📚 IB Computer Science: Calculation Drill | IB 计算机：计算题专项训练
IB Computer Science examinations often include numerical questions that test your ability to apply concepts precisely – from binary arithmetic to algorithm efficiency. This article presents a structured drill covering the most common calculation-based topics, complete with worked examples and step-by-step reasoning.
IB 计算机科学考试中常出现需要精确计算的应用题，涉及二进制运算、算法效率等。本文提供一套结构化的计算题专项训练，涵盖最常见题型，并配有详细示例与步骤讲解，帮助你在计算类题目上稳拿分数。
1. Binary and Hexadecimal Conversion | 二进制与十六进制转换
To convert a binary number to hexadecimal, group the bits into nibbles of four starting from the right (least significant bit). Each group is then replaced by its hex digit equivalent.
将二进制数转换为十六进制，从右侧最低位开始将二进制位每4位分为一组，每一组用对应的十六进制数字替换。
mple: Convert 11010110₂ to hex. Group as 1101 0110. 1101₂ = D₁₆, 0110₂ = 6₁₆, giving D6₁₆.
示例：将 11010110₂ 转为十六进制。分组为 1101 和 0110。1101₂ = D₁₆，0110₂ = 6₁₆，结果为 D6₁₆。
To convert hexadecimal to binary, reverse the process: replace each hex digit with its 4‑bit binary equivalent. For instance, A3₁₆ = 1010 0011₂.
将十六进制转为二进制则反向操作：每个十六进制数字替换为对应的4位二进制。例如 A3₁₆ = 1010 0011₂。
6 r1, 6 ÷ 2 = 3 r0, 3 ÷ 2 = 1 r1, 1 ÷ 2 = 0 r1. Reading remainders upwards gives 110101₂.
十进制转二进制采用除2取余法。53₁₀ 转二进制：53 ÷ 2 = 26 余1，26 ÷ 2 = 13 余0，13 ÷ 2 = 6 余1，6 ÷ 2 = 3 余0，3 ÷ 2 = 1 余1，1 ÷ 2 = 0 余1。余数从下往上读得 110101₂。
2. Two’s Complement Representation | 补码表示与运算
In an 8‑bit two’s complement system, the most significant bit indicates the sign (0 for positive, 1 for negative). The representable range is −128 to +127.
在8位补码系统中，最高位表示符号（0正1负），可表示范围为 −128 至 +127。
bits, then add 1. For −23: magnitude 23 = 00010111₂, invert → 11101000, add 1 → 11101001₂.
求负整数的补码：先写出其绝对值的二进制，所有位取反后加1。如 −23：绝对值 23 = 00010111₂，取反得 11101000，加1得 11101001₂。
h overflow ignored if it falls outside the representable range. For example, −35 + 22 in 8‑ (MSB=1); to check its magnitude, invert and add 1 → 00001101₂ = 13, so the result is −13.
补码加法按普通二进制加法规则进行，超出表示范围的溢出忽略。例如 −35 + 22 在8位系统：−35 为 11011101₂，22 为 00010110₂，和为 11110011₂。结果为负（最高位1），验证其值：取反加1得 00001101₂ = 13，故结果为 −13。
3. Logic Gates and Truth Tables | 逻辑门与真值表
(A AND B) OR (NOT C), a truth table exhaustively lists all input combinations and calculates the output.
对于给定的布尔表达式，如 F = (A AND B) OR (NOT C)，真值表列出所有输入组合并计算相应的输出。
| A | B | C | A AND B | NOT C | F |
| 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 |
The truth table shows that F is 1 when C is 0, regardless of A and B, or when both A and B are 1 while C is 1. This is a typical pattern that can be simplified using a Karnaugh map.
真值表表明，当 C 为 0 时无论 A、B 取值 F 都为 1，此外在 C=1 且 A=B=1 时 F 也为 1。这是可用卡诺图化简的典型模式。
4. Boolean Algebra Simplification | 布尔代数化简
The absorption law states A + A·B = A. Proof: A + A·B = A·(1 + B) = A·1 = A.
吸收律：A + A·B = A。证明：A + A·B = A·(1 + B) = A·1 = A。
A + A·C + B·A + B·C = A + A·C + A·B + B·C. Then factor A: = A·(1 + C + B) + B·C = A + B·C.
化简 (A + B)(A + C)。展开
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