📚 Year 8 CIE Computer Science: Quick Reference Handbook for Formulas and Key Concepts | Year 8 CIE 计算机：公式定理速查手册
This quick reference handbook brings together the most important formulas, conversions, and core principles you will encounter in Year 8 CIE Computer Science. From data units and binary arithmetic to image file size and logic gates, every key idea is presented in a clear, side‑by‑side bilingual format. Use these pages as a revision checklist, a homework helper, or an exam‑day memory jogger. Each section delivers a bite‑sized rule followed by a real‑world example, so you can move from theory to practice without confusion.
这本速查手册汇集了你在八年级 CIE 计算机课程中会遇到的最重要的公式、转换方法和核心原理。从数据单位、二进制运算到图像文件大小和逻辑门，每一个关键概念都用清晰的中英对照形式呈现。你可以把它当作复习清单、作业帮手或考试当天的记忆助手。每个小节先给出简洁的规则，再配上一个实际例子，让你轻松从理论过渡到实践。
1. Data Units and Storage Capacity | 数据单位与存储容量
The smallest unit of data is a bit (b), which stores a single binary value, 0 or 1. Eight bits form one byte (B). Storage capacity is measured by grouping bytes into larger units based on powers of 2. The main units you need to remember are kilobyte, megabyte, gigabyte, and terabyte. When converting between units, multiply by 1024 for each step up, and divide by 1024 for each step down.
数据的最小单位是比特（b），它存储一个二进制值 0 或 1。八个比特组成一个字节（B）。存储容量以 2 的幂次为基础将字节组合成更大的单位。你需要记住的主要单位是千字节、兆字节、吉字节和太字节。单位之间向上转换时每一步乘以 1024，向下转换时每一步除以 1024。
- 1 byte = 8 bits
- 1 KB = 1024 bytes
- 1 MB = 1024 KB
- 1 GB = 1024 MB
- 1 TB = 1024 GB
- 1 字节 = 8 比特
- 1 KB = 1024 字节
- 1 MB = 1024 KB
- 1 GB = 1024 MB
- 1 TB = 1024 GB
Example: To convert 5 GB into MB, multiply by 1024 twice: 5 × 1024 = 5120 MB; 5120 × 1024 = 5,242,880 KB. For a quick check, 5 GB = 5 × 1024 = 5120 MB directly, because each gigabyte contains 1024 megabytes.
示例：将 5 GB 转换为 MB，直接乘以 1024 一次：5 × 1024 = 5120 MB。若要得到 KB，则再乘以 1024：5120 × 1024 = 5,242,880 KB。记住每一级跨越都是 1024 的关系即可。
2. Binary and Denary Conversion | 二进制与十进制转换
Denary (base‑10) uses digits 0–9, while binary (base‑2) uses only 0 and 1. To convert a denary number to binary, repeatedly divide the number by 2 and record the remainders. The binary equivalent is read from the last remainder to the first. To convert binary to denary, write the binary digits under place values that double from right to left, starting at 1, and add the place values where a 1 appears.
十进制使用 0–9 十个数字，二进制只使用 0 和 1。将十进制数转换为二进制时，反复除以 2 并记录余数，从下往上读余数即可得到二进制数。将二进制转换为十进制时，把二进制数按位权展开，位权从右向左依次乘以 1、2、4、8、16……然后将有 1 的位权相加。
Denary to binary: divide by 2, record remainders, read upwards.
十进制转二进制：除以 2 取余数，从下往上读。
Binary to denary: sum (digit × 2position).
二进制转十进制：各位数字 × 2位权 之和。
Example: Convert 29 to binary. 29÷2=14 r1, 14÷2=7 r0, 7÷2=3 r1, 3÷2=1 r1, 1÷2=0 r1. Read remainders upwards: 11101. So 29 in binary is 11101. Convert 10110 back to denary: 1×16 + 0×8 + 1×4 + 1×2 + 0×1 = 16+0+4+2+0 = 22.
示例：将 29 转为二进制。29÷2=14 余 1，14÷2=7 余 0，7÷2=3 余 1，3÷2=1 余 1，1÷2=0 余 1。从下往上读余数：11101，因此 29 的二进制是 11101。将 10110 转回十进制：1×16 + 0×8 + 1×4 + 1×2 + 0×1 = 22。
3. Hexadecimal and Binary Conversion | 十六进制与二进制转换
Hexadecimal (hex) is base‑16, using digits 0–9 and letters A–F. One hex digit represents exactly four binary digits (a nibble). Converting binary to hex is simply a matter of splitting the binary number into groups of four bits, starting from the right, then replacing each group with its hex equivalent. Converting hex to binary does the reverse: expand each hex digit into a 4‑bit binary group.
十六进制是以 16 为基数的计数系统，使用数字 0–9 和字母 A–F。一个十六进制位正好代表四个二进制位（半个字节）。二进制转换为十六进制时，只需从右向左每四位分成一组，然后把每组替换为对应的十六进制符号。反过来，把每个十六进制数字展开为四位二进制数即可。
| Hex | Binary | Denary |
|---|---|---|
| 0 | 0000 | 0 |
| 1 | 0001 | 1 |
| 2 | 0010 | 2 |
| 3 | 0011 | 3 |
| 4 | 0100 | 4 |
| 5 | 0101 | 5 |
| 6 | 0110 | 6 |
| 7 | 0111 | 7 |
| 8 | 1000 | 8 |
| 9 | 1001 | 9 |
| A | 1010 | 10 |
| B | 1011 | 11 |
| C | 1100 | 12 |
| D | 1101 | 13 |
| E | 1110 | 14 |
| F | 1111 | 15 |
Example: Binary 11010110 splits into 1101 0110. From the table, 1101 = D and 0110 = 6, so the hex value is D6. As a check, convert to denary: D6 = 13×16 + 6 = 208 + 6 = 214.
示例：二进制 11010110 分成 1101 和 0110。查表得 1101 = D，0110 = 6，所以十六进制是 D6。验算一下，D6 转为十进制：13×16 + 6 = 214。
4. Binary Addition | 二进制加法
Binary addition follows four simple rules. When the sum of a column is 2 (10 in binary), write 0 and carry 1 to the next column. When the sum is 3 (11 in binary), write 1 and carry 1. Overflow occurs if the result needs more bits than the allotted width, which in an 8‑bit register means a carry out of the leftmost bit.
二进制加法遵循四条简单规则。当某一列的和为 2（即二进制的 10）时，写 0 并进 1；当和为 3（即二进制的 11）时，写 1 并进 1。如果结果需要的位数超过了寄存器宽度，就会发生溢出；在 8 位寄存器中，最高位左侧产生的进位表示溢出。
| A | B | Sum | Carry |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1+1+carry | – | 1 | 1 |
Example: Add 01001101 (77) and 00110011 (51). Working right to left: column 1: 1+1 = 0 carry 1; column 2: 1+0+carry 1 = 0 carry 1; column 3: 0+0+carry 1 = 1 carry 0; continue to get 10000000 (128), which is correct in 8 bits.
示例：将 01001101 (77) 和 00110011 (51) 相加。从右至左：第1列：1+1=0 进1；第2列：1+0+进1 = 0 进1；第3列：0+0+进1 = 1 进0；依次计算得到 10000000 (128)，结果正确。
5. Image File Size Calculation | 图像文件大小计算
The size of an uncompressed bitmap image depends on its resolution (width × height in pixels) and its colour depth, which is the number of bits used to represent each pixel’s colour. The basic formula multiplies the total number of pixels by the colour depth, then converts the result from bits to bytes.
未压缩位图图像的大小取决于图像的分辨率（宽 × 高，以像素计）和颜色深度，即每个像素的颜色用多少个比特表示。基本公式是将总像素数乘以颜色深度，然后将结果从比特转换为字节。
Image file size (bytes) = (width × height × colour depth in bits) ÷ 8
图像文件大小（字节）=（宽 × 高 × 颜色深度，比特）÷ 8
If the colour depth is given as a number of colours, convert it first: colour depth (bits) = log₂(number of colours). For example, 256 colours require 8 bits per pixel, because 2⁸ = 256.
如果给定的是颜色数量，需要先转换为比特：颜色深度（比特）= log₂（颜色数）。例如 256 色需要每个像素 8 比特，因为 2⁸ = 256。
Example: A 1920 × 1080 image with 16 million colours (24 bits per pixel) has a raw file size of (1920 × 1080 × 24) ÷ 8 = 6,220,800 bytes. Divide by 1024 twice gives approximately 5.93 MB.
示例：一张 1920×1080 像素、1600 万色（24 比特/像素）的图像，原始文件大小为 (1920 × 1080 × 24) ÷ 8 = 6,220,800 字节，除以 1024 两次后约为 5.93 MB。
6. Sound File Size Calculation | 声音文件大小计算
Sound is stored digitally by sampling the analogue signal at regular intervals. The file size of an uncompressed audio clip depends on the sample rate (samples per second), the sample resolution (bits per sample), the number of channels (mono = 1, stereo = 2), and the duration in seconds.
声音通过定期采样模拟信号以数字形式存储。未压缩音频片段的大小取决于采样率（每秒采样次数）、采样分辨率（每个样本的比特数）、声道数（单声道 = 1，立体声 = 2）以及以秒为单位的时长。
Sound file size (bits) = sample rate × sample resolution × channels × duration
声音文件大小（比特）= 采样率 × 采样分辨率 × 声道数 × 时长
File size (bytes) = total bits ÷ 8
文件大小（字节）= 总比特数 ÷ 8
Example: A 30‑second stereo recording at 44.1 kHz sample rate and 16‑bit resolution uses: bits = 44100 × 16 × 2 × 30 = 42,336,000 bits. In bytes: 42,336,000 ÷ 8 = 5,292,000 bytes, about 5.05 MB.
示例：一段 30 秒的立体声录音，采样率 44.1 kHz，16 比特分辨率，所需比特数为 44100 × 16 × 2 × 30 = 42,336,000 比特，字节数为 5,292,000 字节，约 5.05 MB。
7. Logic Gate Truth Tables | 逻辑门真值表
Logic gates are the building blocks of digital circuits. The three fundamental gates are NOT, AND, and OR. Their behaviour is fully described by truth tables, which list every possible input combination and the corresponding output. You must also recognise NAND, NOR, and XOR gates, which are combinations of the basic ones.
逻辑门是数字电路的基本构件。三个基本门是非门、与门和或门。它们的特性由真值表完整描述，真值表列出了所有可能的输入组合及对应的输出。你还必须能识别与非门、或非门和异或门，它们由基本门组合而成。
| Gate | Symbol | Truth Table |
|---|---|---|
| NOT | – | A:0→1; A:1→0 |
| AND | · | 00→0; 01→0; 10→0; 11→1 |
| OR | + | 00→0; 01→1; 10→1; 11→1 |
| NAND | – | 00→1; 01→1; 10→1; 11→0 |
| NOR | – | 00→1; 01→0; 10→0; 11→0 |
| XOR | ⊕ | 00→0; 01→1; 10→1; 11→0 |
Example: A simple circuit with inputs A=1 and B=0 going into an AND gate gives output 0; the same inputs into an OR gate give output 1. The XOR gate outputs 1 when the inputs are different.
示例：一个简单电路，输入 A=1、B=0 进入与门时输出为 0；进入或门时输出为 1。异或门在输入不同时输出 1。
8. Boolean Algebra Laws | 布尔代数定律
Boolean algebra helps simplify logic expressions, making circuits cheaper and faster. Key laws include identity, null, idempotent, inverse, commutative, associative, and distributive. A common simplification trick uses De Morgan’s laws, which relate AND and OR via inversion.
布尔代数有助于简化逻辑表达式，从而使电路更便宜、更快。关键定律包括同一律、零律、幂等律、互补律、交换律、结合律和分配律。常用的简化技巧是利用德摩根定律，通过反相连接与和或。
De Morgan’s Law 1: (A · B)’ = A’ + B’
德摩根定律 1：(A · B)’ = A’ + B’
De Morgan’s Law 2: (A + B)’ = A’ · B’
德摩根定律 2：(A + B)’ = A’ · B’
Example: Simplify the expression (A + B) · (A + C). Using the distributive law: A + (B · C). This means the output is true if A is true, or if both B and C are true. The simplified circuit uses fewer gates.
示例：简化表达式 (A + B) · (A + C)。利用分配律得出 A + (B · C)。这意味着只要 A 为真，或者 B 和 C 同时为真，输出就为真。简化后的电路门数更少。
9. Data Transmission Time | 数据传输时间
When transferring a file over a network, the transmission time depends on the file size and the transfer rate. Always make sure the units match: if the file size is in megabytes and the rate is in megabits per second, convert one unit so you are dividing or multiplying consistently.
通过网络传输文件时，传输时间取决于文件大小和传输速率。务必保证单位一致：如果文件大小以兆字节为单位，速率以兆比特每秒为单位，就要对其中一个进行转换，以便进行统一的除法或乘法。
Time (seconds) = File size (bits) ÷ Transfer rate (bits per second)
时间（秒）= 文件大小（比特）÷ 传输速率（比特/秒）
Remember: 1 byte = 8 bits. So to convert megabytes to megabits, multiply by 8. Example: Downloading a 10 MB file on a 2 Mbps connection: 10 MB = 80 Mb; time = 80 Mb ÷ 2 Mbps = 40 seconds.
记住：1 字节 = 8 比特。因此将兆字节转换为兆比特时要乘以 8。示例：从 2 Mbps 的网络上下载一个 10 MB 的文件：10 MB = 80 Mb，时间 = 80 Mb ÷ 2 Mbps = 40 秒。
10. Common Spreadsheet Functions | 电子表格常用函数
Spreadsheets are powerful tools for modelling and data analysis. The functions SUM, AVERAGE, MAX, and MIN perform basic statistical operations on a range of cells. The IF function allows decision‑making based on a condition. The syntax always begins with an equals sign and the function name, followed by parentheses containing the arguments.
电子表格是建模和数据分析的强大工具。SUM、AVERAGE、MAX 和 MIN 函数对单元格区域执行基本的统计运算。IF 函数允许根据条件进行决策。语法始终以等号和函数名开头，括号内包含参数。
- SUM(range) returns the total of the values.
- AVERAGE(range) returns the mean.
- MAX(range) returns the largest value.
- MIN(range) returns the smallest value.
- IF(condition, value_if_true, value_if_false) checks a logical test.
- SUM(区域) 返回数值的总和。
- AVERAGE(区域) 返回平均值。
- MAX(区域) 返回最大值。
- MIN(区域) 返回最小值。
- IF(条件, 真值, 假值) 执行逻辑判断。
Example: To calculate the total of cells A1 to A10, use =SUM(A1:A10). To label scores above 50 as “Pass” and others as “Fail”, use =IF(A1>50, “Pass”, “Fail”).
示例：要计算单元格 A1 到 A10 的总和，使用 =SUM(A1:A10)。要将分数大于 50 的标记为“Pass”，其余标记为“Fail”，使用 =IF(A1>50, “Pass”, “Fail”)。
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