Generally speaking, they are divided into the following categories of application systems:
System Management Platform (SMS)
Clinical Information System (CIS)
Patient Administration System (PAS)
Clinical Workstation System (CWS)
Management Information System (MIS)
Decision Support System (DSS)
Knowledge Management System (KMS)
Public **** Application Services (CAS)
Collaborative Interactive Support Services (TES)
There are also hospitals that are divided into modules by each application system, and the common ones are:
HIS (Hospital Information System (Hospital Information System)
CIS (Clinical Information System) Clinical Information System
LIS (Laboratory Information System) Laboratory (Laboratory) Information System
PACS (Picture Archiving and Communication Systems) Picture Archiving and Communication Systems (PACS) is a system that allows the user to access information from the hospital to the laboratory. Archiving and Communication Systems (PACS) Picture Archiving and Communication Systems (PACS)
RIS (Radiology Information System) Radiology Information Management System (RIMS)
EMR (Electronic Medical Record)
OA (OA) Clinical Information System (CIS)
LIS (Laboratory Information System) p>
OA (Office Automation)
CRM (Customer Relationship Management)
PEIS (Physical Examination Information System) Information System
ORIS (Operation Room Information System) Operating Room Information System
CCIS (Critical Care Information System) Critical Care Information System
WMIS (Wireless Medical Information System) Wireless Medical Information System (WMIS) is a system for managing medical information. WMIS (Wireless Medical Information System)
What are the five major applications of computer networks?The power of the Internet has greatly facilitated the development and popularization of the Internet, which enables a person who has never used a computer before to learn to browse the rich and colorful multimedia information on the Internet in a few minutes. You can go from one website to another, from one country to another, just by clicking on the relevant topics and photos with your mouse, and you can easily roam around the world while sitting at home.
E-mail (E-MAIL): is a combination of computer and communications products, mainly used for the exchange of electronic mail between computer users. The sender of the letter in the computer to enter the content of the letter file, deposited in the sender's mailbox, with the help of communication networks will be placed in the sender's mailbox in the mailbox to the recipient of the letter in the mailbox, when the recipient opens their own computers, with the help of the communication network will be able to open the recipient's mailbox to read the letters sent from the sender. Among the many applications of the Internet, e-mail is undoubtedly one of the most popular. Whether you live at the end of the world or at the tip of the ocean, we can easily communicate with each other through the lovely "E-Mail". Oh! It's my brother's birthday across the ocean, so I quickly send him an E-MAIL card with a paragraph about my family's wishes and expectations, and it flies to him in just a few minutes. In retrospect, the financial account, spent less than a few cents, in the future to contact with my brother no longer need to worry about money! After you have an e-mail address, you can also use the Internet's "news subscription" service. We will be the latest domestic and international news, public service information, entertainment information, computer network industry information, financial and securities information, hot topics and other rich content is classified collection, organization, processing, according to your specified subscription requirements, regularly delivered to your mailbox. With an electronic mailbox, you have a competent secretary to ensure that you keep abreast of the times!
File Transfer (FTP): Many applications, games, songs, and other files on the Internet are ****enabled and can be downloaded for free, so all you need to do is log in to the relevant website and use the FTP protocol to download the files you need.
USENET News: USENET is a worldwide electronic bulletin board for releasing announcements, news, and articles of all kinds for everyone to use. Each of USENET's forums is also known as a newsgroup, and just like a newspaper, each contribution is viewed as a single article that can be read by everyone, and each person who reads it can comment on it according to his or her own point of view.
Remote access (TELNET): Using TELNET, a user on the INTERNET can simulate his or her own computer as a terminal on a remote computer, and then execute programs on it, or use its software and hardware resources.
What are the computer applicationsI. System software
1. Operating systems
2. Programming language design
3. Language processing programs
4. Database management programs
5. System auxiliary processing programs
II.
2, information management software
3, auxiliary design software
4, real-time control software
I now probably know so much, I hope to help you!
What does a computer application system consist of?Generally by the computer hardware system, system software, application software components. The basic computer hardware system consists of operators and controllers, storage, peripheral interfaces and peripheral devices. System software includes operating systems, compiled programs, database management systems, and various high-level languages. Application software consists of general-purpose support software and various application software packages.
The five systems of a computer?Five systems?
The five systems?
The computer hardware system consists of an operator, a controller, a memory, an input device, and an output device
Programming questions on computer applications
Solve: MOV A,R1
MOV R0,A
(2) The content of the external RAM 20H cell is sent to R0. MOV R0,A
(3) The content of the external RAM 20H cell is sent to the internal RAM 20H cell.
Solution: MOV R0,#20H
MOVX A,@R0
MOV @R0,A
(4) The content of the external RAM 1000H cell to the internal RAM 20H cell.
Solution: MOV DPTR,#1000H
MOVX A,@DPTR
MOV 20H,A
(5) The content of the external ROM 2000H cell is sent to the R0.
Solution: MOV DPTR,#2000H
CLR A
MOVC A MOV R0,A
(6) The content of the external RAM 2000H cell is sent to the internal RAM 20H cell.
Solution: MOV DPTR,#2000H
CLR A
MOVC A,@A+DPTR
MOV 20,A
(7) The content of the external RAM 2000H cell is sent to the external RAM 20H cell.
Solution: MOV DPTR,#2000H
CLR A
MOVC A,@A+DPTR
MOV R0, #20H
MOV @R0,A
2-2 It is known that A = 7AH, R0 = 30H, (30H) = A5H, PSW = 81H. ask The result of executing each of the following instructions (each instruction participates in the operation with the information specified in the question).
(1) XCH A,R0 A=30H, R0=7AH, P=0
(2) XCH A,30H A=A5H, (30H)=7AH, P=0
(3) XCH A,@R0 A=A5H, (30H)=7AH, P=0
(4) XCHD A,@R0 A= AAH Cy=0 OV=1 P=0
(9) ADDC A,30H A=20H Cy=1 OV=1 P=1
(10) SUBB A,30H A=D4H Cy=1 OV=1 P=0
(11) SUBB A,#30H A=49H Cy=0 OV=0 P=1
( 12) DA A A=80H Cy=1 P=1
(13) RL A A=F4H Cy=1 P=1
(14) RLC A A=F5H Cy=0 P=0
(15) CJNE A,#30H,00 A=7AH Cy=0 P=1
(16) CJNE A, #30H. 00 A=7AH Cy=1 P=1
2-3 Let the content of the 30H cell of the internal RAM be 40H, i.e., (30H) = 40H, and also know that (40H) = 10H, (10H) = 00H, and the port P1 = CAH, and ask about the content of each of the relevant register cells, storages, and ports after executing the following instructions (i.e., R0, R1, A, B, P1, 40H, 40H, 30H, and 10H cells). 30H and 10H cells).
MOV R0,#30H
MOV A,@R0
MOV R1,A
MOV B,@R1
MOV @R1,P1
MOV P2,P1
MOV 10H,#30H
MOV 30H,10H
Solution: The result after execution of each instruction is shown in the following order: R0=30H, A=40H, R1=40H, B=10H, (40H)=CAH, P2=CAH, (10H)=20H, (30H)=20H.
From this the contents of each cell after execution are shown as follows:
R0=30H, R1=40H, A=40H, B= 10H, P1=CAH, (40H)=CAH, (30H)=20H, (10H)=20H.
2-4 Let R0=17H, A=83H, (17H)=34H, and ask that after the execution of the following instruction, A=?
ANL A,#17H
ORL 17H,A
XRL A,@R0
CPL A
SOLUTION: This question is to familiarize yourself with the logical operation instructions. The execution result of each instruction is as follows:
A←83H∧17H=10000011∧00010111=03H
(17H) ←34H∨03H=00110100∨00000011=37H
A←03H⊕37H=00000011⊕00110111=34H
A←A=00110100=CBH
So after program execution, A=CBH
2-5 Try to write a program to store the contents of three consecutive cells 20H, 21H and 22H of internal RAM into cells 2FH, 2EH and 2DH sequentially.
Solution: This question can be done directly with a transfer instruction:
MOV 2FH,20H
MOV 2EH,21H
MOV 2DH,22H
Of course, it can also be done with a loopback program:
MOV R0,#20H ;first address of source data area
MOV R1,#2FH ;Destination data area first address
MOV R3,#03H ;Data block length
LOOP: MOV A,@R0
MOV @R1,A
INC R0
DEC R1
DJNZ R2,LOOP
SJMP $
2-6 Write a program to subtract two 16-bit numbers: 6F5DH-13B4H. The result is stored in cells 30H and 31H of the internal RAM, with 30H storing the lower 8 bits of the difference.
Solution: The program is as follows:
MOV A,#5DH
CLR C
SUBB A,#0B4H
MOV 30H,A
MOV A,#6FH
SUBB A,#13H
MOV 31H,A
SJMP $
Since the subtraction instruction in the MCS=51 instruction system can only accomplish subtraction with borrowing, it is generally necessary to clear 0 to the incoming bit Cy when starting to subtract.
2-7 Write a program to transfer to the LABEL storage unit if the contents of the accumulator A satisfy the following conditions respectively. Let the storage in A be an unsigned number.
1. A 10
SOLUTION: This problem can be accomplished directly with the Comparison Conditional Transfer instruction, and the relevant program statement is as follows:
CJNE A, #0AH, 00H
JNG LABEL
The Comparison Conditional Transfer instruction locates the value of Cy based on the result of A-0AH. As long as A 10, the subtraction does not need to borrow, i.e., Cy = 0. Another "JNC" instruction determines whether or not to move to LABEL, and the offset 00H indicates that the next statement is executed first, regardless of the result of the comparison.
2, A & gt; 10
Solution: If this is still compared with 0AH, there is no way to distinguish between A & gt; 10 (to be transferred) and A = 10 (not transferred). Of course, you can also add other instructions (such as accumulator judgment zero conditional transfer statement, etc.) to complete the desired function, but it will increase the number of statements, the program is not concise enough. How about comparing directly with 0BH (decimal number 11), after comparing A 11 all make Cy = 0, equivalent to A & gt; 10 to make Cy = 0. Still use two statements to complete the required function
CJNE A, #0BH, 00H
JNG LABEL
3, A 10
Solution: this case is similar to the question (2), can be directly with 11 for the transfer of zero condition statement, etc., but will make the program more concise. ) is similar and can be compared directly with 11, but this time it is transferred when Cy = 1.
CJNE A, #0BH, 00H
JC LABEL
4, A<10
Solution: this is similar to question (1), which can be compared directly with 10:
CJNE A, #0AH, 00H
JC LABEL
2-8 It is known that SP = 25H, PC = 2345H, (24H) = 12H, (25H) = 34H, (26H) = 56H. ask at this point after executing the RET instruction, SP =?PC =?
SOLUTION: In the MCS-51 system, the stack grows upward, so the stack indicator is reduced during the out-of-stack operation. Execute the RET instruction as follows:
PC15-8 (SP) i.e. PC15-8 = (25H) = 34H
SP SP-1 i.e. SP = 24H
PC7-0 (SP) i.e. PC7-0 = (24H) = 12H
SP SP-1 i.e. SP = 23H
The result is: SP = 23H, PC = 3412H
2-9 If SP = 25H, PC = 2345H, and the address where the marker LABEL is located is 3456H, ask what happens to the stacking indicator and the contents of the stack after executing the long call instruction LCALL LABEL, and what is the value of PC equal to?
Solution: The long call instruction accomplishes two functions: it stacks the address of the next instruction into the stack, and it assigns the entry address of the called subroutine to the program counter PC. the process is as follows:
PC PC+3 so PC = 2345H+3 = 2348H
SP SP+1 so SP = 26H
(SP) PC7-0 So (26H) = 48H
SP SP+1 So SP = 27H
(SP) PC15-8 So (27H) = 23H
PC LABEL So PC = 3456H
The result should be SP = 27H, (26H) = 48H, (27H) = 23H and PC = 3456H.
2-10 Can the LCALL instruction in the previous question be directly replaced with an ACALL instruction, and why? If the ACALL instruction is used, what is the range of addresses that can be called?
SOLUTION: The ACALL instruction can only be called in the 2K-bit byte range. Specifically, it is required that after PC+2, the high 5 bits of PC are the same as the high 5 bits of the call address LABEL. Or the PC value and the call address after executing ACALL should be within the same page (2K bit tuple per page address range). In this question:
PC+2 = 2345+2 = 2347H, high 5 bits 00100
And LABEL = 3456H, high 5 bits 00110
The two are not equal, not in the same page, so can not be directly converted to ACALL instruction
If you use the ACALL instruction, the minimum value of the callable address is
2-11 Try to write a program to query whether there is 0AAH in cells 20H-50H of the internal RAM. If there is, set cell 51H to 01H; if it is not found; set cell 51H to 0.
Solution: this problem can be solved in two basic ways. One is to calculate the length of the block of information, and then loop back and compare, the program is as follows:
CLR C
MOV R0, # 20H; R0 for the first address of the block of information
MOV A, # 50H; block of information at the end of the address
SUBB A, R0
INC A
MOV R1, R1 to store the length of the block of information
MOV R1, R1 to store the length of the block of information. A ；R1存资料块长度
LOOP1：CJNE @R0，#0AAH，LOOP2
MOV 51H，#01H ；找到置(51H)=01H
SJMP LOOP3
LOOP2： ING R0
DJNZ R1，LOOP1
MOV 51H, #00H; not found, set (51H) = 01H
LOOP3: SJMP $
The second solution is to directly use the compare condition transfer instruction to determine whether the last comparison has been completed. This is one of the features of the MCS-51 instruction system. The program is as follows:
MOV R0, #20H ；R0存资料块首地址
LOOP1: CJNE @R0, #0AAH, LOOP2 ；比较
MOV 51H, #01H ；找到，置(51H)=01H
SJMP LOOP3
LOOP2: ING R0
DJNZ R0, #51H, LOOP1 ; circle back if not finished comparing
MOV 51H, #00H ; if not found, set (51H) = 01H
LOOP3: SJMP $
2-12 Try to write a program to query the number of 00H cells that appear in cells 20H - 50H of internal RAM. 50H cells, and store the result of the query in cell 51H.
Solution: The solution to this problem is similar to the previous one, and the second solution is as follows:
MOV R0, #1FH
MOV R1, #00H; R1 is used as a counter
LOOP1: ING R0
CJNE @R0, #00H, LOOP2
INC R1; when a 00H is found, the counter is set to 00H. ; find a 00H, counter plus 1
LOOP2: CJNE @R0, #50H, LOOP1 ; circle back if not finished comparing
MOV 51H, R1 ; store the result of comparison
SJMP $
2-13 Try to write a program to find the absolute value of the difference between the number of the two complements of the cells 20H and 21H, i.e. . The result is retained in A. If the result is overflowed, set unit 22H to 0FFH, otherwise, 22H should be 00H.
Solution: Subtraction instruction can be used directly to complete the subtraction, and then based on the sign of the difference to find the absolute value, if the difference is a positive number, the difference is the absolute value of the request. If the difference is negative, then the absolute value can be obtained after the inverse plus one. Of course, after subtraction, we must first determine whether the overflow. If it overflows, you just need to set the overflow flag: make (22H) = 0FFH. The program is as follows:
CLR C
MOV 22H, #0FFH
MOV A, 20H
SUBB A, 21H
JB OV, NEXT
JNB ACC.7, NEXT1; if the difference is positive, then transfer
CPL A; if the difference is negative, then seek to complement the absolute value
JNB ACC.7, NEXT1; the difference is positive, then transfer
CPL A; difference is negative, then seek to complement the absolute value
CPL A; difference is negative, then seek to complement the absolute value
ING A
NEXT1: MOV 22H., #00H
NEXT: SJMP $
2-14 Try to write a program to find the absolute value of the difference between the three complements of the cells 20H, 21H, and 22H, that is . The result is retained in A. If the operation overflows, set cell 23H to 0FFH; otherwise, set cell 23H to 00H.
Solution: The solution to this problem is similar to the previous one. Only after each subtraction to determine whether the overflow, rather than after two subtractions to determine the overflow. Because it is possible that after the first subtraction has overflowed, and then after another subtraction there is no overflow, and then the result is actually incorrect. For example, 64 - (-64) - (-1) = 129. the result is already overflow, but the two-step operation, first by
64 - (-64) = 01000000 - 11000000 = 10000000
This time there is already an overflow, OV = 1. If you do the second subtraction, the result is:
10000000 - 11111111 = 10000000 -11111111 = 10000001
The overflow flag is instead cleared (because it is not possible to overflow a negative number by subtracting it), but the actual result is not correct. So, you need to check the overflow flag after every subtraction operation. The program is as follows:
CLR C
MOV 23H, #0FFH ；First set the overflow flag
MOV A, 20H
SUBB A, 21H ；First subtraction
JB OV, NEXT ；Overflow is the end of the process
CLR C ；Preparing to subtract for the second time
SUBB A, 22H ；Second subtraction
JB OV, NEXT ；Overflow ends
JNB ACC.7, NEXT1 ；Difference is positive then transfer
CPL A ；Seek the inverse and add one
ING A
NEXT1：MOV 23H.,#00H
NEXT: SJMP $
2-15 There is a 16-bit binary number with the high 8 bits stored in cell 21H and the low 8 bits stored in cell 20H. Ask: (1) What logical function is accomplished by executing the following program segment? (2) Can the XCH instruction be replaced by the MOV instruction without changing the logical function of the program? Write the corresponding program segment. (3) Are the results of these two program segments identical? What is the difference?
CLR C
XCH A, 21H
RRC A
XCH A, 21H
XCH A, 20H
RRC A
XCH A, 20H
Solve: (1) The program performs the logical function of shifting the 16-bit numbers of 21H and 20H by one bit. binary numbers to the right by one bit and make the highest bit 0, generally known as logical right shift by one bit.
(2) You can use the MOV instruction to modify the program segment as follows:
CLR C
MOV A, 21H
RRC A
MOV 21H, A
MOV A, 20H
RRC A
MOV 20H, A
(3) There is a difference. ) makes a difference. Although both functions are the same in terms of the 16-bit right shift function, the content of accumulator A remains at the original value after the first program segment is executed, while the value in A is changed after the second program segment is executed.
2-16 Analyze the logical function of the following program segment. If you want to replace one of the swap instructions with a MOV instruction without changing the logical function of the program, how should you modify it? Let two four-bit binary numbers be stored in 20H.
MOV R1,#20H
XCHD A,@R1
ADD A,#01H
XCHD A,@R1
XCH A,@R1
ADD A,#01H
XCH A,@R1
Solve the program. cell by adding 1 to each of the two four binary digits and discarding any rounding that may occur. For example, if (20H)=6AH, the bit (20H)=7BH after the program is executed, and if (20H)=9FH, the bit (20H)=A0H after the program is executed.
There are several modifications that can be made to accomplish the same function without using the swap instruction. Here is just one example:
MOV R1,#20H
XCHD A,@R1
ADD A,#01H ;lower 4 bits plus 1
JB AC,NEXT ;judge the lower 4 bits have no rounding
ADD A,#01H ;higher 4 bits plus 1
NEXT: MOV @R1,A ;Send back to unit 20H
2-17 Analyze the execution result of the following program segment. What would be the difference in the result if the DA A instruction in it was canceled?
CLR C
MOV 20H, #99H
MOV A, 20H
ADD A, #01H
DA A
MOV 20H, A
SOLUTION: The result of execution of the program is that A = 00H, Cy = 1, and (20H) = 00H.
When the DAA instruction is canceled, the result is A = 9AH, Cy = 0, (20H) = 9AH.
2-18 In the following two program segments, two hexadecimal digits are stored in R2. How many times do each of these two segments perform a loopback?
MOV R2, #60H
LOOP: ......
......
DJNZ R2, LOOP
SJMP $
MOV R2
MOV R2, #60H
LOOP: ......
What is the security of the computer application systemIt is the security of the computer system (hardware and software) (anti-intrusion, anti-jamming, data protection, etc.)
Please teach the direction of the design of the computer application system (embedded)I'm an embedded developer, and the majority of the embedded development nowadays are in the embedded field. Now the embedded, the vast majority of the development of the linux platform, so you first need to learn linux, to recommend a few books, these are the first time I learned to read, very good.
The book on linux programming: "Linux Programming (Original Book 2nd Edition)", "[Linux English Original Book Series].WILEY-Beginning_Linux_Programming_Third_Edition", "LINUX Programming White Paper".
The linux kernel: The Do-It-Yourself Homework System, The Linux Kernel and Programming.
What are the latest computer human-computer interaction applications?Google seems to have something on this.
What are the three traditional application areas of computersThe main application areas of computers include the following six areas:
First, information management
Information management is based on database management system, to assist managers to improve decision-making, improve the operation of the computer technology of the strategy. Information processing specifically includes data collection, storage, processing, classification, sorting, retrieval and release of a series of work. Information processing has become the main task of contemporary computer. It is the foundation of modern management. According to statistics, more than 80% of the computer is mainly used in information management, the leading direction of computer applications. Information management has been widely used with office automation, enterprise computer-aided management and decision-making, intelligence retrieval, libraries, film and television animation design, accounting computerization and other industries.
The application of computers has penetrated into all areas of society, is increasingly changing the traditional way of work, study and life, and promote the society of scientific computing
Scientific computing is the earliest application of computers, refers to the use of computers to complete the numerical computation of scientific research and engineering technology issues raised. In modern scientific and technological work, the task of scientific computing is a large number and complex. The use of computers with high computing speed, large storage capacity and the ability of continuous computing, can solve a variety of scientific computing problems that can not be completed manually. For example, engineering design, earthquake prediction, weather forecasting, rocket launching, etc. need to be undertaken by the computer huge and complex calculations.
Second, process control
Process control is the use of computers to collect real-time data, analyze the information, according to the optimal value of the control object quickly and automatically adjust or automatically control. The use of computers for process control, not only can greatly improve the automation level of control, but also improve the timeliness and accuracy of control, thereby improving labor conditions, improve production and qualification rate. Therefore, the computer process control has been widely used in machinery, metallurgy, petroleum, chemical, electric power and other sectors.
Third, auxiliary technology
Computer-aided technology, including CAD, CAM and CAI.
1, ComputerAidedDesign (ComputerAidedDesign, referred to as CAD)
Computer-aided design is the use of computer systems to assist the designer to carry out the engineering or product design, in order to achieve the best design results of a technology. CAD technology has been applied to aircraft design, ship design, architectural design, mechanical design, large-scale integrated circuit design. The use of computer-aided design can shorten the design time, improve efficiency, save manpower, material and financial resources, and more importantly, improve the quality of design.
2, Computer Aided Manufacturing (ComputerAidedManufacturing, CAM)
Computer Aided Manufacturing is the use of computer systems for the processing of products to control the process, the input information is the parts of the process routes and engineering content, the output information is the tool's trajectory. The integration of CAD and CAM technology can realize the automation of the production of design products, this technology has been made into a computer-integrated manufacturing system. Some countries have CAD and computer-aided manufacturing (ComputerAidedManufacturing), computer-aided testing (ComputerAidedTest) and computer-aided engineering (ComputerAidedEngineering) to form an integrated system, so that the design, manufacturing, testing and management of organic Composition as a whole, the formation of a high degree of automation system, thus giving rise to automated production lines and "unmanned factories".
3, Computer Aided Instruction (ComputerAidedInstruction, referred to as CAI)
Computer-aided instruction is the use of computer systems for classroom teaching. Teaching courseware can be made with PowerPoint or Flash, etc. CAI can not only reduce the burden of teachers, but also teaching content vivid, realistic image, dynamic demonstration of experimental principles or operation process to stimulate students' interest in learning, improve the quality of teaching and learning for the cultivation of modern high-quality talent to provide an effective method.
Fourth, translation
In 1947, the United States mathematician, engineer Warren Weaver and British physicist, engineer Andrew Booth put forward a computer translation (referred to as "machine translation") of the idea of machine translation from this into the historical arena, and went through a long and winding road of development. Machine translation has been listed as one of the world's top ten scientific and technological problems in the 21st century. At the same time, machine translation technology also has a huge demand for applications.
Machine translation eliminates the barriers between different words and languages, which can be called a high-tech move for the benefit of mankind. However, the quality of machine translation has long been a problem, and is still far from the ideal goal. Chinese mathematician and linguist Prof. Zhou Haizhong thinks that it is impossible for machine translation to reach the level of "trust, attainment and elegance" when human beings have not yet understood how the brain carries out fuzzy recognition and logical judgment of language. I am afraid that this point of view out of the constraints on the quality of the translation of the bottleneck.
V. Multimedia Application
With the development of electronic technology, especially communication and computer technology, people have the ability to synthesize various media such as text, audio, video, animation, graphic and image, which constitutes a brand-new concept - "multimedia" (Multimedia). Multimedia" (Multimedia). In the fields of healthcare, education, business, banking, insurance, administration, military, industry, broadcasting, communication and publishing, the application of multimedia is developing rapidly.
Six, computer network
Computer network is composed of a number of independent and have the ability to exchange information computer interconnections, in order to realize the resource **** enjoy the system. The application of computers in the network so that communication between human beings across the time and space barriers. Computer network has become the material basis for human beings to establish an information society, which brings great convenience and speed to our work, such as the use of bank credit cards on a national scale, the use of train and airplane ticket system, and so on. You can browse, retrieve information, send and receive e-mails, read books and newspapers, play online games, shop for goods, participate in discussions on many issues, and realize remote medical services on the world's largest Internet network, the Inter.