IB and Edexcel Computer Science: Calculation Drill | IB 与 Edexcel 计算机：计算题专项训练

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

This article provides a focused drill on the calculation-style questions commonly found in IB and Edexcel Computer Science exams. From binary conversions to algorithm efficiency, mastering these quantitative techniques is essential for top scores. Each section introduces a key topic, step-by-step methods, and worked examples.

本文针对 IB 和 Edexcel 计算机科学考试中常见的计算题型进行专项训练，涵盖二进制转换、算法效率等量化技巧。掌握这些计算方法是获取高分的关键。每节介绍一个核心主题，提供分步方法与具体示例。

1. Number Systems & Conversions | 数制与转换

The binary (base-2), decimal (base-10), and hexadecimal (base-16) systems are fundamental. To convert a binary number to decimal, sum the products of each bit and its place value (power of 2). For example, binary 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8+0+2+1 = 11₁₀.

二进制（基2）、十进制（基10）和十六进制（基16）是基础。将二进制数转换为十进制，需将每一位乘以对应的位权（2的幂次）并求和。例如，二进制 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8+0+2+1 = 11₁₀。

To convert decimal to binary, repeatedly divide the number by 2, recording remainders. Read the remainders from bottom to top. For 25₁₀: 25÷2=12 r1, 12÷2=6 r0, 6÷2=3 r0, 3÷2=1 r1, 1÷2=0 r1 → 11001₂. For hexadecimal, group binary digits in sets of four. e.g., 1101 0110₂ = D6₁₆.

十进制转二进制：反复除以2并记录余数，从下往上读余数。例如 25₁₀：25÷2=12 余1，12÷2=6 余0，6÷2=3 余0，3÷2=1 余1，1÷2=0 余1 → 11001₂。二进制转十六进制：将二进制数四位一组分组。如 1101 0110₂ = D6₁₆。

2. Integer Representation: Two’s Complement | 整数表示：补码

Two’s complement is the standard method for representing signed integers. In an n-bit system, the most significant bit is the sign bit (0 for positive, 1 for negative). To find the two’s complement of a negative number, invert all bits of its positive magnitude and add 1. Example: in 8 bits, represent -13. Positive 13 is 0000 1101. Invert to 1111 0010, add 1 → 1111 0011. So -13₁₀ = 1111 0011 (two’s complement).

补码是表示有符号整数的标准方法。在 n 位系统中，最高位为符号位（0 正，1 负）。求负数的补码：将其绝对值的二进制按位取反后加 1。例如，用 8 位表示 -13。正数 13 为 0000 1101。取反得 1111 0010，加 1 → 1111 0011。因此 -13₁₀ = 1111 0011（补码）。

To convert a negative two’s complement binary back to decimal, note the sign bit is 1, indicating negative. Invert the bits and add 1 to find the magnitude, then apply a negative sign. For 1111 0011, invert → 0000 1100, add 1 → 0000 1101 = 13, so the value is -13. The range for n-bit two’s complement is -2ⁿ⁻¹ to 2ⁿ⁻¹ – 1.

将负数的补码转回十进制：符号位为 1 表示负数，对数值位取反加 1 得到绝对值，再加负号。如 1111 0011，取反得 0000 1100，加 1 → 0000 1101 = 13，因此数值为 -13。n 位补码的表示范围为 -2ⁿ⁻¹ 到 2ⁿ⁻¹ – 1。

3. Floating-Point Representation | 浮点数表示

Floating-point numbers follow the IEEE 754 standard: sign (1 bit), exponent (8 bits for single precision, bias 127), and mantissa (23 bits, with implicit leading 1). To convert a decimal like 9.75 to binary floating point: binary of 9.75 = 1001.11₂ = 1.00111 × 2³. Sign = 0, exponent = 3 + 127 = 130 → 10000010₂, mantissa = 0011100…0. The 32-bit representation is 0 10000010 00111000000000000000000.

浮点数遵循 IEEE 754 标准：符号位（1 位）、指数位（单精度 8 位，偏移量 127）和尾数（23 位，有一个隐含的 1）。将十进制数 9.75 转换为二进制浮点数：9.75 的二进制为 1001.11₂ = 1.00111 × 2³。符号 = 0，指数 = 3 + 127 = 130 → 10000010₂，尾数 = 001110…0。32 位表示为 0 10000010 00111000000000000000000。

In the exam, you may need to calculate the decimal value from a binary IEEE 754 pattern. Parse sign, exponent, mantissa. Exponent actual = exponent field – 127. Value = (-1)^sign × (1 + mantissa fraction) × 2^(actual exponent). Practice with edge cases like denormalized numbers.

考试中可能需要根据 IEEE 754 二进制模式计算十进制值。解析符号、指数、尾数。实际指数 = 指数字段 – 127。值 = (-1)^符号 × (1 + 尾数小数部分) × 2^(实际指数)。同时练习非规格化数等特殊情况。

4. Boolean Algebra & Logic Gate Simplification | 布尔代数与逻辑门化简

Boolean algebra uses operators AND (·), OR (+), NOT (‾). Simplifying expressions reduces gate count. Apply laws like De Morgan’s, distribution, absorption. Example: Simplify A·B + A·(B+C). = A·B + A·B + A·C = A·B + A·C = A·(B+C). Use truth tables to verify equivalence.

布尔代数使用与 (·)、或 (+)、非 (‾

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