Interdisciplinary Problem-Solving for Year 13 SQA Computer Science | Year 13 SQA 计算机：跨学科综合题型训练

📚 Interdisciplinary Problem-Solving for Year 13 SQA Computer Science | Year 13 SQA 计算机：跨学科综合题型训练

Many advanced computing challenges sit at the crossroads of multiple disciplines, and the SQA Advanced Higher Computing Science course increasingly rewards students who can synthesise knowledge from mathematics, physics, biology, economics, and beyond. This article presents a series of integrated scenarios, worked examples, and guided exercises designed to build the cross-subject fluency essential for top-band answers. Each section pairs a computing concept with a non-computing context, shows how to model the problem, and then unpacks a full solution in Python-like pseudocode or structured discussion.

许多高级计算挑战处于多学科交叉点，SQA 高等计算机科学课程越来越青睐能够综合运用数学、物理、生物、经济学等知识的考生。本文提供一系列融合情境、范例和引导式练习，旨在培养跨学科思维，帮助你在考试中获得高分。每个小节都将一个计算概念与一个非计算机领域背景配对，展示如何建模，并用类 Python 伪代码或结构化论述剖析完整解答。

1. Modelling Population Dynamics with 2D Arrays | 用二维数组模拟种群动态

A conservation biologist asks you to create a cellular automaton that predicts the spread of a plant species across a 100 by 100 grid. Each cell can be empty, occupied by a young plant, or occupied by a mature plant. Every year, a mature plant can seed a neighbouring cell (up, down, left, right) with probability 0.3, provided that cell is empty. Young plants become mature after one year. Mature plants have a 20% chance of dying. Write a function simulate(grid, years) that returns the number of mature plants after a given number of years. Explain how the data structure choice influences memory efficiency.

一位保护生物学家请你创建一个元胞自动机，预测某种植物在 100×100 网格上的扩散。每个单元格可以是空、被幼株占据或被成熟株占据。每年，成熟株有 0.3 的概率在其相邻的空单元格（上下左右）中播种。幼株一年后成熟。成熟株每年有 20% 的概率死亡。编写函数 simulate(grid, years)，返回指定年份后的成熟株数量，并说明数据结构选择如何影响内存效率。

Modelling strategy: Use a 2D list of integers (0 for empty, 1 for young, 2 for mature). The grid size 100×100 means 10,000 elements, which a list of lists can comfortably hold. An alternative with dictionaries only storing occupied cells could be more memory-efficient for sparse populations but slower for dense ones.

建模策略：使用整数二维列表（0 表示空，1 表示幼株，2 表示成熟株）。100×100 的网格共 10,000 个元素，列表的列表完全可以容纳。若种群稀疏，采用字典仅存储被占据的单元格会更省内存，但密集时速度较慢。

Pseudocode for one year step:

for each cell (row, col):
    if grid[row][col] == 1:
        newGrid[row][col] = 2         # young -> mature
    elif grid[row][col] == 2:
        if random() < 0.2:
            newGrid[row][col] = 0     # death
        else:
            newGrid[row][col] = 2     # survive
        for each neighbour (nRow, nCol):
            if grid[nRow][nCol] == 0 and random() < 0.3:
                newGrid[nRow][nCol] = 1   # seed

This modular approach clearly separates biological rules from the code structure, making it easy to adjust parameters.

这种模块化方法将生物学规则与代码结构清晰分离，便于调整参数。

2. Graph Traversal for Transport Networks | 图的遍历在交通网络中的应用

A logistics company operates a hub-and-spoke delivery system. The network can be modelled as a weighted graph where nodes are depots and edges represent roads with travel time in minutes. The company wants to find the quickest route from the central hub (node 'A') to every other depot. Define an adjacency list representation and implement Dijkstra's algorithm. Discuss why a breadth-first search would be inappropriate for this scenario.

一家物流公司采用轴辐式配送系统。网络可建模为带权图，节点为仓库，边表示道路，权重为以分钟计的行程时间。公司希望找到从中心枢纽（节点 'A'）到所有其他仓库的最快路线。定义邻接表表示法并实现 Dijkstra 算法。讨论为何广度优先搜索在此场景中不适用。

Graph representation:

graph = {
    'A': [('B', 12), ('C', 7)],
    'B': [('D', 9), ('E', 15)],
    'C': [('D', 10), ('F', 14)],
    'D': [('E', 4), ('F', 2)],
    'E': [('G', 6)],
    'F': [('G', 8)],
    'G': []
}

图的表示：邻接表将每个顶点映射到一个（邻居, 权重）列表。与列表/矩阵表示相比，邻接表在稀疏图上节省空间。

Dijkstra pseudocode:

dist = {node: ∞ for node in graph}
dist['A'] = 0
priority_queue = [('A', 0)]
while queue not empty:
    current, d = pop smallest
    if d > dist[current]: continue
    for neighbour, weight in graph[current]:
        new_dist = d + weight
        if new_dist < dist[neighbour]:
            dist[neighbour] = new_dist
            push (neighbour, new_dist)

BFS ignores edge weights, assuming all edges cost the same, which would misrepresent real road travel times. Dijkstra's algorithm respects varying weights and is optimal for non-negative edges.

BFS 忽略边权重，假设所有边成本相同，这无法体现真实的道路通行时间。Dijkstra 算法考虑权重变化，对非负边是最优的。

3. SQL Queries for School Timetabling | 学校排课中的 SQL 查询

A school database contains the following tables: Student(StudentID, Name, YearGroup), Subject(SubjectID, Name), and Choice(StudentID, SubjectID). Each student selects up to five subjects. Write an SQL statement that lists the names of all Year 13 students taking both Computer Science and Physics. Additionally, create a query that identifies subjects chosen by fewer than 10 students. Explain how database normalisation helps maintain consistency when a subject name changes.

某学校数据库包含表：Student(StudentID, Name, YearGroup)、Subject(SubjectID, Name) 和 Choice(StudentID, SubjectID)。每位学生最多选择 5 门科目。写出一个 SQL 语句，列出同时选修计算机科学和物理的所有 Year 13 学生姓名。另外，创建一个查询找出选修人数少于 10 人的科目。解释数据库规范化如何在科目名称更改时帮助维护一致性。

Query 1 - Intersection approach:

SELECT s.Name
FROM Student s
JOIN Choice c1 ON s.StudentID = c1.StudentID
JOIN Subject sub1 ON c1.SubjectID = sub1.SubjectID AND sub1.Name = 'Computer Science'
JOIN Choice c2 ON s.StudentID = c2.StudentID
JOIN Subject sub2 ON c2.SubjectID = sub2.SubjectID AND sub2.Name = 'Physics'
WHERE s.YearGroup = 13;

查询 1 —— 交集方法：通过自连接并要求两门不同科目都匹配，确保只返回同时选修两者的学生。

Query 2 - HAVING clause:

SELECT sub.Name, COUNT(*) AS Enrolment
FROM Subject sub
JOIN Choice c ON sub.SubjectID = c.SubjectID
GROUP BY sub.SubjectID, sub.Name
HAVING COUNT(*) < 10;

Normalisation separates subject names into the Subject table. If a subject is renamed, only one row needs updating, preserving referential integrity across all student choices.

规范化将科目名称分离到 Subject 表。若科目更名，只需更新一行，即可保持所有学生选课记录的参照完整性。

4. Boolean Algebra and Digital Circuit Design | 布尔代数与数字电路设计

A security system for a physics laboratory enables a laser when two out of three authorised scientists swipe their ID cards (A, B, C). Write the Boolean expression for the output L (laser enabled) and simplify it using a Karnaugh map. Then design the logic circuit using only NAND gates. Explain how the simplified expression reduces component cost.

物理实验室的安保系统要求三位授权科学家 A、B、C 中至少有两人刷卡时，激光器 L 才启用。写出输出 L 的布尔表达式并用卡诺图化简。然后用仅使用与非门设计逻辑电路。解释化简后的表达式如何降低元件成本。

The minterms are: ABC, AB¬C, A¬BC, ¬ABC. Sum of products: L = AB + BC + AC. A Karnaugh map confirms no further algebraic simplification is needed. Two-level AND-OR realisation requires three 2-input AND gates and one 3-input OR gate.

最小项为：ABC、AB¬C、A¬BC、¬ABC。乘积项之和：L = AB + BC + AC。卡诺图确认无需进一步代数简化。两级与-或实现需三个 2 输入与门和一个 3 输入或门。

Using NAND equivalence: double negation gives L = ¬(¬(AB) · ¬(BC) · ¬(AC)). This uses three 2-input NAND gates plus one 3-input NAND gate, reducing inventory to a single gate type and potentially lowering fabrication costs.

使用与非等价：双重否定得 L = ¬(¬(AB) · ¬(BC) · ¬(AC))。这使用三个 2 输入与非门和一个 3 输入与非门，将元件简化为单一门类型，可降低制造成本。

5. Recursion and Fractal Geometry in Nature | 递归与自然界的分形几何

Fractals appear in ferns, snowflakes, and coastlines. A simplified Barnsley fern can be generated by an iterative function system, but recursion offers an elegant alternative for drawing a fractal tree. Define a recursive function tree(branch_length, angle, depth) that draws a tree using turtle graphics. The base case stops when depth reaches 0. The recursive rule: draw the branch, then call tree for two smaller branches at ±angle with reduced length. Discuss how recursion depth affects both visual detail and execution time.

分形出现在蕨类、雪花和海岸线中。简化的 Barnsley 蕨可通过迭代函数系统生成，但递归为绘制分形树提供了优雅的替代方案。定义一个递归函数 tree(branch_length, angle, depth)，使用海龟绘图。基例在深度为 0 时停止。递归规则：绘制树枝，然后以 ±angle 调用两个较短的树枝。讨论递归深度如何影响视觉细节和执行时间。

Pseudocode:

function tree(length, angle, depth):
    if depth == 0: return
    forward(length)
    left(angle)
    tree(length * 0.7, angle, depth - 1)
    right(2 * angle)
    tree(length * 0.7, angle, depth - 1)
    left(angle)
    backward(length)

Each increase in depth multiplies the number of branches by about 2, so time complexity is O(2^n). Excessive depth can cause stack overflow or unacceptable delays, illustrating the practical limits of recursion.

深度每增加一级，分支数约乘以 2，因此时间复杂度为 O(2^n)。深度过大可能导致堆栈溢出或无法接受的延迟，这说明了递归的实际限制。

6. Machine Learning Classifier for Disease Diagnosis | 用于疾病诊断的机器学习分类器

A medical dataset contains two features: white blood cell count (WBC) and body temperature (°C). Each patient is labelled as 'healthy' or 'infection'. You decide to implement a k-nearest neighbour classifier with k=3. Describe the algorithm steps. Given the training data below, classify a new patient with WBC=8.2 and temperature=37.9. Explain the impact of choosing k too small or too large, linking your answer to the bias-variance trade-off.

某医学数据集包含两个特征：白细胞计数 (WBC) 和体温 (°C)。每位患者被标记为 '健康' 或 '感染'。你决定实现 k=3 的 k 近邻分类器。描述算法步骤。根据以下训练数据，对白细胞计数=8.2、体温=37.9 的新患者进行分类。解释 k 值选择过小或过大的影响，并联系偏差-方差权衡。

WBC (×10⁹/L)	Temp (°C)	Label
5.1	36.5	Healthy
9.3	38.1	Infection
7.8	37.6	Infection
4.9	36.8	Healthy
10.1	38.5	Infection

Step 1: Calculate Euclidean distance to each training point. Step 2: Select the 3 nearest neighbours. Step 3: Majority vote determines the label. Euclidean distance to (8.2,37.9): to (5.1,36.5) ≈ 3.38; to (9.3,38.1) ≈ 1.12; to (7.8,37.6) ≈ 0.48; to (4.9,36.8) ≈ 3.48; to (10.1,38.5) ≈ 1.99. The three smallest distances are 0.48 (Infection), 1.12 (Infection), 1.99 (Infection) → classified as Infection.

步骤 1：计算到每个训练点的欧氏距离。步骤 2：选出 3 个最近邻。步骤 3：多数投票决定标签。到 (8.2,37.9) 的欧氏距离：到 (5.1,36.5) ≈ 3.38；到 (9.3,38.1) ≈ 1.12；到 (7.8,37.6) ≈ 0.48；到 (4.9,36.8) ≈ 3.48；到 (10.1,38.5) ≈ 1.99。三个最小距离为 0.48（感染）、1.12（感染）、1.99（感染）→ 分类为感染。

A very small k (e.g., k=1) leads to high variance, as the decision boundary follows noise. A large k smooths the boundary but may introduce bias by including points from distant regions. The bias-variance trade-off directly applies: small k → low bias, high variance; large k → high bias, low variance.

k 非常小（如 k=1）会导致高方差，因为决策边界会跟随噪声。k 较大会平滑边界，但可能因纳入远距离点而引入偏差。偏差-方差权衡直接适用：小 k → 低偏差、高方差；大 k → 高偏差、低方差。

7. Encryption and Number Theory | 加密与数论

The RSA encryption algorithm relies on properties of prime numbers and modular arithmetic. A student wants to generate a public/private key pair using primes p=61 and q=53. Calculate n, φ(n), choose e=17, and determine the private key d. Show the encryption of the message M=65. Explain why the difficulty of factoring large semiprimes ensures security.

RSA 加密算法依赖于素数和模运算的性质。一名学生想用素数 p=61 和 q=53 生成公钥/私钥对。计算 n、φ(n)，选择 e=17，并确定私钥 d。展示对消息 M=65 的加密。解释为何对大合数分解的困难性保证了安全性。

n = 61 × 53 = 3233. φ(n) = (61-1)(53-1) = 60×52 = 3120. We need d such that e·d ≡ 1 mod φ(n). 17 × 2753 = 46801, and 46801 mod 3120 = 1, so d = 2753. Encryption: C = 65^17 mod 3233. Using repeated squaring: 65^2=4225 mod 3233 = 992; 65^4=992^2 mod 3233 = 984064 mod 3233 = 1232; 65^8=1232^2 mod 3233 = 1517824 mod 3233 = 889; 65^16=889^2 mod 3233 = 790321 mod 3233 = 1040. Then 65^17 = 1040×65 mod 3233 = 67600 mod 3233 = 2790. So ciphertext = 2790.

n = 61 × 53 = 3233。φ(n) = (61-1)(53-1) = 60×52 = 3120。需要 d 满足 e·d ≡ 1 mod φ(n)。17 × 2753 = 46801，46801 mod 3120 = 1，故 d = 2753。加密：C = 65^17 mod 3233。使用平方乘：65^2=4225 mod 3233 = 992；65^4=992^2 mod 3233 = 984064 mod 3233 = 1232；65^8=1232^2 mod 3233 = 1517824 mod 3233 = 889；65^16=889^2 mod 3233 = 790321 mod 3233 = 1040。然后 65^17 = 1040×65 mod 3233 = 67600 mod 3233 = 2790。密文 = 2790。

Factoring n=3233 into 61×53 is trivial for a human, but for 600-digit primes it is computationally infeasible with current technology. This one-way function ensures that deriving the private key from the public key is practically impossible.

将 n=3233 分解为 61×53 对人类来说很简单，但对于 600 位的素数，当前技术下计算上不可行。这种单向函数确保从公钥推导出私钥在实践中不可能。

8. Parallel Processing and Climate Modelling | 并行处理与气候建模

Climate simulations divide the atmosphere into a 3D grid and solve fluid dynamics equations at each node for each time step. These calculations are inherently parallel because the update at one grid point depends only on its immediate neighbours. Outline how you would parallelise a simplified 2D temperature diffusion model using multiple processors. Discuss Amdahl's Law and explain why sequential dependency on boundary conditions may limit speedup.

气候模拟将大气划分为三维网格，并在每个时间步对每个节点求解流体动力学方程。这些计算本质上是可并行的，因为一个网格点的更新仅依赖于其近邻。概述如何利用多处理器并行化简化后的二维温度扩散模型。讨论阿姆达尔定律，并解释为何边界条件的顺序依赖可能限制加速比。

A parallelisation strategy for a grid of size N×N using P processors: divide the grid into strip partitions (each processor handles N/P rows). At each iteration, each processor updates its interior rows independently. After updating, neighbouring processors exchange boundary rows (ghost cells) to provide the required neighbour values for the next iteration. This uses the Message Passing Interface (MPI) or shared memory with barriers.

使用 P 个处理器对 N×N 网格的并行化策略：将网格划分为条带分区（每个处理器处理 N/P 行）。每次迭代，每个处理器独立更新其内部行。更新后，相邻处理器交换边界行（幽灵单元），为下一次迭代提供所需的邻居值。这使用消息传递接口 (MPI) 或带屏障的共享内存。

Amdahl's Law: Speedup ≤ 1 / (S + (1-S)/P), where S is the sequential fraction. Communication and synchronisation of boundary data is inherently sequential; as P increases, the communication overhead becomes the bottleneck, limiting speedup even if the computational part is perfectly parallel.

阿姆达尔定律：加速比 ≤ 1 / (S + (1-S)/P)，其中 S 为串行部分比例。边界数据的通信和同步本质上是串行的；随着 P 增大，通信开销成为瓶颈，即使计算部分完美并行，也会限制加速比。

9. Heuristic Search for Puzzle Solving | 解决谜题的启发式搜索

The 8-puzzle consists of a 3×3 grid with tiles numbered 1 to 8 and one blank space. A tile adjacent to the blank can slide into it. The goal is to reach a specific configuration from a given start state. Implement the A* search algorithm using the Manhattan distance heuristic. Explain why this heuristic is admissible and how it prunes the search space compared to breadth-first search.

八数码问题由一个 3×3 网格组成，包含编号 1 至 8 的方块和一个空格。与空格相邻的方块可滑入空格。目标是从给定的初始状态到达特定布局。使用曼哈顿距离启发式实现 A* 搜索算法。解释为什么该启发式是可接受的，以及相比广度优先搜索它是如何剪枝搜索空间的。

Each state is a tuple of 9 values. Manhattan distance sums, for each tile except the blank, the absolute difference in row and column between current and goal positions. It never overestimates the true number of moves because each tile must move at least its Manhattan distance steps. Thus the heuristic is admissible, guaranteeing A* finds an optimal solution.

每个状态是一个包含 9 个值的元组。曼哈顿距离对每个方块（空格除外）计算其当前位置与目标位置行、列差的绝对值之和。它绝不会高估实际移动步数，因为每个方块至少需要移动其曼哈顿距离那么多步。因此启发式是可接受的，保证 A* 找到最优解。

A* with this heuristic expands far fewer nodes than BFS. BFS treats all moves as equal cost and expands in all directions, while A* prioritises states that appear closer to the goal. The heuristic directs the search, dramatically reducing the explored state space.

使用该启发式的 A* 比 BFS 扩展的节点少得多。BFS 将所有移动视为等代价，向所有方向扩展，而 A* 优先处理看似更接近目标的状态。启发式引导搜索，大幅缩减探索的状态空间。

10. Data Encryption and Steganography in Multimedia | 多媒体中的数据加密与隐写术

A digital artist wants to embed a copyright watermark in the least significant bits (LSB) of an image's pixel values. Each pixel is stored as three bytes (R, G, B). The watermark is a binary string. Design a function embed_watermark(image, watermark) that modifies the LSB of each colour channel to hide the bits. Discuss the trade-off between capacity and image quality. How could you detect tampering if the image is later compressed with a lossy format like JPEG?

一位数字艺术家想在图像像素值的最低有效位 (LSB) 中嵌入版权水印。每个像素存储为三个字节 (R, G, B)。水印是一个二进制字符串。设计一个函数 embed_watermark(image, watermark)，修改每个颜色通道的 LSB 来隐藏比特。讨论容量与图像质量之间的权衡。如果图像后续经 JPEG 等有损格式压缩，如何检测篡改？

Pseudocode:

function embed_watermark(image, watermark):
    index = 0
    for each pixel in image:
        for channel in [R, G, B]:
            if index < len(watermark):
                set LSB of channel to watermark[index]
                index += 1
    return image

Modifying the LSB changes pixel values by at most 1, which is imperceptible to the human eye. However, storing 1 bit per channel means hiding a large watermark requires many pixels, reducing capacity for very detailed images. Lossy compression (JPEG) discards high-frequency data, which often includes LSBs, destroying the watermark. To detect tampering, one can embed a fragile checksum or use robust watermarking in frequency-domain coefficients.

修改 LSB 最多使像素值变化 1，人眼无法察觉。但每个通道存储 1 比特，意味着隐藏一个大水印需要大量像素，对细节丰富的图像会降低容量。有损压缩 (JPEG) 丢弃高频数据，通常包括 LSB，会破坏水印。为检测篡改，可嵌入脆弱校验和，或在频域系数中使用鲁棒水印。

11. Regular Expressions and Language Processing | 正则表达式与语言处理

A linguistics researcher wants to extract all words from a text that start with a capital letter and are followed by one or more lowercase letters, excluding words that contain digits. Write a regular expression pattern that matches such tokens. Demonstrate your pattern on the sentence: "Dr. Smith met Queen Elizabeth II at Buckingham Palace in 2023." Explain how the pattern could be adapted to find all proper nouns that appear immediately before a Roman numeral.

一位语言学研究者想从文本中提取所有以大写字母开头、后跟一个或多个小写字母且不含数字的单词。写出匹配该类标记的正则表达式。在句子 "Dr. Smith met Queen Elizabeth II at Buckingham Palace in 2023." 上演示你的模式。解释如何调整该模式以查找紧接在罗马数字之前的所有专有名词。

Pattern: \b[A-Z][a-z]+\b. This matches a word boundary, then an uppercase letter, one or more lowercase letters, and another word boundary. Applied to the sentence, it matches "Smith", "Queen", "Elizabeth", "Buckingham", "Palace". "Dr" fails because there is no following lowercase letter; "II" fails because no lowercase; "2023" fails because digit.

模式：\b[A-Z][a-z]+\b。匹配词边界，然后一个大写字母，一个或多个小写字母，再一个词边界。应用于该句，匹配 "Smith"、"Queen"、"Elizabeth"、"Buckingham"、"Palace"。"Dr" 失败因为无后续小写字母；"II" 失败因为无小写；"2023" 失败因为数字。

To find proper nouns immediately before a Roman numeral, use a positive lookahead: \b[A-Z][a-z]+\b(?=\s[IVXLCDM]+\b). This matches a capitalised word only if followed by a space and a sequence of Roman numeral characters forming a word.

要查找紧接在罗马数字之前的专有名词，使用正向前瞻：\b[A-Z][a-z]+\b(?=\s[IVXLCDM]+\b)。仅当大写单词后跟一个空格和构成单词的罗马数字字符序列时才匹配。

12. Ethical Algorithm Design and Bias | 算法设计伦理与偏见

An AI system used for university admissions predicts student success based on historical data containing demographic features. Discuss three ways bias can enter such a model despite the absence of explicitly discriminatory variables. Propose technical and procedural mitigations, referencing concepts like feature selection, fairness constraints, and transparency. Connect your answer to the SQA curriculum's emphasis on social, legal, and ethical implications.

某大学招生使用的 AI 系统基于包含人口统计特征的历史数据来预测学生成功。讨论尽管没有明确的歧视性变量，偏见仍可能以哪三种方式进入此类模型。提出技术和程序上的缓解措施，引用特征选择、公平性约束和透明度等概念。将你的回答与 SQA 课程对社会、法律和伦理影响的强调联系起来。

Sources of bias: 1. Proxy variables: Postcode can correlate with race or socioeconomic status, acting as a proxy even if ethnicity is removed. 2. Historical bias: If past admissions favoured certain groups, the training labels reflect that inequality, which the model learns to replicate. 3. Sampling bias: Underrepresentation of minority groups in the training set leads to poorer predictive performance for those groups.

偏见来源：1. 代理变量：邮政编码可能与种族或社会经济地位相关，即使移除种族，它仍可作为代理。2. 历史偏见：如果过去的招生偏向某些群体，训练标签反映了这种不平等，模型学会复制它。3. 采样偏差：训练集中少数群体代表性不足，导致对这些群体的预测性能较差。

Mitigations: Conduct a fairness audit using metrics like demographic parity or equalised odds. Apply feature engineering to remove or transform proxies. Use adversarial debiasing during training. Procedurally, involve diverse stakeholders in model design and publish model cards for transparency. SQA requires understanding these ethical dimensions, as computing professionals must safeguard against discriminatory outcomes.

缓解措施：使用人口均等或均等几率等指标进行公平性审计。应用特征工程去除或转换代理变量。训练期间使用对抗性去偏。程序上，让多元利益相关者参与模型设计，并发布模型卡以提高透明度。SQA 要求理解这些伦理维度，因为计算专业人员必须防止歧视性后果。

Published by TutorHao | Computing Revision Series | aleveler.com

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Interdisciplinary Problem-Solving for Year 13 SQA Computer Science | Year 13 SQA 计算机：跨学科综合题型训练