CCEA Year 11 Computer Science Formula & Theorem Quick Reference | CCEA 11年级计算机科学公式定理速查手册

📚 CCEA Year 11 Computer Science Formula & Theorem Quick Reference | CCEA 11年级计算机科学公式定理速查手册

This quick reference handbook brings together the essential formulas, theorems, and calculation methods you will encounter in CCEA Year 11 Computer Science. From binary conversions and file size estimation to Boolean logic and algorithm efficiency, use these concise explanations and worked examples to strengthen your revision and boost exam confidence.

这本速查手册汇集了CCEA 11年级计算机科学课程中您将遇到的关键公式、定理和计算方法。从二进制转换、文件大小估算，到布尔逻辑和算法效率，利用这些简洁的解释和示例来巩固复习、增强考试信心。

1. Binary to Decimal Conversion | 二进制转十进制

To convert a binary number to decimal, multiply each bit by 2 raised to the power of its position (starting from 0 on the right) and sum the results.

要将二进制数转换为十进制，请将每个位乘以2的位权次幂（从右端开始，位置编号为0），然后求和。

Formula: Decimal = dₙ₋₁ × 2ⁿ⁻¹ + … + d₂ × 2² + d₁ × 2¹ + d₀ × 2⁰

公式：十进制值 = dₙ₋₁ × 2ⁿ⁻¹ + … + d₂ × 2² + d₁ × 2¹ + d₀ × 2⁰

Example: 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11₁₀

示例：1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11₁₀

Always label the base (subscript 2 for binary, 10 for decimal) to avoid confusion in exam answers.

为了在考试答题时避免混淆，务必标注基数（二进制下标2，十进制下标10）。

2. Decimal to Binary Conversion | 十进制转二进制

Use repeated division by 2, recording the remainder each time. The binary result is the sequence of remainders read from bottom to top.

使用连续除以2的方法，记录每次的余数。从下往上读出的余数序列就是二进制结果。

Algorithm: While the number is greater than 0, divide by 2, note quotient and remainder; continue dividing the quotient.

算法：当数字大于0时，反复除以2，记录商和余数；继续除商。

Example: Convert 13 to binary. 13÷2 = 6 rem 1, 6÷2 = 3 rem 0, 3÷2 = 1 rem 1, 1÷2 = 0 rem 1. Read remainders upwards: 1101₂.

示例：将13转换为二进制。13÷2=6余1，6÷2=3余0，3÷2=1余1，1÷2=0余1。从下往上读余数：1101₂。

Checking with the conversion formula confirms accuracy: 1101₂ = 8+4+0+1 = 13. Always verify your answer in exams if time permits.

用转换公式验证即可确认准确性：1101₂ = 8+4+0+1 = 13。考试时若时间允许，一定要验算。

3. Hexadecimal Conversions | 十六进制转换

Hexadecimal (base 16) uses digits 0-9 and letters A-F (A=10, B=11, …, F=15). Each hex digit represents a group of 4 binary bits (nibble).

十六进制（基数为16）使用数字0-9和字母A-F（A=10，B=11，…，F=15）。每个十六进制位代表4个二进制位（半字节）。

Binary to hex: split the binary number into groups of 4 bits from the right, convert each group. Hex to binary: replace each hex digit with its 4-bit binary equivalent.

二进制转十六进制：从右到左将二进制数4位一组进行分割，分别转换每组。十六进制转二进制：将每个十六进制数字替换为其4位二进制等价形式。

Table: 0:0000, 1:0001, 2:0010, 3:0011, 4:0100, 5:0101, 6:0110, 7:0111, 8:1000, 9:1001, A:1010, B:1011, C:1100, D:1101, E:1110, F:1111

对照表：0:0000，1:0001，2:0010，3:0011，4:0100，5:0101，6:0110，7:0111，8:1000，9:1001，A:1010，B:1011，C:1100，D:1101，E:1110，F:1111

Hex to decimal: multiply each digit by 16^position and add. Decimal to hex: repeated division by 16, using remainders (convert 10-15 to A-F).

十六进制转十进制：将每位数字乘以16的位权次幂然后求和。十进制转十六进制：连续除以16，使用余数（将10-15转换为A-F）。

4. Data Storage Units & File Size Conversion | 数据存储单位与文件大小换算

All file size calculations rely on converting bits to bytes and applying the standard binary prefixes.

所有文件大小的计算都依赖于将位转换为字节并应用标准的二进制前缀。

Key equivalence: 1 byte = 8 bits. Larger units: 1 KB (kibibyte) = 1024 bytes, 1 MB = 1024 KB, 1 GB = 1024 MB, 1 TB = 1024 GB. In CCEA exams, 1 KB is normally taken as 1024 bytes unless stated otherwise.

关键等量关系：1字节 = 8位。更大的单位：1 KB（千字节）= 1024字节，1 MB = 1024 KB，1 GB = 1024 MB，1 TB = 1024 GB。在CCEA考试中，一般将1 KB视作1024字节，除非另有说明。

Conversion factor: to go from a smaller unit to a larger unit, divide by 1024; from larger to smaller, multiply by 1024.

转换因子：从小单位转换为大单位，除以1024；从大单位转换为小单位，乘以1024。

Use this table for quick reference:

快速查阅此表：

Unit	Equivalent in bytes	中文
1 bit (b)	0.125 bytes	1位 = 0.125字节
1 byte (B)	8 bits	1字节 = 8位
1 KB	1024 B	1千字节 = 1024字节
1 MB	1024 KB	1兆字节 = 1024千字节
1 GB	1024 MB	1吉字节 = 1024兆字节

5. Image File Size Calculation | 图像文件大小计算

The size of a bitmap image depends on its resolution (pixel dimensions) and colour depth (bits per pixel).

位图图像的大小取决于其分辨率（像素维度）和色彩深度（每像素位数）。

Formula: File size (bits) = width (pixels) × height (pixels) × colour depth (bits per pixel)

公式：文件大小（位）= 宽度（像素）× 高度（像素）× 色彩深度（每像素位数）

To express in bytes, divide the bit total by 8. For kilobytes, divide further by 1024.

要以字节表示，将总位数除以8。要进一步换算为千字节，再除以1024。

Example: a 300×200 pixel image with 24-bit colour. Bits = 300 × 200 × 24 = 1,440,000 bits. Bytes = 1,440,000 ÷ 8 = 180,000 B = about 175.78 KB.

示例：一幅300×200像素、24位色彩的图像。位数 = 300 × 200 × 24 = 1,440,000位。字节 = 1,440,000 ÷ 8 = 180,000 B ≈ 175.78 KB。

Increasing colour depth or resolution increases file size proportionally. Watch out for metadata overhead; in CCEA questions, you often ignore header data unless specified.

提高色彩深度或分辨率会按比例增大文件大小。注意元数据开销；在CCEA考题中，除非明确说明，通常忽略文件头数据。

6. Sound File Size Calculation | 声音文件大小计算

Digital audio file size is determined by sampling rate, sample resolution (bit depth), duration, and number of channels.

数字音频文件大小由采样率、采样分辨率（位深度）、持续时间和声道数决定。

Formula: File size (bits) = sampling rate (Hz) × bit depth × duration (seconds) × number of channels

公式：文件大小（位）= 采样率（Hz）× 位深度 × 持续时间（秒）× 声道数

Then convert bits to bytes, KB, MB as needed. For stereo, channels = 2; for mono, channels = 1.

然后根据需要将位转换为字节、KB、MB。立体声时声道数=2，单声道时声道数=1。

Example: CD-quality audio (44.1 kHz, 16-bit, stereo) for 60 seconds: bits = 44,100 × 16 × 60 × 2 = 84,672,000 bits = 10,584,000 bytes ≈ 10.09 MB.

示例：CD品质音频（44.1 kHz，16位，立体声）持续60秒：位数 = 44,100 × 16 × 60 × 2 = 84,672,000位 = 10,584,000字节 ≈ 10.09 MB。

Remember that 1 kHz = 1000 Hz. In some exam contexts, ‘sample rate’ is given in kHz – multiply by 1000 before using the formula.

记住1 kHz = 1000 Hz。在有些试题语境中，“采样率”以kHz给出——使用公式前先乘以1000。

7. Data Transfer Time Calculation | 数据传输时间计算

Transfer time depends on file size and the transfer rate (bit rate). It is a direct application of the speed–distance–time relationship.

传输时间取决于文件大小和传输速率（比特率）。它是速度–距离–时间关系的直接应用。

Formula: Time (seconds) = File size (bits) ÷ Transfer rate (bits per second, bps)

公式：时间（秒）= 文件大小（位）÷ 传输速率（位每秒，bps）

Make sure both measurements are in the same unit. Convert file size to bits and bit rate to bps if necessary.

确保两个度量单位一致。如有必要，将文件大小转换为位，将比特率转换为bps。

Example: A 2 MB file over a 4 Mbps connection. File size = 2 × 1024 × 1024 × 8 = 16,777,216 bits. Rate = 4,000,000 bps. Time = 16,777,216 ÷ 4,000,000 ≈ 4.19 seconds.

示例：通过4 Mbps连接传输一个2 MB的文件。文件大小 = 2 × 1024 × 1024 × 8 = 16,777,216位。速率 = 4,000,000 bps。时间 = 16,777,216 ÷ 4,000,000 ≈ 4.19秒。

For network speeds, 1 Mbps = 1,000,000 bps (decimal) is usually assumed, but check the question for the specific convention.

对于网络速度，通常假定1 Mbps = 1,000,000 bps（十进制），但要留意考题所采用的具体约定。

8. Boolean Logic Identities & Theorems | 布尔逻辑恒等式与定理

Boolean algebra forms the backbone of digital circuit simplification. Knowing the key identities helps you reduce expressions and solve logic gate problems.

布尔代数构成了数字电路简化的基础。了解关键恒等式有助于化简表达式和解决逻辑门问题。

Basic laws (A, B, C represent Boolean variables, · is AND, + is OR, ‘ is NOT):

基本定律（A, B, C代表布尔变量，·表示与，+表示或，’表示非）：

Commutative / 交换律: A·B = B·A, A+B = B+A
Associative / 结合律: (A·B)·C = A·(B·C), (A+B)+C = A+(B+C)
Distributive / 分配律: A·(B+C) = A·B + A·C, A+(B·C) = (A+B)·(A+C)
Identity / 同一律: A·1 = A, A+0 = A
Complement / 互补律: A·A’ = 0, A+A’ = 1
Idempotent / 幂等律: A·A = A, A+A = A
Double negation / 双重否定律: (A’)’ = A
Absorption / 吸收律: A+(A·B) = A, A·(A+B) = A
De Morgan’s Laws / 德摩根定律: (A+B)’ = A’·B’, (A·B)’ = A’+B’

These identities are crucial for simplifying logic circuits. In CCEA exams, you might be asked to draw a simplified logic diagram from a given expression.

这些恒等式对于简化逻辑电路至关重要。在CCEA考试中，可能会要求您根据给定的表达式画出简化的逻辑图。

Truth tables for AND, OR, NOT should be memorised. Practice using De Morgan’s laws to transform NAND and NOR gates.

应熟记与门、或门、非门的真值表。通过练习运用德摩根定律来转换与非门和或非门。

9. Addressable Memory & Address Bus Width | 可寻址内存与地址总线宽度

The number of unique memory locations a CPU can address is determined by the width of the address bus. This directly limits maximum system memory.

CPU能够寻址的唯一内存单元数量由地址总线宽度决定。这直接限制了最大系统内存。

Formula: Number of addressable locations = 2ⁿ, where n is the width of the address bus in bits.

公式：可寻址单元数 = 2ⁿ，其中n是地址总线的宽度（以位为单位）。

If each memory location stores 1 byte, then the total memory capacity = 2ⁿ bytes.

如果每个内存单元存储1个字节，则总内存容量 = 2ⁿ 字节。

Example: a 16-bit address bus can address 2¹⁶ = 65,536 different locations (64 KB). A 32-bit address bus gives 2³² = 4,294,967,296 locations (4 GB).

示例：16位地址总线可寻址 2¹⁶ = 65,536 个不同单元（64 KB）。32位地址总线则可寻址 2³² = 4,294,967,296 个单元（4 GB）。

In some systems, word size may be larger than a byte, so multiply the number of locations by word size to get total capacity in bits.

在某些系统中，字长可能大于1字节，因此要用单元数乘以字长来得到总容量（以位为单位）。

10. Compression Ratio Calculation | 压缩比计算

Compression reduces file size for storage or transmission. The compression ratio quantifies the effectiveness.

压缩是为了存储或传输而缩减文件大小。压缩比定量衡量其效果。

Formula: Compression ratio = Uncompressed size ÷ Compressed size

公式：压缩比 = 未压缩大小 ÷ 压缩后大小

A ratio of 5:1 means the compressed file is one fifth of the original size. Space saving percentage = (1 – 1/ratio) × 100%.

5:1的比率意味着压缩后的文件是原始文件大小的五分之一。空间节省百分比 = (1 – 1/ratio) × 100% 。

Example: an original image is 2 MB, compressed to 500 KB. Ratio = 2048 KB ÷ 500 KB = 4.096:1, approximately 4:1. Space saving ≈ 75%.

示例：原始图像为2 MB，压缩后为500 KB。比率 = 2048 KB ÷ 500 KB = 4.096:1，约4:1。空间节省约75%。

CCEA often asks you to calculate the compressed file size given the original size and compression ratio. Rearrange accordingly: Compressed size = Original size ÷ Ratio.

CCEA常会要求您根据原始大小和压缩比计算压缩后的文件大小。相应地变换公式：压缩后大小 = 原始大小 ÷ 压缩比。

11. Algorithm Efficiency: Search & Sort Comparisons | 算法效率：搜索与排序比较

Understanding the maximum number of comparisons helps evaluate algorithm performance. For a dataset of n items:

了解最大比较次数有助于评估算法性能。对于一个有n个项的数据集：

Linear search / 线性搜索: maximum comparisons = n (worst case, item is at the end or not present).
线性搜索：最大比较次数 = n（最坏情况，目标项在末尾或不存在）。
Binary search / 二分搜索: maximum comparisons ≈ ⌈log₂(n)⌉ (on a sorted list). To calculate, repeatedly ask: 2^k ≥ n; k is the maximum steps.
二分搜索：最大比较次数 ≈ ⌈log₂(n)⌉（在有序列表上）。计算时反复判断：2^k ≥ n；k即为最大步数。

Example: 1000 items, linear search up to 1000 comparisons; binary search at most ceil(log₂(1000)) = 10 steps (since 2¹⁰ = 1024).

示例：1000个项，线性搜索最多1000次比较；二分搜索最多ceil(log₂(1000)) = 10步（因为2¹⁰ = 1024）。

For sorting: Bubble sort requires (n-1) passes, but the number of comparisons is n(n-1)/2 in the worst case. However, for CCEA Year 11, focus on identifying which algorithm is more efficient for a given scenario.

对排序而言：冒泡排序需要(n-1)趟，但最坏情况下的比较次数为n(n-1)/2。不过CCEA 11年级的重点是判断在给定场景下哪种算法效率更高。

12. Quick Exam Tips & Common Pitfalls | 考试快速提示与常见误区

Always write out the formula before substituting values. Showing the steps earns method marks even if the final answer slips.

务必在代入数值前写出公式。展示解题步骤能赢得过程分，即使最终答案出错。

Double-check unit conversions: bits to bytes divide by 8; bytes to KB divide by 1024. Do not mix 1000 and 1024 unless the question explicitly states it.

仔细复核单位转换：位转字节除以8；字节转KB除以1024。除非考题明确说明，否则不要混用1000和1024。

For Boolean simplification, verify your result with a truth table. It is the safest way to confirm identities like De Morgan’s.

对于布尔化简，用真值表验证结果。这是确认诸如德摩根定律这类恒等式最稳妥的方法。

Practice reading binary and hexadecimal quickly – it saves time on the data representation questions. Remember the nibble grouping rule.

练习快速认读二进制和十六进制——这会在数据表示题上节省时间。记住半字节分组规则。

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CCEA Year 11 Computer Science Formula & Theorem Quick Reference | CCEA 11年级计算机科学公式定理速查手册