📚 Year 12 Edexcel Computer Science: Unit Test Mock Paper Analysis | Year 12 Edexcel 计算机：单元测试模拟卷解析
This post walks through a comprehensive mock exam paper for the Edexcel AS Computer Science Unit 1. Each question is broken down with clear explanations and model answers to help you master the key concepts tested in the real examination.
本文对一份 Edexcel AS 计算机科学单元1模拟试卷进行全面解析，逐题给出详尽解释和标准答案，助你掌握真实考试的核心考点。
1. Overview of the Mock Paper | 模拟卷结构概览
The mock paper consists of two sections: Section A contains 10 multiple-choice questions worth 1 mark each; Section B features short-answer and extended-response questions worth a total of 40 marks. The overall time allowed is 1 hour 30 minutes. Key topics include data representation, computer architecture, networking, algorithms, and computational thinking.
模拟卷由两部分组成：A部分含10道单选题，每题1分；B部分为简答和扩展题，共40分。总时长1小时30分钟。重点考查数据表示、计算机体系结构、网络、算法和计算思维等模块。
2. Q1: Two's Complement Conversion | 补码转换
The question asks to represent the denary number -37 as an 8-bit two's complement binary number. (2 marks)
题目要求将十进制数 -37 表示为8位二进制补码。(2分)
Step 1: Write +37 in binary using 8 bits. 37 = 32 + 4 + 1 → 00100101₂.
步骤1：写出+37的8位二进制：37 = 32+4+1 → 00100101₂。
Step 2: Invert all bits (one's complement): 11011010₂.
步骤2：将所有位取反（反码）：11011010₂。
Add 1 to the least significant bit: 11011010 + 1 = 11011011₂. Therefore, -37 in 8-bit two's complement is 11011011₂.
步骤3：在最低位加1：11011010 + 1 = 11011011₂。因此，-37的8位补码表示为11011011₂。
3. Q2: Logic Gate Truth Table | 逻辑门真值表
ve the truth table for P = (A AND B) OR (NOT C) for all possible inputs A, B, C. (4 marks)
问题：为表达式 P = (A AND B) OR (NOT C) 列出所有输入组合的真值表。(4分)
| A | B | C | A AND B | NOT C | P |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 |
Explanation: The truth table evaluates each sub-expression and the final output. This is a common exam question testing Boolean logic.
解析：真值表逐步计算子表达式和最终输出，是考查布尔逻辑的常见题型。
4. Q3: TCP/IP Protocol Stack | 传输控制协议/网际协议栈
Question: Explain how the TCP/IP protocol stack is used when a user enters a URL in a web browser to retrieve a webpage. (4 marks)
问题：说明当用户在网页浏览器中输入网址获取页面时，TCP/IP协议栈如何工作。(4分)
The application layer prepares an HTTP GET request and passes it down. The transport layer (TCP) segments the data, adds source and destination port numbers (port 80 for HTTP), and provides reliable delivery.
应用层准备好HTTP GET请求并向下传递；传输层（TCP）将数据分段，添加源和目的端口号（HTTP用80端口），并提供可靠传输。
The internet layer (IP) encapsulates the segment into an IP packet, adding the source and destination IP addresses. The link layer then frames the packet, adding the MAC addresses for the next hop, and transmits it over the physical medium.
网络层（IP）将段封装为IP数据包，添加源和目的IP地址；链路层再将数据包封装成帧，添加下一跳的MAC地址，并通过物理介质传输。
At the server side, the process is reversed: link layer removes the frame header, network layer extracts the IP packet, transport layer reassembles segments, and the application layer interprets the HTTP request to send back the webpage.
在服务器端，过程反向进行：链路层去掉帧头，网络层提取IP包，传输层重组分段，应用层解释HTTP请求并返回网页。
5. Q4: Fetch-Decode-Execute Cycle | 取指-译码-执行周期
Question: Describe the fetch-decode-execute cycle of a CPU, including the roles of the program counter (PC), memory address register (MAR), memory data register (MDR), current instruction register (CIR), and the arithmetic logic unit (ALU). (6 marks)
问题：描述CPU的取指-译码-执行周期，包括程序计数器(PC)、内存地址寄存器(MAR)、内存数据寄存器(MDR)、当前指令寄存器(CIR)和算术逻辑单元(ALU)的作用。(6分)
Fetch: The PC holds the address of the next instruction to be executed. This address is copied to the MAR, and the PC is incremented. The control unit sends a read signal to memory, and the instruction stored at that address is loaded into the MDR, then transferred to the CIR.
取指：PC存放待执行指令的地址，该地址被复制到MAR，同时PC递增；控制单元发出读信号，内存中该地址处的指令被加载到MDR，再传送到CIR。
Decode: The control unit decodes the instruction in the CIR to determine what operation is required and which operands are involved.
译码：控制单元对CIR中的指令进行译码，确定所需操作和操作数。
Execute: Depending on the instruction, the ALU may perform arithmetic or logic operations, data may be read from or written to memory, or the PC may be updated with a branch address. The result is stored back in a register or memory. The cycle then repeats.
执行：根据指令，ALU可能进行算术或逻辑运算，数据可能被读/写内存，或者PC被更新为跳转地址；结果存回寄存器或内存，然后周期重复。
6. Q5: Sorting Algorithms Comparison | 排序算法比较
Question: Compare bubble sort and merge sort in terms of time complexity and space complexity. (4 marks)
问题：从时间复杂度和空间复杂度两方面比较冒泡排序与归并排序。(4分)
Bubble sort has an average and worst-case time complexity of O(n²) because it uses nested loops to swap adjacent elements. Its space complexity is O(1) as it operates in-place.
冒泡排序的平均和最坏时间复杂度为 O(n²)，因为它使用嵌套循环交换相邻元素；空间复杂度是 O(1)，因其原地操作。
Merge sort achieves a time complexity of O(n log n) in all cases by dividing the array into halves, sorting recursively, and merging. However, it requires O(n) additional space for the temporary arrays used during merging.
归并排序通过将数组二分为子数组、递归排序并合并，在所有情况下都达到 O(n log n) 的时间复杂度；但合并时需要使用临时数组，因此需要 O(n) 的额外空间。
For large datasets, merge sort is far more time-efficient than bubble sort, but it uses significantly more memory.
对于大数据集，归并排序的时间效率远优于冒泡排序，但内存占用显著增加。
7. Q6: Stack Operations Tracing | 栈操作跟踪
Question: A stack is implemented using an array with a maximum capacity of 10. Initially the stack is empty. Show the state of the stack and the top pointer after each of the following operations: push(5), push(8), pop(), push(3), push(7). (3 marks)
问题：使用容量为10的数组实现栈，初始为空。请显示执行 push(5), push(8), pop(), push(3), push(7) 各步之后栈的状态和栈顶指针。(3分)
Initially: stack = [], top = -1.
初始：栈为空，top = -1。
returns to 0. After push(3): stack[1] = 3, top = 1. After push(7): stack[2] = 7, top = 2.
push(5)后：stack[0]=5, top=0；push(8)后：stack[1]=8, top=1；pop()后：移除8, top回退至0；push(3)后：stack[1]=3, top=1；push(7)后：stack[2]=7, top=2。
The final stack contains [5, 3, 7] with top pointing to index 2.
最终栈内容为[5, 3, 7]，栈顶指针指向索引2。
8. Q7: Pseudocode Trace Table | 伪代码跟踪表
endwhile. Produce a trace table showing the value of y at the start of each iteration. (4 marks)
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