[Software Designer] knowledge points and questions

Http://edu.csai.cn/zt/ztda/

2010, bottom
Http://training.51cto.com/art/201011/230700.htm (exam)
Http://training.51cto.com/art/201011/232662.htm (answer)

Underhttp://training.51cto.com/art/201011/231194.htm

Memory composition?
2 K * 4 memory chips constitute 16 K * 8 memory, how to address?
1. DMA
Direct Memory Access (direct storage of memory)
This is a high-speed data transmission operation that allows you to directly read and write data between external devices and memory, neither through the CPU nor requires CPU intervention. The entire data transmission operation is performed under the control of a "DMA controller. In addition to processing data at the beginning and end of data transmission, the CPU can perform other work during transmission. In this way, the CPU and input and output are in parallel most of the time. Therefore, the efficiency of the entire computer system is greatly improved.

2. Word Length is one of the main technical indicators of CPU
Indicates the number of binary digits that the CPU can process in parallel at a time.

3. The font length of a computer is 32 bits, and its memory capacity is 1 GB. What is the address range if the memory space is edited by words?
A computer has 32 characters (4 bytes );
The memory capacity is 1 GB, which can be edited by words, that is (1/4) G = 256 m.

256 m is 256*1024*1024;
Address lines required: 8, 10, 10, and 28 address lines.
The address has a total of 28-bit binary numbers, which are written in hexadecimal notation and are seven digits.

So the correct answer to this question is: A. 0000000 ~ Fffffff

4. ARP: Address Resolution Protocol

5. A simple undirected graph containing n vertices and e-edge. There are _ (6) _ zero elements in the storage structure of the adjacent matrix.
(6) A. E B .2e C. n2-e D. n2-2e

6. A binary tree with the minimum length of the weighted path is called the Harman tree, that is, the optimal binary tree.

7. node degree, number of subnodes owned by the node

8. Tree degree and graph degree (inbound degree)

9. pre-order traversal is also called root traversal and first-order traversal. It can be recorded as the left and right sides of the root.
In-order traversal, also called root traversal, the order is left subtree, root, right subtree
Post-order traversal, also known as post-root traversal. The traversal sequence is left subtree, right subtree, and root

10.
A linear table stored in sequence has 123 elements, which are divided into three parts according to the requirements of block search. If the index table uses the sequential Search Method to Determine the sub-block, and the sequential search method is also used to determine the sub-block, in the case of equal probability, the average length of successful multipart search is _ (11 )__.
The 123 elements are divided into three parts: A, B, and C, with 41 elements in each part.

For the elements in block A, the first step in the search process is to find block a first, and then find an element in block.

As it is a sequential search, only one step is required to find block.

Then find the specified element in block.
Because it is a sequential query, it takes two steps to find the first element.
And so on, it takes 41 steps to find 41st elements.

Therefore, the length of each element in block A is 2, 3, 4, · 42.

For Block B, the principle is the same, but finding Block B itself requires one more step than finding block A, because it is sequential search.
Therefore, the length of each element in Block B is 3, 4, 5, · 43.

Similarly, the length of each element in Block C is 4, 5, 6, · 44.

So the average search length is
2 + 3 + 3 +... + 42
+ 3 + 4 + 5 +... + 43
+ 4 + 5 + 6 +... + 44
Divide by the total number of elements 123
The final result is 23.

11.

Cbfffh (835583)-a4000h (671744) + 1 = 28000 H (163840)

By byte address => the size of a storage unit is 1 byte.

Both a4000h and cbfffh represent hexadecimal data, and h Represents hexadecimal data. Therefore, the total number of bytes from a4000h to cbfffh is cbfffh (835583)-a4000h (671744) + 1 = 28000 H (163840). In some equations, the ending number of h indicates the hexadecimal number, and the number in parentheses is the corresponding decimal number, therefore, they have a total of 163840 bytes in the middle, and 163840/1024 = 160, resulting in a total of 160*1024 bytes from a4000h to cbfffh.

Adding 1 to a large address is equal to cc000h, and then subtracting the large address from the small address is:
CC000H-A4000H = 28000 H (12-10 = 2, 12-4 = 8)
H stands for hexadecimal

In upper case, B Indicates byte, and in lower case, B Indicates bit. So, 1B = 8b

12.
The interrupt response time refers:
The time from sending the interrupt request to entering the Interrupt Processing

13.
Single instruction stream multi-data stream computer (SIMD) refers to the execution of one instruction simultaneously by one control component and multiple processing units. Therefore, SIMD executes the same command at the same time in synchronous mode.

14. First, let's take a look at the magnetic record principle. When magnetic recording data, the magnetic surface memory is read and written through the head. In the head coil, pulse current of a certain direction and size is passed in, and the magnet of the magnetic head is magnetized to establish a magnetic field of a certain direction and intensity. When the magnetic media passes through the head, it is magnetized from the gap of the head. Due to the residual magnetic effect, when the magnetic field disappears, the surface of the magnetic medium still has the residual magnetic field. The current direction in the coil is different, and the direction of the magnetic medium is different, which is used to represent "1" and "0 ". As the current changes and the relative movement between magnetic media heads, the binary information sequence can be converted into the magnetization unit sequence of the media surface. The reading process is a inverse process, that is, the recorded magnetization unit sequence is restored to the current pulse sequence. Therefore, it is not difficult to come to the answer: when a single head writes data to the magnetic coating of the disk, it is written in serial mode.

15.
The 64-bit cache uses a group-connected image. The size of the block is 128 bytes, and each 4 blocks is a group. If the master capacity is 4096 blocks and is edited by words, the primary storage address is (____) and the primary storage area number is.

Solution:

The addresses associated with the group are: area code, group code, block code, and intra-block address.

The size of each partition of the primary storage is equal to the size of the entire cache. Therefore, the number of partitions required for the primary storage is 4096/64 = 64. Because 26 = 64, 6 locations are required to represent the Partition Number.

Each 4 parts is a group, so the total number of groups is 64/4 = 16. Because 24 = 16, 4 digits are required to represent the Group number.

Each group contains four blocks. Therefore, two digits are required for the block number.

The block address consists of 128 bytes, 27 = 128, so the block must be expressed in seven bits.

Therefore, the number of digits of the primary address is 6 + 4 + 2 + 7 = 19.

Number of digits of the primary storage area code = 6

16. DFD --> Data Flow digoal Data Flow Diagram
: Arrow, indicating data flow;
0: circle or elliptic, indicating processing;
=: Parallel Bars (with one opening and one closed) indicate data storage;
□: Box, indicating the source or end point of the data.

17.
Path testing is a technology used to design test cases based on the path. It is often used in state transition testing.

The basic path test method is based on the control flow diagram of the program. By analyzing the loop complexity of the control structure, the basic executable path set is exported to design the test case method. The designed test case should ensure that each executable statement of the program in the test should be executed at least once.

18. project management tools
Gantt Chart: a chart that describes the progress of a project using graphs (especially bar charts. Each bar symbol represents a different meaning. For example, the bar symbol or color of a key task may be different from that of a Non-key task. The symbols of the summary task (activity or stage) may be different from those of other tasks.
PERT diagram: Uses flowcharts to represent the current dependencies of all tasks. PERT refers to the plan evaluation and review technology, which is a network diagram.
Project management tools generally support the PERT and Gantt diagrams. PERT is the plan evaluation and review technology. This technology applies the Network Method to the review and inspection of the work plan arrangement. Generally, an edge with an arrow is used to indicate an activity. The start and end nodes of the edge indicate the start event and end event of the activity. The length of the edge indicates the workload or duration of the activity, the order of each node reflects the time sequence constraints of each activity. The PERT network diagram can be used to find the key path and relaxation time, and adjust the activities and resource allocation in the plan. A Gantt chart is a two-dimensional crosstab chart that is widely used for schedule management of various engineering activities. The abscissa of an image is the timeline. Each activity is represented by a horizontal line segment. the abscissa value corresponding to its start and end points is the start and end time of the activity. Although new project management methods and technologies change the way people are used to working, it takes some time to learn and master new tools, however, automatic project management tools have many advantages over manual management.

19. The processing logic can be expressed in structured languages, decision tables, decision trees, and other forms. They can also be expressed in combination.

20. objectoriented (OO) is also called the rapid prototyping method.

21. To convert a Data Flow Diagram (DFD) into a software structure, you must first study the DFD type. Various software systems, regardless of the size and complexity of DFD, can be divided into conversion type and transaction type.
The transformed DFD is composed of input, transform, and output. The process of transforming data processing is generally divided into three steps: getting data, transforming data, and giving data, these three steps reflect the basic idea of transformed DFD. Transformation is the main processing of the system. The data stream at the transformation input end is the logical input of the system, and the output end is the logical output.
If a processing splits its input stream into many divergent data streams, forming many processing paths, and selects one path based on the input values for execution, the DFD of this feature is called a transactional data flow chart, and this processing is called a transaction processing center.

22. http://technet.microsoft.com/zh-cn/library/cc771298 (ws.10). aspx
The tunnel in VPN is formed by the tunnel protocol. There are three main tunnel protocols used by VPN: point-to-point tunnel protocol (PPTP), L2 tunnel protocol (L2TP), and IPSec.

23.
In a UNIX operating system, if the number of command parameters you type is 1, run the cat $1 command. If the number of command parameters you type is 2, run the cat> $2 <$1 command. Fill in the vacancy section of the shell program shown below.

Case (25) in
1) Cat $1 ;;
2) Cat >>2 2 <$1 ;;
*) Echo 'default... '
Esac

(25) A. $ B. $ @ C. $ # D. $ *

In a UNIX operating system, shell defines variables $, $ @, $ #, and $ * as follows:

$ Indicates the number of process IDs of the current command.

$ @ Is basically the same as $ *, but when escaped using double quotation marks, "$ @" can still be divided into multiple parameters, but "$ *" is merged into one parameter.

$ # Indicates the number of location parameters, excluding the command name.

$ * Indicates all location parameters, that is, equivalent to $1, $2, $3 ,...

The correct answer to question (25) is C.

24.
Process PA constantly writes data to the pipeline. Process Pb reads data from the pipeline and processes the data, as shown in figure 4. If the PV operation is used to achieve pipeline communication between the process Pa and Pb, and ensure the correctness of concurrent execution of the two processes, at least _ (26) _ is required )__.
 
Figure 4 Process Diagram

(26) A.1 semaphores. The initial values of semaphores are 0 and B. The initial values of semaphores are 0 and 1.
C.3 semaphores. The initial values of semaphores are 0, 0, and 1. d. 4 semaphores. The initial values of semaphores are 0, 0, 1, and 1.

25.
● In the diagram provided by UML, you can use _ (30) _ to model the logical database mode; __( 31) _ to model interfaces, classes, and collaborative behavior, the time sequence of object behavior is emphasized. __( 32) _ is used for system function modeling, and the control flow of objects is emphasized.
(30) A. use case diagram B. Component diagram C. Activity diagram D. Class Diagram
(31) A. Collaboration Diagram B. State Diagram C. Sequence diagram D. Object Diagram
(32) A. status diagram B. use case diagram C. Activity diagram D. Class Diagram
 
Reference answer:
(30) D (31) B (32) c

26
● In A Complete Binary Tree, the root sequence number is 1 ,__ (33) _ to determine whether the two contacts with the sequence number p and q are on the same layer.
(33) A. llog2p "= llog2q" B. log2q = log2q
C. llog2p "+ 1 = llog2q" D. llog2p "= + llog2q" + 1
 
Reference answer:
A

27.
Heap is defined as follows:
Sequence of n elements {k [1], K [2],……, K [N]} is called heap only when the following relationship is met.
K [I] <= K [2I] & K [I] <= K [2I + 1]
Or
K [I]> = K [2I] & L [I]> = K [2I + 1]
I = 1, 2, 3 ,...... Rounded down of n/2

Specifically, if we regard the one-dimensional array corresponding to the secondary sequence (I .e. the storage structure of the sequence using one-dimensional array) as a full binary tree, the meaning of the heap indicates that, the values of all non-terminal nodes in the Complete Binary Tree are not greater than (or not less than) The values of left and right child nodes.

28.
Binary sort tree is also called Binary Search Tree ). It is defined as a binary sorting tree, an empty tree, or a binary tree that meets the following requirements:
① If its left subtree is not empty, the values of all nodes on the left subtree are smaller than the values of the root node;
② If its right subtree is not empty, the value of all nodes on the right subtree is greater than the value of the root node;
③ The left and right sub-trees are a binary sorting tree.
Therefore, the binary tree is a binary tree that meets the requirements of the BST.

29.
A balanced binary tree is called a balanced binary tree. Its strict definition is as follows:
An empty tree is a balanced binary tree. If T is a non-empty Binary Tree, its left and right subtree are TL and TR, make HL and HR respectively the depth of left and right subtree. When and only when
① TL and TR are balanced binary trees;
② | HL-HR | ≤ 1;
T is a balanced binary tree.

30. Large top heaps
The heap top element is the largest, and its leaves are smaller than it. The same rule is also made for the leaves as the root.

31.
The length of a generalized table is the number of commas in the first brace.
We can see that there is only one element, namely (A, B, (), c), d), E, (f), g)
So the length is 1.
The depth is the number of braces, and the depth is 4

32.
● _ (38) _ is the interface standard for multimedia content description.
(38) A.MPEG-1 B .MPEG-2 C.MPEG-4 D.MPEG-7
Reference answer:
D

33.
● The formula for calculating the Uncompressed Digital audio data transmission rate is _ (39 )__.
(39) A. sampling frequency (HZ) × bit × number of channels × 1/8
B. sampling frequency (HZ) × Quantizing digits (BIT) × Channels
C. sampling frequency (HZ) × Quantizing digits (BIT) × 1/8
D. sampling frequency (HZ) × Number of quantifiers (BIT) × number of channels × 1/16
 
Reference answer:
B

34.
● The color space used by the color printer is _ (40 )__.
(40) A. RGB Color Space B. cmy Color Space C. YUV color space D. HSV Color Space
Reference answer:
B

35.
● The time redundancy information in MPEG videos can be compressed and encoded using the _ (41) _ method.
(41) A. Inter-Frame Prediction and transformation coding B. Hoffman coding and Motion Compensation
C. Transform encoding and stroke encoding D. Inter-Frame Prediction and Motion Compensation
 
Reference answer:
D

36.
Solution: Based on the function dependency set, all attributes are classified into four types.
1. L Class: All appear in the left half of the function dependency
2. R: All appear in the right half of the function dependency.
3. lr: It appears on both the left and right sides of the function dependency.
4. N: not included in function dependency
Possible candidate keys include L, LR, and N.
For Class L, find its closure. If it contains all attributes, it indicates that it is a candidate key and a unique candidate key.
For the LR class, find its closure. If it contains all the attributes, It is the candidate key. If it does not include, combine one of the attributes.
For N classes, add them directly to the candidate key.

37.
● Relational mode R (u, F), where u = {w, x, y, z}, F = {wx → y, W → X, X → Z, Y → w }. The candidate values of relational mode R are _ (45) _, and _ (46) _. They are lossless connections and maintain the decomposition of function dependencies.
(45) A. W and y B. WY C. wx D. WZ
(46) A. P = {r1 (WY), R2 (xz)} B. P = {r1 (WZ), R2 (xy )}
C. p = {r1 (wxy), R2 (xz)} D. P = {r1 (wx), R2 (YZ )}
 
Reference answer:
(45) A (46) c

38.
The OMT method is currently one of the most mature and practical methods. It models the system from three aspects. Each model reflects the characteristics of the system from one side. The three models are: object model, dynamic model, and functional model.

39.
● In the worst case, the lower bound of the computing time of a Sort Algorithm Based on keyword comparison is O (nlogn ). In the following sorting algorithm, the worst case is that the calculation time can reach O (nlogn) by _ (59) __. the design method used by this algorithm is _ (60) __.
(59) A. merge algorithm B. insert algorithm C. Select algorithm D. Bubble Algorithm
(60) A. Divide and conquer method B. Greedy method C. Dynamic Programming Method D. Backtracking Method
 
Reference answer:
(59) A (60)

40.
● The transmission medium specified in the Ethernet 100base-tx standard is _ (61 )__.
(61) A.3 class UTP B .5 class UTP C. Single-Mode Optical Fiber D. Multimode Optical Fiber
 
Reference answer:
B
41.
● Many network communications require multicast. Among the following options, the applications that do not use the multicast protocol are _ (62 )__. In IPv4, the _ (63) _ Class address is used as the multicast address.
(62) A. vod B. netmeeting C. Computers D. FTP
(63) A. a B. B c.d d. e

42.
● Make the twisted pair into a crossover line (one end is in the order of ELA/tia568a and the other end is in the order of ELA/tia568b). The two devices connected to the twisted pair can be _ (64 )__.
(64) A. Nic and nic B. Nic and switch
C. Ethernet ports of NICs and hubs D. Ethernet ports of switches and uplinks of lower-level switches
Reference answer:
A