ccna training guide

i am the 51CTO lecturer Kanhaijiang, at the Institute 11.11( People's IT Learning Festival) Come on the occasion, and everyone to share my CCNA customs clearance experience. The text is coming ~ ~ ~in writing this title, I seem to have returned to the original to learn Cisco Qizaotanhei, but a fun time. Obviously, I am definitely not a network genius, no extraordinary talent, even at that time even a computer belonging to their own elusive, so in the

March 2016 2 August-April 1st Ivy Alliance Training Center (Avtech Institute of Technology), for a mobile company IT staff of 5 days of CCNA Routing and Exchange practical training class successfully implemented! This training arrangement is held in the beautiful Shenzhen.CCNA Certified professionals have the knowledge

CCNA Training Course (3) -- Vlan VTP relay single-arm route generation Tree Protocol This is the third day of the CCNA training course. There are many knowledge points on this day, and the experiment content is gradually becoming more complicated.VLAN A vswitch separates the conflict domains and the broadcast domains.

Tags: ccnaThe more complex the interconnection network environment is, the more likely the network is to fail in terms of connectivity, and the cause of the failure is hard to confirm. Before a network failure occurs, the network administrator should constantly monitor network changes and make timely adjustments to prevent problems. When a network fault occurs, appropriate methods and tools should be used to determine the root cause of the fault, and correct troubleshooting plans should be made.

ICMP Echos to 192.168.20.1, timeout is 2 seconds: .!!!! Success rate is 80 percent (4/5), round-trip min/avg/max = 24/48/84 MS From the above, we can see that through the deployment of layer-3 switching technology, the host between VLANs can communicate with each other. This experiment is complete. [PL1]Disable layer-3 routing to simulate hosts. [PL2]Define the default gateway. [PL3]Enable the layer-3 routing function. By default, the layer-3 Switch disables the routing function. To enable int

Algorithm competition entry-classic-Training GuideBasic InformationAuthor: Liu rujia Chen Feng [Translator's introduction]Series name: algorithm art and informatics CompetitionPress: Tsinghua University PressISBN: 9787302291077Mounting time:Published on: February 1, October 2012Start: 16Page number: 1Version: 1-1Category: Computer> Computer science theory and basic knowledge> computing theory> AlgorithmMore about, algorithm competition getting started

Getting started with algorithms Classic Training Guide Chapter I example 5This problem personal feeling is still very technical content, if pure simulation can kill.The situation in which an ant is turned around in a collision is finally seen as "over-worn", and this "transformation thought" is very powerful. If only I had this skill one day.Personal feeling if you can't see it, the problem hangs.So when de

operatorConsttask x)Const - { - returnj>x.j;//because we're going to go from big to small, so here's a change . the } - }; - - intMain () + { - intcases=0, N; + while(SCANF ("%d",N)) A { atcases++; - if(n==0) Break; - task KASE[MAXN]; - for(intI=0; i"%d%d",kase[i].b,KASE[I].J); -Sort (kase,kase+n); - intm=0, ans=0; in for(intI=0; i) - { tom+=kase[i].b; +Ans=max (ans,m+KASE[I].J); - } thecout" Case"": "Endl; * }

certain number of the result of the M1 K, The number of figures and the result of the remainder K is m2. So for numbers with higher digits but the first few digits are deterministic, we can build a recursive relationship to narrow down the size of the problem. For example, for the figure 22xxx, which represents an integer on [22000,22999], there are three "free bits", that is, the bit can be taken over [0,9]. It can be seen that the solution on this interval corresponds to another smaller state

constructor directly when defining the point.};typedef point Vector;//this is because vectors have ordered parameters of two dimensions.Vectoroperator+ (Vector a,vector B) {returnVector (a.x+b.x, a.y+b.y);} Vectoroperator-(point A,point B) {returnVector (a.x-b.x, a.y-b.y);} Vectoroperator* (Vector A,DoubleP) {returnVector (A.x*p, a.y*p);} Vectoroperator/(Vector A,DoubleP) {returnVector (a.x/p, a.y/p);}DoubleCross (vector A, vector B)//Vector cross product{ returna.x * B.Y-A.Y *b.x;} Point Ge

("%d%d", n,m) = =2N) { for(intI=0; i//Point Set for(intI=0; i//Input Edge { intu,v; scanf ("%d%d",u,v); G[u].push_back (v); G[v].push_back (U); } FIND_BCC (n); //calculate the number of two connected componentsprintf"Point-two connected components%d \ n", bcc_cnt); for(intI=1; i//outputs each of the two connected components. Maybe point A is output in the first two-connected component, and it appears in the 2nd double-connected component because i

(ant a,ant B)//Sort by Location{ returna.posB.pos;}Const Chardire[][Ten]={"L","Turning","R"};//array Save output resultintORDER[MAXN];//the function of order array is to preserve the relative position of each ant's final state;intMain () {intT1; CIN>>T1; for(intKase=1; kase) {cout"Case #"":"Endl; intL,t,n; CIN>>l>>t>>N; for(intI=0; i) { intx; CharC; CIN>>x>>C; intD= (c=='L'?-1:1); Before[i]=Ant{i,x,d}; After[i]=ant{0, x+d*t,d};//The direction is still d, because there mus

not open## Controls who can get stuff from the this server.#Require all grantedLook at the set-row statistics used by httpd.confEgrep-v "^.*#|^$" httpd.conf|nlExtra is an Apache extension profile directory[Email protected] extra]# tree.├──httpd-autoindex.conf├──httpd-dav.conf├──httpd-default.conf├──httpd-info.conf├──httpd-languages.conf├──httpd-manual.conf├──httpd-mpm.conf├──httpd-multilang-errordoc.conf├──httpd-ssl.conf├──httpd-userdir.conf├──httpd-vhosts.conf└──proxy-html.confWhich is mainly

help with Vim :help split . You can refer to the previous article of the site Vim split screen. :split→ Create split screen ( :vsplit create vertical split screen) : dir is the direction, can be hjkl or one of ←↓↑→, it is used to switch split screen. (or ): Maximized size ( (or ): Increase size Conclusion Above are the 90% most commonly used commands by the author. I suggest you learn 1 to 2 new commands every day. After three weeks,

The next section continues to say SersyncThis section is about the actual experiment of Sersync.Installation of primary server Sersync, settingAnd then a synchronous demo.Let people know more about the principle of synchronous automation softwareIn which the real-time synchronization is explainedWhen the amount of synchronization is highThe client may not be able to update in real time20170705l07-09-03 old boy Linux Operational training-sersync real-t

can assume that the red side with the dot i has an AI bar, and the black Edge has a (N-1-ai) bar. Then the same point of the two different color sides must be able to form a unique triangle ,So, according to the multiplication principle, we can get the number of non-monochromatic triangles for point IThere are a total of n points, and each edge is counted two times. So we need to sum up and divide by 2 to obtain the final result:Finally we can get the number of monochrome triangles:Over~ the sm

Tags: nginx use environment Nginx service with PAC Linux OPS include fileMulti-instance setup for NginxLet's start with a description of the parameters appended to the Nginx-S back plus reload is the same effect that reloading means and Apache graceful-V Lowercase v Displays the version number after exiting-V-Uppercase with the version number of the Nginx and configuration environment-T is the meaning of test, check the configuration file is correct-C After the address of the configuration file,

Sersync pressure test in the next sectionThis section focuses on the pressure test text of the previous section to do a real stress testTest out exactly what the sersync pressure limit is.From 10 to 100 per second# Tree |wc-l #查看写了多少文件# uptime #查看负载Here is the test script#!/bin/shcount=100While TrueDoFor ((i=1;iDo/BIN/CP 10k.jpg/backup/test/$i/10k_echo $ (date) $RANDOM |md5sum |cut-c 1-8.jpgDoneSleep 1For ((i=1;iDo/BIN/CP 30k.jpg/backup/test/$i/10k_echo $ (date) $RANDOM |md5sum |cut-c 1-8.jpgDon

(XL) Lt1-2The problem is still not very difficult, but note the operator overloads so that sort, as well as the construction members, and the sort encounter vector iterator.Wrote two copies of the code.Attach your own common method first:#include #include#includeusing namespacestd;Const intmaxn= ++5;structsoldier{intb; intJ; BOOL operatorConstSoldier a)Const{ returnJ>A.J; }}; Soldier A[MAXN];intMain () {intN,kase=0; while(SCANF ("%d", n) = =1N) { for(intI=0; i"%d%d",a[i].b,A[I].J

connected components. the necessary and sufficient condition of the binary graph is that there is no odd circle, which can be obviously obtained: if a two-connected component is a binary graph, then the inside point must not be on the odd circle. If a two-connected component is not a binary graph, will the dots in it be on the odd circle?The answer is certain in the odd circle, the first two connected components is not a binary map, then there must be a singular ring C. There are two states fo