linx 1010

Linx monitoring and sharing-disk space monitoring + email reminder, linx -- Share the first monitoring script. mount point disk space usage monitoring: If the used space reaches 75%, an alarm is triggered and an alarm email is sent. Install Netease's email client on your mobile phone to receive real-time notifications. For mail configuration, see the previous article: http://blog.csdn.net/rookie_ceo/article

; $mailtmp echo "[$DT] Server: $IP disk: ${arr[$i]} usage has reached $warn_pct%, please Timely processing. "#>>" $performance _path "/performance_disk_$ (date +%y%m%d). Log echo" [$DT] Server: $IP disk: ${arr[$i]} usage Have reached $warn_pct%, please deal with it in time. "GT;GT $mailtmp pp=${tt///\\n} echo-e $pp >> $mailtmp dt= ' date + '%y-%m-% D%h:%m:%s "' Echo-e" \nbest wishes! \ n------------------------------------\nca system Monitor \n$dt ">> $mailtmp cat $mailtmp | Mailx-s "$warn _name

, starting to execute the process, and so forth. 2, independent of each other, the memory is isolated using virtual memory, each process has a private address space, as if our program exclusive and use of memory.V. Process life cycle for a program, the process has three states: run, Hang, die. 1, set up a task (create a process) 2, ready to run (waiting for the CPU to allocate time slices) 3, is running (in the actual running state, if not allocated to the actual CPU chip, continue to wait) 4, s

Linx monitoring sharing-Key Process Monitoring (mysqld) + email reminder, linxmysqld Key Process Monitoring Mysqld: The process exits and an alarm is triggered. If you do not exit, you will be notified of monitoring mysqld (cpu (> = 100%) | mem (> = 80%) and the mysql status information will be collected and analyzed; number of connections (> = max_connections * 0.8) start to remind; number of opened handles (> = 8192*0.8 ?) Start reminder. To be impr

Android system development (8)-basic concepts of linx Processes1. in the traditional sense, the proc file system is used to store information on Block devices. The/proc directory is a virtual file system, where the data is stored in the memory, therefore, this directory does not occupy any hard disk space. It mainly includes the following system information: memory management system process features Data File System Device Driver System Bus Power Mana

Method to mount another Linx Server File System on A Linux Server objective: to access the file system specified on server A on server B, configure Server A to edit/etc/exports and add: /home 192.168.1.1 (rw) # The IP address is the address of server B. The directory is the directory to be shared and nfs service is started:/etc/init. d/nfs start can be mounted to server B with mount 192.168.1.1: /hoome/mnt enter the/mnt directory of server B and you w

alert value (CPU >= 100% or mem >=80%), please note. "#>>" $performance _path "/performance_$warn_name_$ (date +%y%m%d). Log echo" [$DT] Server: $IP mysqld process CPU And mem exceeds the alert value (CPU >= 100% or mem >=80%), please note. ">> $mailtmp echo-e" [$DT] $p 1 ">> $mailtmp cat $infotmp >> $ma Iltmp dt= ' date + '%y-%m-%d%h:%m:%s '' Echo-e ' \nbest wishes! \ n------------------------------------\nca system Monitor \n$dt ">> $mailtmp cat $mailtmp | Mailx-s "$warn _name[Alarm]: server

1010. One-element polynomial derivation (25) time limit of MS Memory Limit 32000 KB Code length limit 8000 B The Standard of the Judgment procedure The design function asks for the derivative of the unary polynomial. Input format: Enter the polynomial non 0 coefficients and indices (integers with an absolute value of not more than 1000) in an exponential degradation manner. Numbers are separated by spaces. Output format: The coefficients and indices

Bzoj 1010 The second slope optimization has a new understanding of the slope optimization. Happy (Foggy The question is actually about slope optimization. Slope optimization. Slope optimization. Okay, that's a naked question. The node that we selected when we considered updating f[i] If K is more superior than J Then there are: f[j]+ (j-i+sum[i]-sum[j]-l) ^2 f[j]+ (Sum[i]-sum[j]) ^2-2* (i-j+l) * (Sum[i]-sum[j]) + (i-j+l) ^2 Make t1=i-j+l,t2=i-k+

[bzoj]1010 Test Instructions: Professor P was going to see the Olympics, but he couldn't give up his toys, so he decided to ship all the toys to Beijing. He uses his own compressor to pressIt can turn any item into a pile and then into a special one-dimensional container. Professor P has an N-N toy numbered 1...N 1...N, and the first toy passesAfter compression becomes a one-dimensional ci ci. For ease of finishing, Professor P requires that the num

Timus 1010. The discrete function requires that the given discrete function be used to locate the greatest dip in the line of the corresponding point in the Cartesian coordinate system. 1010. discrete function Time Limit: 1.0 secondMemory limit: 16 MBThere is a discrete function. It is specified for integer arguments from 1 N(2 ≤ N ≤ 100000). Each value of the function is longint (signed long in C ++ ). Yo

1010. Radix (25) Time Limit 400 ms memory limit 32000 kb code length limit 16000 B discriminant program standard author Chen, Yue Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is "yes", if 6 is a decimal number and 110 is a binary number. Now for any pair of positive integers N1 and N2, your task is to find the radix of one number while that of the other is given. Input specification: Each inp

HDU-1010-Tempter of the Bone Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 1010 I have been tossing this question for a long time. I wrote BFS at the beginning, but it was always WA. Later I found that this question was required to arrive at a given point in time. It was neither early nor late, BFS finds the shortest time and does not answer the question each time. Therefore, it switches to DFS. The pruni

ERROR 1010 (HY000): Error dropping database (can't rmdir '. \ test \', errno: 17), hy000errno Mysql5.5.40 has been deployed in the production system. Due to development requirements, dump the data in the production database and import it to the test environment created by the development! After the test is completed, delete the temporary data as needed, Delete database directly Mysql> drop database test; ERROR: Error

After a day, I felt like I was not doing anything. I wanted to make up for my questions. I was thinking about what to do. My roommate said I 'd try Pat 1010. I skipped this question, it seems that this pass rate (0.07) is a bit exaggerated. The question is "know one number". Calculate the base number of the other number to make the two numbers equal. Naturally, considering that binary search is used to determine the base number, numbers represent [0-9

dp/slope optimization According to the description of the topic it is easy to set out the Regulation equation: $$ f[i]=min\{f[j]+ (s[i]-s[j]+i-j-1-l) ^2 \}$$where $ $s [i]=\sum_{k=1}^{i} C[k] $$And $x$ is the $s[i]-s[j]+i-j-1$.This $x$ 's expression is really not good-looking, we are easy to find that $i-j$ can actually be combined with $s[i]-s[j]$, even if $c [i]=c[i]+1$, then $s[i]=\sum_{k=1}^{i} (c[i]+1) =\sum_{k=1}^{i}c [I]+i $, so $x=s[i]-s[j]-1$. Then merge that $-1$ with $l$, that i

1010.Zero EscapeThe number root of a number is only related to the value after its mod~9 m o d 9, as long as the backpack can complete the calculation of the personnel assignment. Concrete Proof: Digital root =\sum_{i=0}^{w}a_i∑?I=0?W??A?I??, Number =\sum_{i=0}^{w}10^i*a_i∑?I=0?w?? 10? I?? ∗a? I?? Number-Number root =\SUM_{I=0}^{W} (10^i-1) *a_i∑?I=0?W??(10?I??−1)∗A? i ??, this number in modmod99 domain 0 Time complexity o

determine whether there is a result, so that the DFS steps can be reduced; - return ; the } * } $ Panax Notoginseng } - intMain () the { + while(cin>>n>>m>>tn!=0|| m!=0|| t!=0) A { thetm=0; + for(intI=0; i) - { $ for(intj=0; j) $ { -Cin>>A[i][j]; - if(a[i][j]=='S') the { -p=i;q=J;Wuyi } the if(a[i][j]=='D') - { Wudx=i;dy=J

Starting from S to stopping D, and arriving at t at the specified time (not early or late ). Note that each '.' can only go once. Analysis: DFS, but common DFS and TL are used, so pruning is required. We can think that if we can arrive at, but there are other points around the D point that we can step on, and we can observe that, from a point (not a D point) if the distance from point D (ABS (X-dx) + ABS (Y-dy) is an odd number, the odd number step is required, and the even number is the even n

intlen=t.length; - intCout=0; -string[]num=NewString[2]; theNum[0]= ""; -Num[1]= "";//Be sure to initialize to an empty string, otherwise add a number, such as 1, will become null1; - for(inti=0;i){ - if(T[i].equals ("+")){ +cout++; - Continue; + } A Else if(T[i].equals ("=")){ at Continue; - } - intA=Ht.get (T[i]); -num[cout]=num[cout]+A;