Year 11 CCEA Computer Science: Interdisciplinary Comprehensive Question Training | CCEA 11年级计算机：跨学科综合题型训练

📚 Year 11 CCEA Computer Science: Interdisciplinary Comprehensive Question Training | CCEA 11年级计算机：跨学科综合题型训练

In the CCEA Year 11 Computer Science curriculum, real-world problems frequently bridge mathematics, physics, business and ethics. This article presents a series of cross-disciplinary question drills to strengthen your ability to apply computational thinking in unfamiliar contexts. Each section introduces a scenario, the core computing concept, and a step-by-step walkthrough – perfect for exam practice.

在 CCEA 11 年级计算机课程中，现实问题往往跨联数学、物理、商业和伦理。本文提供一系列跨学科题型训练，帮助你在陌生情境中运用计算思维。每一节都包含一个情景、核心计算机概念以及分步解析，是备考的理想练习。

1. Binary Arithmetic and Data Representation | 二进制运算与数据表示

Scenario: An air quality monitor stores particle concentrations as 8‑bit unsigned integers. Two successive readings are 10101101₂ and 01110011₂. The system adds them to calculate total particles. (a) Perform the binary addition. (b) Express the sum in denary and hexadecimal. (c) State whether an overflow flag is raised, given that the maximum value is 255.

情景：空气质量监测仪将粒子浓度存储为 8 位无符号整数。连续两次读数分别是 10101101₂ 和 01110011₂。系统将它们相加得出总粒子数。(a) 执行二进制加法。(b) 将结果表示为十进制和十六进制。(c) 已知最大值为 255，判断是否触发溢出标志。

Solution (a): Align the bits and add column by column, carrying 1 whenever a column sums to 2 (10₂) or 3 (11₂).

解答 (a)：对齐各列，逐列相加，当一列之和为 2（10₂）或 3（11₂）时向高位进 1。

   1 0 1 0 1 1 0 1
+  0 1 1 1 0 0 1 1
--------------------
 1 0 0 1 0 0 0 0 0

Working from the least significant bit, the additions produce a ninth bit, giving the 9‑bit result 100100000₂.

从最低有效位开始逐位相加，产生第九位，得到 9 位结果 100100000₂。

Solution (b): Denary value = 1×256 + 0×128 + 0×64 + 1×32 + 0×16 + 0×8 + 0×4 + 0×2 + 0×1 = 288₁₀. To convert to hexadecimal, group the binary number into nibbles: 1 0010 0000₂ → 1 2 0₁₆. Therefore the sum is 120₁₆.

解答 (b)：十进制值 = 1×256 + 0×128 + 0×64 + 1×32 + 0×16 + 0×8 + 0×4 + 0×2 + 0×1 = 288₁₀。转为十六进制时，将二进制按 4 位分组：1 0010 0000₂ → 1 2 0₁₆。因此和为 120₁₆。

10101101₂ + 01110011₂ = 100100000₂ (288₁₀ = 120₁₆)

Solution (c): Because 288 > 255, the result cannot be stored in 8 bits. The processor would set the overflow flag (or carry flag) and typically retain only the lower 8 bits (00100000₂), which would be inaccurate. Overflow is detected.

解答 (c)：由于 288 > 255，结果无法用 8 位存储。处理器会置起溢出标志（或进位标志），通常只保留低 8 位（00100000₂），这会导致结果错误。因此发生了溢出。

2. Boolean Logic and Truth Tables | 布尔逻辑与真值表

Scenario: A safety circuit uses two sensors A and B. The alarm must sound if (A is triggered AND B is triggered) OR (A is NOT triggered AND B is triggered). Simplify the expression using a truth table and suggest a more efficient logic gate.

情景：一个安全电路使用两个传感器 A 和 B。当 (A 触发 且 B 触发) 或 (A 未触发 且 B 触发) 时报警器必须响起。用真值表简化该表达式，并建议一种更高效的逻辑门。

The Boolean expression is (A ∧ B) ∨ (¬A ∧ B). Let us build its truth table:

布尔表达式为 (A ∧ B) ∨ (¬A ∧ B)。构造真值表：

The output column matches B exactly. Hence the expression simplifies to just B. This is a powerful demonstration of Boolean algebra: the alarm simply needs to check sensor B, regardless of A. The circuit can be built with a single wire (buffer) from B.

输出列与 B 完全一致。因此表达式简化为 B。这充分体现了布尔代数的威力：报警器只需检查传感器 B，与 A 无关。电路只需从 B 接一根线（缓冲器）即可。

3. Algorithms and Flowcharts | 算法与流程图

Scenario: A maths teacher asks you to design an algorithm that calculates the sum of the first n natural numbers (1 + 2 + … + n). Produce pseudocode and then compare your solution with the arithmetic formula n(n+1)/2 to discuss efficiency.

情景：数学老师请你设计一个算法，计算前 n 个自然数的和（1 + 2 + … + n）。写出伪代码，然后与算术公式 n(n+1)/2 对比，讨论效率。

Pseudocode (iterative):

伪代码（迭代）：

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