📚 Interdisciplinary Practice for Cambridge IGCSE Computer Science | 剑桥IGCSE计算机科学跨学科综合题型训练
Cambridge IGCSE Computer Science challenges students to connect computational thinking with other subjects such as Mathematics, Physics, Statistics, and Geography. This article provides a set of interdisciplinary problem-solving exercises that mirror the kind of integrated questions you might encounter in your Year 10 assessments. Each section blends core computing concepts with real-world applications from another discipline, helping you to develop a deeper understanding and to prepare effectively for exams.
剑桥 IGCSE 计算机科学要求学习者将计算思维与数学、物理、统计学和地理学等其他学科联系起来。本文提供了一系列跨学科综合题型训练，模拟你在十年级测评中可能遇到的综合性题目。每个小节都将核心计算机概念与另一学科的实际应用相融合，帮助你深入理解并有效备考。
1. Binary Arithmetic and Mathematics | 二进制运算与数学
Computers rely on binary arithmetic for all operations. This exercise links binary addition and two's complement representation with decimal verification, strengthening your mathematical reasoning.
计算机的所有运算都依赖二进制算术。本练习将二进制加法与补码表示同十进制验证相结合，以此强化你的数学推理能力。
Problem: Perform the binary addition 1011₂ + 1101₂. Show all carry bits and give the final binary result. Verify by converting both addends and the sum to decimal. Then, using an 8-bit two's complement system, find the representation of -6.
问题：完成二进制加法 1011₂ + 1101₂，写出所有进位并给出最终的二进制结果。将两个加数及和转换为十进制进行验证。接着，在 8 位补码系统中，写出 -6 的表示形式。
column 1+0+carry1=2 (write 0, carry 1); next column 0+1+carry1=2 (write 0, carry 1); leftmost column 1+1+carry1=3, which is 11₂. The result is 11000₂. In decimal, 11 + 13 = 24, and 11000₂ equals 16+8 = 24. To represent -6 in 8-bit two' omplement: start with +6 as 00000110, invert bits to 11111001, and add 1, giving 11111010.
首先对齐二进制数。1011₂（十进制 11）加 1101₂（十进制 13）。逐位相加：1+1=2（写 0 进 1）；下一位 1+0+进位 1=2（写 0 进 1）；再下一位 0+1+进位 1=2（写 0 进 1）；最左位 1+1+进位 1=3，即 11₂。结果为 11000₂。十进制验证 11+13=24，11000₂ 等于 16+8=24。要用 8 位补码表示 -6：先写出 +6 为 00000110，按位取反得 11111001，再加 1 得 11111010。
This task shows how binary computation is underpinned by place-value mathematics and how negative numbers are handled through modular arithmetic, a direct bridge to higher-level algebra.
这项任务展示了二进制计算如何以位值数学为基础，以及负数如何通过模运算处理，这是通往高等代数的直接桥梁。
2. Logic Gates and Boolean Simplification | 逻辑门与布尔化简
Logic circuits are the foundation of digital electronics. This problem asks you to work with Boolean expressions, drawing upon algebraic laws that you also use in Mathematics.
逻辑电路是数字电子学的基础。本题要求你运用布尔表达式进行化简，并运用你在数学课上学到的代数定律。
Problem: A AND B). (i) Draw the truth table for the expression. (ii) Using Boolean algebra, simplify Q to its simplest form and name the law that justifies your key step.
问题：某逻辑电路有输入 A 和 B，输出 Q = (A AND B) OR (NOT A AND B)。(i) 画出该表达式的真值表。(ii) 利用布尔代数将 Q 化简为最简形式，并说明关键步骤依据的定律。
=0. When A=1, B=1: 1 AND 1=1, 0 AND 1=0, Q=1. Q matches B exactly. For simplification, write Q = A·B + ¬A·B. Factor out B: Q = B·(A + ¬A). By the complement law, A + ¬ the identity law. The output Q simply equals B, so the circuit can be replaced by a wire.
真值表有 4 行。A=0, B=0 时，(0 AND 0)=0，(1 AND 0)=0，Q=0。A=0, B=1 时：0 AND 1=0，1 AND 1=1，Q=1。A=1, B=0 时：1 AND 0=0，0 AND 0=0，Q=0。A=1, B=1 时：1 AND 1=1，0 AND 1=0，Q=1。Q 与 B 完全一致。化简过程：Q = A·B + ¬A·B。提取公因子 B：Q = B·(A + ¬A)。由互补律 A + ¬A = 1，得 Q = B·1 = B。这是恒等律的体现。输出 Q 实际上就等于 B，因此该电路可用一根导线替代。
This demonstrates that logical reasoning in computer science mirrors algebraic manipulation in Mathematics: both rely on systematic application of laws to reduce complexity.
这说明了计算机科学中的逻辑推理与数学中的代数变换是相通的：二者都依赖系统地应用定律来降低复杂度。
3. Data Storage Calculation and Units | 数据存储计算与单位
Understanding file size requires working with sampling theory and unit conversions, bridging Computer Science with Physics and Mathematics.
计算文件大小需要掌握采样理论和单位换算，这连接了计算机科学、物理和数学。
Problem: A stereo music track is recorded with a sample rate of 44.1 kHz and a 16-bit resolution. The song lasts 3 minutes and 30 seconds. Calculate the uncompressed file size in megabytes (MB). Show all steps and unit conversions clearly.
问题：一段立体声音乐以 44.1 kHz 采样率、16 位分辨率录制，时长为 3 分 30 秒。计算未压缩的文件大小（以 MB 为单位）。请清晰展示全部步骤和单位换算。
. For stereo, there are 2 channels. Each sample uses 16 bits = 2 bytes. The total number of bits = sample rate × seconds × bit depth × channels = 44100 × 210 × 16 × 2. Let' . To go to kilobytes, divide by 1024: approx 36,180.47 KB. For megabytes, divide again by 1024: approximately 35.33 MB.
首先，将时长转换为秒：3 分 30 秒 = 210 秒。采样率为 44100 Hz，即每通道每秒 44100 个样本。立体声有 2 个通道，每个样本 16 位 = 2 字节。总位数 = 采样率 × 秒数 × 位深 × 通道数 = 44100 × 210 × 16 × 2。先计算 44100 × 210 = 9,261,000。再乘 32（16×2）得 296,352,000 位。转换为字节：除以 8，得到 37,044,000 字节。转换为千字节：除以 1024，约 36,180.47 KB。再除以 1024 得到约 35.33 MB。
This calculation connects the physical measurement of sound waves with digital storage, highlighting how sample rate and bit depth directly affect file size—a concept rooted in both physics and mathematics.
这一计算将声波的物理测量与数字存储联系起来，凸显采样率和位深如何直接影响文件大小——这一概念植根于物理与数学之中。
4. Algorithm Design and Number Theory | 算法设计与数论
Writing algorithms often draws directly from mathematical procedures. Here, you will design a pseudocode solution for a classic number theory problem: finding prime factors.
编写算法经常直接借鉴数学步骤。这里，你将为一个经典的数论问题设计伪代码解：求质因数。
Problem: eated as many times as they occur). For example, for N = 12, the output should be 2, 2, 3.
问题：编写伪代码，接受一个大于 1 的正整数 N，输出其所有的质因数（重复次数与实际分解相同）。例如，输入 N = 12，应输出 2, 2, 3。
A possible solution: DECLARE N, factor AS INTEGER INPUT N factor ← 2 WHILE N > 1 DO WHILE N MOD factor = 0 DO OUTPUT factor N ← N / factor ENDWHILE factor ← factor + 1 ENDWHILE
一个可行方案： 声明 N, factor 为整数 输入 N factor ← 2 当 N > 1 执行 当 N MOD factor = 0 执行 输出 factor N ← N / factor 结束当 factor ← factor + 1 结束当
This algorithm systematically divides N by increasing integers, identical to the mathematical method of prime factorization by trial division. It reinforces the algorithmic thinking taught in Computer Science while applying number theory concepts from Mathematics.
该算法通过递增的整数系统地对 N 进行除法运算，与数学中通过试除法进行质因数分解的方法完全相同。它在运用数学中的数论概念的同时，强化了计算机科学所教授的计算思维。
5. Sensors, ADC and Physical Measurement | 传感器、模数转换与物理测量
Monitoring systems often convert analogue signals from sensors into digital data. This question integrates the physics of voltage and temperature with the computing principles of ADC.
监控系统常常将传感器的模拟信号转换为数字数据。本题将电压与温度的物理知识和模数转换的计算原理融合在一起。
Problem: A linear temperature sensor produces 0.5 V at 0°C and 2.5 V at 100°C. An 8-bit ADC with a reference voltage of 5.0 V is used to read the sensor. The digital reading obtained is 153 (decimal). Determine the measured temperature to one decimal place.
问题：线性温度传感器在 0°C 时输出 0.5 V，在 100°C 时输出 2.5 V。使用参考电压为 5.0 V 的 8 位 ADC 读取该传感器，得到的数字读数为 153（十进制）。计算所测温度，保留一位小数。
8 V. Since the sensor is linear, the relationship is: temperature = ( (V - 0.5) / (2.5 - 0.5) ) × 100°C = (V - 0.5) × 50°C/V. Plugging in V = 2.988 V gives (2.988 - 0.5) × 50 = 2.488 × 50 = 124.4°C. Thus the measured temperature is 124.4°C.
8 位 ADC 共有 2⁸ = 256 个可能的值，步长为 5.0 V / 256 ≈ 0.01953 V。读数 153 对应的电压为 153 × 0.01953 V ≈ 2.988 V。传感器线性关系为：温度 = ( (V - 0.5) / (2.5 - 0.5) ) × 100°C = (V - 0.5) × 50°C/V。代入 V = 2.988 V 得 (2.988 - 0.5) × 50 = 2.488 × 50 = 124.4°C。因此测量温度为 124.4°C。
This exercise illustrates how raw binary outputs are interpreted through calibration curves, directly applying the physics of sensing and the mathematics of linear equations.
本练习阐明了如何通过校准曲线解读原始二进制数据，直接应用了传感物理和线性方程数学。
6. Database Queries and Statistical Analysis | 数据库查询与统计分析
Retrieving information from a database often involves aggregation and filtering, which closely resembles the statistical operations taught in Mathematics.
从数据库中检索信息通常涉及聚合和筛选，这与数学课所教授的统计操作非常相似。
Problem: A school database has a table EXAM with fields StudentID, Subject, Score. Write an SQL query that displays each subject and its average score, but only for subjects where the average score exceeds 70. Order the results by average score descending.
问题：学校数据库中有一个 EXAM 表，包含字段 StudentID、Subject、Score。编写一条 SQL 查询，显示每个科目及其平均分，但仅显示平均分超过 70 的科目，并按平均分降序排列。
The required query uses GROUP BY with the HAVING clause: SELECT Subject, AVG(Score) AS AverageScore FROM EXAM GROUP BY Subject HAVING AVG(Score) > 70 ORDER BY AverageScore DESC;
所需查询使用 GROUP BY 和 HAVING 子句： SELECT Subject, AVG(Score) AS AverageScore FROM EXAM GROUP BY Subject HAVING AVG(Score) > 70 ORDER BY AverageScore DESC;
This task reflects how database aggregation functions mirror statistical measures such as mean. Filtering with HAVING is equivalent to applying a condition to a grouped data set, a common requirement in both data analysis and computational problem solving.
这一任务反映出数据库聚合函数如何与平均值等统计量相对应。使用 HAVING 进行筛选相当于对分组数据集施加条件，这是数据分析和计算问题解决中的常见需求。
7. Network Transmission and Distance Problems | 网络传输与距离问题
Data traveling over long distances is subject to both transmission delay and propagation delay. This scenario brings together the physics of signal speed and the data rates of networking.
长距离传输数据时会同时产生传输时延和传播时延。这一场景结合了信号速度的物理原理和网络的数据速率。
Problem: ing protocol overheads, calculate the total time from the first bit being sent to the last bit being received. Express the result in seconds, accurate to two decimal places.
问题：一个 100 MB 的文件通过 100 Mbps 的光纤链路传输。光纤长度为 500 km，光在光纤中的速度约为 2 × 10⁸ m/s。忽略协议开销，计算从第一个比特发出到最后一个比特被接收的总时间。结果以秒为单位，精确到两位小数。
s. Propagation delay = distance / speed = 500,000 m / 200,000,000 m/s = 0.0025 s. Total time = 8.388608 + 0.0025 ≈ 8.39 seconds.
文件大小换算为位：100 MB = 100 × 1024 × 1024 × 8 位 = 838,860,800 位。传输时延 = 总位数 / 比特率 = 838,860,800 / 100,000,000 = 8.388608 s。传播时延 = 距离 / 速度 = 500,000 m / 200,000,000 m/s = 0.0025 s。总时间 = 8.388608 + 0.0025 ≈ 8.39 秒。
This problem underscores the impact of geographical distance on network performance, combining digital data rates with the physical constraint of light propagation—a direct link to both geography and physics.
本题凸显了地理距离对网络性能的影响，将数字数据速率与光的物理传播限制结合起来——这是地理学和物理学的直接交叉。
8. Encryption, Caesar Cipher and Modular Maths | 加密、凯撒密码与模运算
Classical ciphers like Caesar's use modular arithmetic, a mathematical concept that also appears in computing for hashing and cyclical buffers.
像凯撒密码这样的古典密码使用模运算，这一数学概念在计算中也会出现在哈希和循环缓冲区中。
Problem: Using a Caesar cipher with a shift of 3, encrypt the plaintext "HELLO". Then decrypt the ciphertext "KHOOR" back to its original meaning. Show the mathematical mapping using modular arithmetic.
问题：使用移位为 3 的凯撒密码，加密明文 "HELLO"。然后将密文 "KHOOR" 解密还原。用模运算展示数学映射过程。
Assign A=0, B=1, ..., Z=25.
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