📚 Year 10 CCEA Computer Science Formula & Theorem Quick Reference | CCEA Year 10 计算机公式定理速查手册
This quick reference handbook compiles the essential formulas, conversion methods, logic laws, and computational theorems required for the CCEA Year 10 Computer Science course. From data storage units to algorithm efficiency, every key calculation is presented with concise explanations and paired examples in English and Chinese to support bilingual revision.
本速查手册汇总了 CCEA Year 10 计算机科学课程所需的核心公式、转换方法、逻辑定律和计算定理。从数据存储单位到算法效率，每一个关键计算都配有简短的中英文解释和示例，助力双语复习。
1. Data Storage Units | 数据存储单位换算
All digital data is stored as sequences of bits (binary digits). Since a single bit can only represent a 0 or 1, larger units are built upon powers of two. The most common progression is bit, byte, kilobyte, megabyte, gigabyte, and terabyte. In computing, storage is defined using binary prefixes where 1 kilobyte equals 2¹⁰ bytes, not 1000 bytes.
所有数字数据都以位（二进制数字）序列存储。由于一个位只能表示0或1，更大的单位建立在2的幂次基础上。最常见的递增单位是位、字节、千字节、兆字节、吉字节和太字节。在计算领域，存储容量使用二进制前缀，1千字节等于2¹⁰字节，而不是1000字节。
1 byte = 8 bits
1 KB = 2¹⁰ bytes = 1024 bytes
1 MB = 1024 KB
1 GB = 1024 MB
1 TB = 1024 GB
24 × 1024 KB = 2,097,152 KB. Use the binary scale for all CCEA exam calculations unless explicitly told to use decimal units.
在这些单位之间转换时，向上或向下换算总是乘以或除以1024。例如，一个2 GB的文件相当于2 × 1024 × 1024 KB = 2,097,152 KB。除非题目明确要求使用十进制单位，否则所有CCEA考试中的计算都使用二进制换算。
2. Image File Size Calculation | 图像文件大小计算
An uncompressed bitmap image stores the colour of every pixel individually. The total number of pixels is the product of width and height, and each pixel uses a fixed number of bits—the colour depth. Common colour depths are 1 bit (black and white), 8 bits (256 colours), and 24 bits (true colour, 16.7 million colours). The file size in bytes is obtained by dividing the total bits by 8.
未压缩的位图图像会单独存储每个像素的颜色。像素总数是宽度与高度的乘积，每个像素使用固定数量的位数——颜色深度。常见的颜色深度有1位（黑白）、8位（256色）和24位（真彩色，1670万色）。文件大小的字节数由总位数除以8得到。
Image size (bytes) = (width × height × colour depth in bits) ÷ 8
For an image with dimensions 800 × 600 and a colour depth of 24 bits: Size = (800 × 600 × 24) ÷ 8 = 1,440,000 bytes ≈ 1.37 MB. Increasing the colour depth or resolution directly increases the file size, making this formula vital for estimating storage requirements.
对于尺寸为800 × 600、颜色深度为24位的图像：大小 = (800 × 600 × 24) ÷ 8 = 1,440,000字节 ≈ 1.37 MB。增加颜色深度或分辨率会直接增大文件大小，因此该公式对于估算存储需求至关重要。
3. Sound File Size Calculation | 声音文件大小计算
Digital audio is recorded by sampling the sound wave at regular intervals. The key parameters are the sample rate (measured in Hz, e.g. 44,100 Hz for CD quality), the bit depth (number of bits per sample, e.g. 16), the number of channels (1 for mono, 2 for stereo), and the duration in seconds. The uncompressed file size is the product of these factors divided by 8 to convert bits into bytes.
数字音频通过对声波进行等间隔采样来录制。关键参数包括采样率（以Hz为单位，例如CD质量为44,100 Hz）、位深度（每样本位数，如16）、声道数（单声道为1，立体声为2）和持续时间（秒）。未压缩文件的大小为这些因子的乘积除以8，将位转换为字节。
Sound size (bytes) = (sample rate × bit depth × channels × duration) ÷ 8
A 60-second CD-quality stereo recording has: Size = (44100 × 16 × 2 × 60) ÷ 8 = 10,584,000 bytes ≈ 10.1 MB. Higher sample rates and bit depths improve fidelity but produce larger files, a trade-off frequently examined in CCEA papers.
一段60秒的CD质量立体声录音：大小 = (44100 × 16 × 2 × 60) ÷ 8 = 10,584,000字节 ≈ 10.1 MB。更高的采样率和位深度能提升保真度，但会产生更大的文件，这是CCEA试卷中经常考察的权衡点。
4. Binary-Decimal Conversions | 二进制与十进制转换
of 2. To convert a binary number to decimal, multiply each bit by its place value (2⁷, 2⁶, …, 2⁰) and sum the results. The rightmost bit holds the lowest place value (2⁰ = 1).
二进制数只使用数字0和1，每一位的位置代表2的幂。要将二进制转换为十进制，可将每个位乘以其位值（2⁷、2⁶……2⁰）并求和。最右边的位具有最低的位值（2⁰ = 1）。
1101₂ = 1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8 + 4 + 0 + 1 = 13₁₀
o form the binary equivalent. For 13: 13÷2=6 rem 1, 6÷2=3 rem 0, 3÷2=1 rem 1, 1÷2=0 rem 1, giving 1101₂. Always verify by converting back.
要将十进制转换为二进制，反复将十进制数除以2，记录余数。从下往上读取余数以形成二进制数。对于13：13÷2=6余1，6÷2=3余0，3÷2=1余1，1÷2=0余1，得到1101₂。务必通过反向转换来验证。
5. Hexadecimal Conversions | 十六进制转换公式
Hexadecimal (base 16) uses digits 0–9 and letters A–F to represent values 10–15. Each hex digit corresponds exactly to a 4-bit binary nibble, which makes hex a compact way to write binary. Converting between binary and hex is simply grouping bits into fours from the right.
十六进制（基数为16）使用数字0–9和字母A–F表示10–15。每个十六进制位恰好对应一个4位的二进制半字节，这使得十六进制成为书写二进制的紧凑形式。二进制与十六进制之间的转换只需从右起将位每四位一组。
2A₁₆ = 2×16¹ + A×16⁰ = 2×16 + 10×1 = 32 + 10 = 42₁₀
d the binary number so it has a multiple of 4 bits, then replace each group: 1010₂ = A₁₆, 0010₂ = 2₁₆, so 10100010₂ = A2₁₆. This shortcut is frequently tested.
将十六进制转换为十进制时，将每位数字乘以16的对应次幂，从右侧的0次幂开始。将二进制转换为十六进制时，填补二进制数使其位数为4的倍数，然后替换每一组：1010₂ = A₁₆，0010₂ = 2₁₆，因此10100010₂ = A2₁₆。这种快捷转换经常在考试中出现。
6. Logic Gates & Boolean Algebra | 逻辑门与布尔代数
Logic gates implement Boolean functions on binary inputs. The fundamental gates are AND, OR, NOT, NAND, NOR, and XOR. Their behaviour is defined by truth tables that list output for every combination of inputs. In Boolean notation, AND is represented by '∧', OR by '∨', and NOT by '¬'.
逻辑门对二进制输入实现布尔函数。基本门包括与门、或门、非门、与非门、或非门和异或门。它们的行为由真值表定义，列出了每种输入组合下的输出。在布尔表示法中，与用 '∧' 表示，或用 '∨' 表示，非用 '¬' 表示。
| A | B | A ∧ B | A ∨ B | ¬A | NAND | NOR | XOR |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Boolean algebra provides laws to simplify logic expressions. The table below summarises the most important identities and laws, including De Morgan's theorems, which are essential for circuit minimisation and for tasks such as converting between NAND/NOR implementations.
布尔代数提供了化简逻辑表达式的定律。下表总结了最重要的恒等式和定律，包括德摩根定理，这对于电路最小化以及转换与非/或非门实现等任务至关重要。
| Law / 定律 | AND form | OR form |
|---|---|---|
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