IB & AQA Computer Science Calculation Drill | IB与AQA计算机：计算题专项训练

📚 IB & AQA Computer Science Calculation Drill | IB与AQA计算机：计算题专项训练

Mastering calculation-based questions is key to excelling in IB and AQA Computer Science exams. This guide covers essential computational skills, from number systems and floating-point arithmetic to networking and algorithm analysis, with step-by-step examples.

掌握计算题对于IB和AQA计算机科学考试至关重要。本文涵盖核心计算技能，包括数制、浮点运算、网络和算法分析，并提供分步例题。

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1. Number Base Conversions | 进制转换

Converting between binary, decimal, and hexadecimal is fundamental. Binary to decimal uses positional weights (powers of 2). Decimal to hex employs repeated division by 16, reading remainders backwards.

二进制、十进制和十六进制间的转换是基础。二进制转十进制使用位权(2的幂)。十进制转十六进制采用反复除以16，逆序读余数。

Example: convert 11010101₂ to decimal.

示例：将11010101₂转为十进制。

Step 1: Write place values 128, 64, 32, 16, 8, 4, 2, 1 under each bit.

步骤1：在每个二进制位下写出位值128, 64, 32, 16, 8, 4, 2, 1。

Step 2: Sum values where the bit is 1: 128 + 64 + 16 + 4 + 1 = 213₁₀.

步骤2：将位为1的位值相加：128 + 64 + 16 + 4 + 1 = 213₁₀。

Convert 200₁₀ to hex: 200 ÷ 16 = 12 remainder 8. 12 is C, so C8₁₆. In binary, C8₁₆ = 1100 1000₂.

将200₁₀转为十六进制：200 ÷ 16 = 12 余 8，12为C，得到C8₁₆。二进制为1100 1000₂。

128	64	32	16	8	4	2	1
1	1	0	1	0	1	0	1

Every hex digit maps to 4 bits, so conversions like 3F₁₆ = 0011 1111₂ are quick.

每个十六进制位对应4个二进制位，例如3F₁₆ = 0011 1111₂，转换快捷。

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2. Binary Arithmetic & Two’s Complement | 二进制运算与补码

Addition follows column-wise carrying. For 1011₂ + 1101₂, align bits and add, carrying 1 when sum ≥ 2. The result is 11000₂ (24₁₀).

加法按位对齐进位。计算1011₂ + 1101₂，对应位相加，和≥2时进位，结果为11000₂ (24₁₀)。

Subtraction uses two’s complement. To compute 5 – 3 in 8-bit: 5 is 00000101, -3 is two’s complement of 00000011 → flip bits 11111100, add 1 → 11111101.

减法使用二补数。8位下计算5 – 3：5为00000101，-3是00000011的补码 → 各位取反11111100，加1得11111101。

Add: 00000101 + 11111101 = (1)00000010. Ignore the carry beyond 8 bits; result 00000010 = 2₁₀.

相加：00000101 + 11111101 = (1)00000010，忽略超出8位的进位，结果为00000010 = 2₁₀。

Overflow occurs when operands have same sign and the result’s sign differs. With 4-bit signed numbers: 0111 (7) + 0001 (1) = 1000 (-8), which is overflow.

溢出发生在操作数同号而结果异号时。4位有符号数：0111 (7) + 0001 (1) = 1000 (-8)，发生溢出。

Key formula: -X = (2ⁿ – X) in n-bit two’s complement.

核心公式：n位补码下，-X = (2ⁿ – X)。

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3. Floating-Point Representation | 浮点数表示

IEEE 754 single precision uses 32 bits: 1 sign bit, 8 exponent bits (bias 127), and 23 mantissa bits. Convert -13.75 to this format.

IEEE 754单精度使用32位：1位符号，8位指数(偏移127)，23位尾数。将-13.75转为该格式。

Step 1: Sign bit = 1 for negative.

步骤1：负号，符号位 = 1。

Step 2: Write 13.75 in binary. 13₁₀ = 1101₂, 0.75×2=1.5→1, 0.5×2=1.0→1, so 0.75 = .11₂, giving 1101.11₂.

步骤2：写出13.75的二进制。13₁₀ = 1101₂，0.75×2=1.5取1，0.5×2=1.0取1，得.11₂，合为1101.11₂。

Step 3: Normalise to 1.10111 × 2³. True exponent = 3. Stored exponent = 3 + 127 = 130 = 10000010₂.

步骤3：规范化为1.10111 × 2³。真实指数 = 3，存储指数 = 3 + 127 = 130 = 10000010₂。

Step 4: Mantissa = 10111000000000000000000 (23 bits, drop the leading 1).

步骤4：尾数 = 10111000000000000000000（23位，去掉前面的1）。

Final representation: 1 10000010 10111000000000000000000.

最终表示：1 10000010 10111000000000000000000。

Decimal → binary fraction: multiply fractional part by 2 repeatedly, record integer parts.

十进制小数转二进制：反复乘2取整。

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4. Logic Gates & Truth Tables | 逻辑门与真值表

Given a circuit: Y = (A AND B) OR (NOT C). Calculate Y for the inputs A=1, B=0, C=1.

给定电路：Y = (A AND B) OR (NOT C)。计算输入A=1, B=0, C=1时Y的值。

Step 1: A AND B = 1 AND 0 = 0.

步骤1：A AND B = 1 AND 0 = 0。

Step 2: NOT C = NOT 1 = 0.

步骤2：NOT C = NOT 1 = 0。

Step 3: 0 OR 0 = 0, so Y = 0.

步骤3：0 OR 0 = 0，因此Y = 0。

To build a full truth table, list all 2³ = 8 input combinations and evaluate stepwise. This skill is frequent in exam calculation tasks.

构建完整真值表需列出所有2³=8种输入组合并逐项求值。这种题型在考试计算中很常见。

You can also simplify expressions using Boolean algebra. For instance, A + A’B = A + B (absorption). Apply it to reduce gate counts.

也可用布尔代数化简表达式，例如 A + A’B = A + B（吸收律），用于减少门数。

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5. Bitwise Operations & Masking | 位运算与掩码

Bitwise AND, OR, XOR, NOT, and shifts operate on individual bits. Apply mask 00001111 to byte 10110110 using AND.

按位与、或、异或、非以及移位操作针对单个位。对字节10110110应用掩码000

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