inductive method

n-1 position of the number generation arrangement.2. A number generation arrangement of J (J from 1) to n-1 position, in which the number I (starting from 2) is sequentially taken out from 2,,.., N and the number on position J is exchanged, and the number generated in position j+1 to n-1 position is arranged.3 .... When j = N, generates an arrangement and prints4. Retrace the previous step, take the number of words i+1 and J on the digital exchange, and the sequence after the j+1 to generate a

Python Deep properties for 03 objects

e) print (num.neg.__doc__) del X.negThe above num is a number, and neg is an attribute that is used to denote a negative number of numbers. When a number is determined, its negative numbers are always determined, and when we modify a negative number, the value itself should change. These two points are implemented by Getneg and Setneg. Delneg, however, is that if you delete an attribute neg, the action you should take is to delete the property value. The last parameter ("I ' M negative") of the

Array-03. Sort by bubbling (20)

The bubbling sort, which sorts n integers from small to large, works like this: compare adjacent two elements from beginning to end, and swap them if the preceding elements are larger than the following elements immediately after them. By scanning again, the last element must be the largest element. The first N-1 elements are then scanned again in the same way. And so on, the order of n number is finished by processing two elements at the end.This is required for any given K (Input format:The in

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