Control-point Attribute
High-level protocols can use this property to enable device-specific processes, such as commands or instructions for a given procedure on a device, to complete 2.7 Protocol Methods
ATT uses protocol methods to discover, read, write, notify, and indicate attributesMethods can be divided into the following types-Request-Response-Command-Notification-Indication-Confirmation
Some att pdus contain a certified signature (authentication Signature)Allow PDU send side to not requ
Before the SUSE Linux network card renaming problem, is installed on the virtual machine RAC, by replicating the virtual machine need to complete. Unlike Redhat Linux and Oralce Linux, the following records them for reference.
1, the new node network configuration
#启动节点之后没有任何IP配置信息
bo2dbp:~ # Ifconfig
Lo Link encap:local Loopback
inet addr:127.0.0.1 mask:255.0.0.0
Inet6 addr::: 1/128 scope:host
Up loopback RUNNING mtu:16436 metric:1
RX packets
The network cannot be accessed after Linux is installed.After installing the new Linux system on the virtual machine, ping www.baidu.com and find that it cannot be ping.Type ifconfig query configuration:
Eno16777728:flags=4163
Ether 00:0c:29:3e:ad:2d Txqueuelen (Ethernet)
RX Packets 0 Bytes 0 (0.0 B)
RX errors 0 dropped 0 overruns 0 frame 0
BackpressureThe data stream in Rx is emitted from one place to another. The speed at which each place processes data is different. What happens if the producer launches the data faster than the consumer can handle? In a synchronous operation, this is not a problem, for example://produceobservable.create (o, {o.onnext (1 ) O.onnext (2 ) ; O.oncompleted () ; }) Consumeproducer.subscribe (I--{try {thread.sleep (1000 ) System.out .println (
the multi-Job Scheduling queue function is enabled, and they are distributed across all available processing cores.
Assume that these NICs are named "enp1s0f0" and "enp1s0f1":
[Root @ bkjia ~] # Tuna -- irq 'enp1s0f * '-- socket 0 -- spread -- show_irqs# Users affinity69 enp1s0f0 0 igb70 enp1s0f0-rx-0 1 igb71 enp1s0f0-rx-1 2 igb72 enp1s0f0-rx-2 3 igb73 enp1s0f0-
In Ubuntu, I configure virtual IP addresses to achieve high availability. Next, let's take a look at how to set virtual IP addresses in Ubuntu12.04. First, let's take a look at my Ubuntu IP address: www.2cto.com [plain] chenshu @ sloop2 :~ $ Ifconfigeth0... configure the virtual IP address in Ubuntu. the purpose of configuring the virtual IP address is to achieve high availability. Next, let's take a look at how to set the virtual IP address in Ubuntu12.04. First, let's take a look at my Ubuntu
A sub-station of Youku has a command execution vulnerability that can be SHELL and can penetrate through the Intranet.
Target: http://channel.3g.youku.com/ykmks/login.doWhoami: rootWebPath: /opt/www/ykmks/webapps/ykmksOS.Name: LinuxOS.Version: 2.6.18-194.el5Java.Home: /opt/jdk/jreJava.Version: 1.6.0_13OS.arch: amd64User.Name: rootUser.Home: /rootUser.Dir: /opt/www/ykmks/webapps/ykmksJava.Class.Path: /opt/tomcat/bin/bootstrap.jarJava.IO.Tmpdir: /opt/tomcat/temp
eth0 Link encap:Ethernet H
Enumerate f (N) ... Check if f (f (n) *k) equals the value of the enumeration.1#include 2#include 3#include 4#include 5 #definell Long Long6 using namespacestd;7 Const intmaxn=50233;8 intI,ans;9 ll A,b,k;Ten One intRaCharRx; AInlineintRead () { -Rx=getchar (), ra=0; - while(rx'0'|| Rx>'9') rx=GetChar (); the
Stupid greedy.Discover a duang that only occurs two times ... So, after sorting, throw it in the middle.Mom, the home page is all Pascal players.1#include 2#include 3#include 4#include 5 #definell Long Long6 using namespacestd;7 Const intmaxn=303;8 intA[MAXN];9 intI,j,k,n,m,ans;Ten One intRaCharRx; AInlineintRead () { -Rx=getchar (), ra=0; - while(rx'0'|| Rx
Mark the points to indicate that the points passing through the path from the root to the point are +1Finally count the number of marks in the tree.1#include 2#include 3#include 4 using namespacestd;5 Const intmaxn=100233;6 structzs{intToo,pre;} e[maxn1];intTOT,LAST[MAXN];7 intADD[MAXN],BEL[MAXN],FA[MAXN],SZ[MAXN],DEP[MAXN];8 intI,j,k,n,m,ans;9 Ten One intRaCharRx; AInlineintRead () { -Rx=getchar (), ra=0; - while(
At first I thought it was a bit of power and I never wrote Qaq.After reading the question found is the edge right ...Path xor of U to V and = ROOT to u path xor and xor or root to V path XORJust take a trie or a line tree and record it.1#include 2#include 3#include 4#include 5 #definell Long Long6 using namespacestd;7 Const intmaxn=100233;8 structzs{intToo,pre,dis;} e[maxn1];intTOT,LAST[MAXN];9 intA[MAXN],VAL[MAXN];Ten intch[maxn* +][2],tt; One intI,j,k,n,m,ans; A BOOLs[ -]; - - intRaCharRx; t
The surface cannot look straight into the series.The line of the tree.The operations involved are: Interval assignment of 0, calculate the number of 1 in the interval, interval assignment is 1, the maximum number of consecutive 1 in the interval.1#include 2#include 3#include 4 using namespacestd;5 Const intmaxn=200233,mxnode=maxn1;6 intLc[mxnode],rc[mxnode],sz[mxnode],num0[mxnode],mxl0[mxnode],mxr0[mxnode],mx0[mxnode],tot;7 intTag[mxnode];8 intI,j,k,n,m,l,r,mxr,num,ans;9 BOOLFirst ;Ten One int
At first, the face was crazy.Later I thought of maintaining a left and right hand two pointers L and R. Represents the different kinds of digital It is clear that the leftmost, legal L increases with R and does not decrease.By the way discretization, remember the number of different kinds of numbers to calculate the answer.Time complexity O (n)1#include 2#include 3#include 4#include 5 using namespacestd;6 Const intmaxn=1e5+233;7 structzs{intV,id;} A[MAXN];8 intMP[MAXN],SM[MAXN];9 inti,j,n,m,ans,
Compared to the egg, we can dye a background and dye something else on the background.By CCZ Master of the puzzle can get. The maximum node depth in the spanning tree is the number of times it takes to indent the link in the same color in the target State, and to raise the point.If the maximum depth is white, remember-1.1#include 2#include 3#include 4#include 5 using namespacestd;6 Const intmaxn=2523, xx[4]={1,-1,0,0},yy[4]={0,0,1,-1};7 structzs{8 intToo,pre;9}e[123333];intTOT,LAST[MAXN];Ten
left and right endpoint l,r; Find two thirds points m1,m2 (lThis problem, first, the relationship is very easy to find, found to be a single-peak function, then three points to find the most value canBut here's the three-point set of three points, also very good understandingFor the outer three m1,m2, if the size of the comparison, it is necessary to make three points inside to determine, this is three points set three pointsCode#include #include#include#include#includeusing namespacestd;intRea
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