Question bank brushing questions:
Written in front:
This is the question bank I prepared for the school transfer exam.
It is also a question bank for university computer, and it is also suitable for the final exam of the university computer course.
A selection of important topics has been selected.
Table of contents
Question bank brushing questions:
Topic 1: (Next is the related topics of binary system/information representation)
Topic 5: (Next is computer system-related topics)
Question 14: (The next three questions are all recitation questions)
Topic 27: (Next is computer network-related topics)
Topic 33: (Next is a comprehensive review topic)
Topic 49: (The following is the big question)
Topic 1: (Next is the related topics of binary system/information representation)
Parse:
bitrate = 11025 * 8 * 2 = 176400
Data volume = (176400 * 1) / 8 = 22050 (b)
22050 / 1024 = 21.53 (2 decimal places)
Variation exercises:
bitrate = 44100 * 16 * 2 = 1411200
Data volume = 1411200 * 300 / 8 = 52920000 (b)
Storage capacity = 52920000 / 1024 /1024 = 50.5MB (b->KB->MB) (one decimal place is reserved by default)
Question-making skills:
Recite the formula:
Bit rate = sampling frequency * quantization bits * number of channels (stereo is 2)
Audio data volume = (bit rate * duration (seconds)) / 8 (unit is byte)
1 byte (b) (byte) = 8 bits (bit)
1024 bytes (b) = 1 KB 1024 KB = 1 MB 1024 MB = 1 GB
Topic 2:
Parse:
Area code transfer internal code: (Note: the first two digits and the last two digits of the area code are calculated separately)
1. Convert the area code into hexadecimal
2. Then add 2020H into the international code
3. Add 8080H to make machine internal code
Note: (B means binary, D means decimal, Q means octal, H means hexadecimal)
41D 78D -> 29H 4EH
29H 4EH + 20H 20H + 80H 80H = C9H EEH = C9EE
Topic 3:
Parse:
Decimal to binary conversion of floating point numbers:
Multiply by 2 and round up, arrange in order
57 -> 11 1001
875 * 2 = 1750 ... 1
750 * 2 = 1500 ... 1
500 * 2 = 1000 ... 1
Solution: 11 1001.111
Topic 4:
Parse:
(1024 * 768 * 24) / 8 = 98304(b)
98304(b) / 1024 / 1024 = 2.25 MB
Question-making skills:
1. If the picture is black and white, the pixel depth is 8, if it is color, the pixel depth is 24
2. Rear formula: storage capacity (b) = (image resolution * pixel depth) / 8
Topic 5: (Next is computer system-related topics)
Parse:
The answer is registers.
Topic 6:
Parse:
Three technical approaches to improve the parallelism of computer systems:
Topic 7:
Parse:
The first is a text editor
The second is the MySQL database (as a back-end development, how can you not know...)
The third is the Linux system (note the same as above)
The fourth is the legendary IE browser
Topic 8:
Parse:
Topic 9:
Parse:
In fact, the closest to the computer hardware system is the operating system, which is a large piece of software.
Topic 10:
Parse:
corresponding to:
Corresponding to the value or address of the data , the result of the operation is also the value of the data, and the type cannot be used as an operand , and the calculation method can be
Topic 11:
Parse:
A: wrong
B: correct
C: correct
D: Error (multiple runs of a program will correspond to multiple processes)
Topic 12:
Parse:
1. It is a distribution version of the Linux system
2. Is Microsoft Corporation
3. It is the Android system. In fact, the bottom layer of the Android system is also Linux.
4. Solaris is a system too
Topic 13:
Parse:
Yes, stored program and program control.
Question 14: (The next three questions are all recitation questions)
Topic 15:
Topic 16:
Topic 17:
Parse:
It is placed in the memory, not the arithmetic unit.
Topic 18:
In fact, Cache is a cache.
From fast to slow: register > Cache > main memory > auxiliary memory > external memory
Topic 19:
Parse:
A specific address, not the operating system.
Topic 20:
Parse:
Topic 21:
Parse:
Yes, definitely put in IR.
Topic 22:
Parse:
ALU is Arithmetic Logic Unit.
The third one is right.
The fourth one is also right.
Topic 23:
Parse:
Yes, computer instructions contain opcodes and address codes.
Topic 24:
Parse:
in addition:
ROM is read-only memory.
CMOS 是互补金属氧化物半导体，计算机里的一个重要芯片。
Cache 是高速缓存啦。
题目25：
解析：
补充：
总结：
数据总线和控制总线是双向的，地址总线是单向的(从CPU到I/O口)。
题目26：
解析：
题目27：（接下是计算机网络相关题目）
解析：
首先：
A类地址：255.xxx.xxx.xxx，B类地址：255.255.xxx.xxx，C类地址：255.255.255.xxx
题目是C类：
子网掩码规则：11111111.11111111.11111111.子网网络号+主机号（一共8位）
子网网络号看子公司数量<2的几次方就有几个1：9 < 2^4 所以子网网络号是4个1 ：1111
主机号看最大计算机台数，还剩4个位：2^4 - 2 = 14 > 12 所以还够用，结尾补4个0：0000
所以：11111111.11111111.11111111.11110000 = 255.255.255.240
选A。
题目27：
解析：
DNS是应用层协议。
题目28：
解析：
1. 子网掩码的每一段都不会大于255
2. 子网掩码的二进制是连续的1和0
所以：（它们不是连续的1和0）
第三个：11111111 11111110 1110 0000 00000000
第四个：11111111 11111111 01000000 00000000
题目29：
解析：
没有同步。
题目30：
解析：
子网二进制：11111111.11111111.11111111.11100000
是C类：子网位数是位数：11100000 有3个1，所以是3，
2^3 = 8 所以子网数目最多位8个。
主机数：2^5 - 2 = 30 所以至多30个。
题目31：
解析：
他是A类，
网络号是他的一个位置的数：10
子网号是他中间的两个位置：1.86
主机号是他的最后的位置的数：29
题目32：
解析：
这是个B类地址。
248：11111000 所以子网有 2^5 = 32 个；
There are 11 zeros in total, so 2^11 - 2 = 2046 hosts can be connected .
Calculation of host address :
Convert the ip address and subnet mask to binary:
IP address: 10101100 00010000 00010000 00000000
Subnet mask: 11111111 11111111 11111000 00000000
network address = ip address & subnet mask = 10101100 00010000 00010000 00000000
Broadcast address = (network address + 1) + (original number of hosts - 1) (2048-1=2047)
= 10101100 00010000 00010000 00000001 + 00000111 11111110
= 10101100 00010000 00010111 11111111
There should be 2048 hosts, and the -2 operation is because the two hosts are used as network numbers and broadcasts respectively.
So the host range is (network address+1) ~ (broadcast address-1)
Namely: 10101100.00010000.00010000.00000001 ~ 10101100.00010000.00010111.11111110
Convert to decimal: 172.16.16.1 ~ 172.16.23.254
Topic 32:
Parse:
Network layer functions:
Transport layer functions:
Typical protocol:
Topic 33: (Next is a comprehensive review topic)
No resolution.
Topic 34:
No resolution.
Topic 35:
Parse:
The core component of a computer is the CPU.
Topic 36:
Parse:
Topic 37:
No resolution.
Topic 38:
Parse:
There is no cache in the CPU, there are registers,
The cache is in memory.
Topic 39:
Parse:
In fact, I still don't quite understand it, but let's memorize it.
Summarize:
The highest bit of a class A address is 0, the highest bit of a class B address is 10, and the highest bit of a class C address is 110.
Topic 40:
Parse:
Summarize:
There is a medium:
Twisted pair is cheap and easy;
Coaxial cable complex, anti-interference;
Large amount of optical fiber, anti-interference, light, good security.
No medium:
Short wave quality is poor and stable;
Large amount of microwave ground, unstable (easy to be affected);
Microwave satellites are large and far away, with high delay.
Topic 41:
Parse:
The OSI reference model is not used now, it is too complicated to use, just refer to it.
Topic 42:
Parse:
Then memorize it:
Calculation means -> calculation process -> calculation execution
Instrumentation -> Formalization -> Automation
Topic 43:
Parse:
Computational thinking is human thinking, not computer thinking!
Topic 44:
Parse:
Then...then recite it.
Or carry the core back?
Topic 45:
Parse:
RAM and ROM correspond to memory and hard disk respectively.
The main difference between them is the sentence of option B (recited).
Topic 46:
Parse:
Topic 47:
Parse:
Calculators and controllers.
Topic 48:
Parse:
In fact, I don't quite understand this question, but the binary value of the first paragraph of option B is 238,
If it does not belong to the three types of ABC addresses, I will treat it as not a legal host address.
Topic 49: (The following is the big question)
Topic 50:
Topic 51:
Topic 52:
Topic 53:
Topic 54:
Topic 55:
Summarize:
To be honest, are the last few big questions serious? The difficulty is too exaggerated, and I really can't memorize them. . .