📚 Year 11 CIE Computer Science: Formula & Theorem Quick Reference Guide | CIE计算机公式定理速查手册
This quick reference guide gathers the essential formulas, laws, and calculation methods required for the CIE IGCSE Computer Science syllabus (0478). It covers data representation, logic, error detection, encryption, and transmission to help you prepare for your examinations efficiently.
本速查手册汇集了CIE IGCSE计算机科学（0478）课程所需的关键公式、定律和计算方法，涵盖数据表示、逻辑、错误检测、加密和传输，帮助你高效备考。
1. Data Storage Units & File Size | 数据存储单位与文件大小计算
In computing, storage is based on the binary system. File sizes are measured in bytes, and each larger unit equals 1024 of the smaller one.
计算机存储基于二进制系统。文件大小以字节为单位，每个较大单位的容量是前一个单位的1024倍。
1 kilobyte (KB) = 1024 bytes
1千字节 (KB) = 1024 字节
1 megabyte (MB) = 1024 KB = 1,048,576 bytes
1兆字节 (MB) = 1024 KB = 1,048,576 字节
1 gigabyte (GB) = 1024 MB
1吉字节 (GB) = 1024 MB
For a text file, the basic formula is:
对于文本文件，基本公式为：
File size (bits) = number of characters × bits per character
文件大小（比特） = 字符数 × 每个字符的比特数
To convert bits to bytes, divide by 8:
将比特转换为字节，除以8：
File size (bytes) = file size in bits ÷ 8
文件大小（字节） = 文件大小（比特） ÷ 8
2. Image File Size Calculation | 图像文件大小计算
lution and colour depth. Resolution is the total number of pixels, usually width × height.
位图图像的大小取决于分辨率和颜色深度。分辨率是像素总数，通常为宽×高。
Image file size (bits) = width (pixels) × height (pixels) × colour depth (bits per pixel)
图像文件大小（比特） = 宽（像素） × 高（像素） × 颜色深度（每像素比特数）
To express the result in bytes, divide by 8. Colour depth is often 1 bit (monochrome), 8 bits (256 colours), or 24 bits (true colour).
结果以字节表示时需除以8。颜色深度通常为1比特（单色）、8比特（256色）或24比特（真彩色）。
Image file size (bytes) = (width × height × colour depth) ÷ 8
图像文件大小（字节） = （宽 × 高 × 颜色深度） ÷ 8
3. Sound File Size Calculation | 声音文件大小计算
Digital audio size is determined by sample rate, bit depth, duration, and the number of audio channels.
数字音频大小由采样率、位深度、时长和声道数决定。
ize (bits) = sample rate (Hz) × bit depth (bits) × duration (seconds) × number of channels
声音文件大小（比特） = 采样率（赫兹） × 位深度（比特） × 时长（秒） × 声道数
Common sample rates are 44.1 kHz and 48 kHz. Stereo audio uses 2 channels, while mono uses 1 channel.
常见的采样率为44.1 kHz和48 kHz。立体声使用2个声道，单声道使用1个声道。
Sound file size (bytes) = (sample rate × bit depth × duration × channels) ÷ 8
声音文件大小（字节） = （采样率 × 位深度 × 时长 × 声道数） ÷ 8
4. Binary & Hexadecimal Conversions | 二进制与十六进制转换
Computers use binary (base-2) internally, but hexadecimal (base-16) is often used as a shorter representation.
计算机内部使用二进制（基数为2），但十六进制（基数为16）常作为更简短的表示方法。
Each hexadecimal digit represents 4 bits. The conversion table is:
每个十六进制位代表4个二进制位。转换表如下：
| Hexadecimal | Binary | Decimal |
|---|---|---|
| 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 |
To convert a hexadecimal number to binary, replace each hex digit with its 4-bit equivalent. To convert binary to hex, group the bits into sets of four (starting from the right) and then use the table.
将十六进制数转换为二进制时，将每个十六进制位替换为对应的4位二进制数。将二进制转换为十六进制时，从右开始每4位一组，再查表即可。
5. Binary Addition & Overflow | 二进制加法与溢出
Binary addition follows simple rules similar to decimal addition, but with only 0 and 1.
二进制加法遵循与十进制加法类似的简单规则，但只有0和1。
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0, carry 1 to the next column
1 + 1 + 1 = 1, carry 1
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0，向下一列进位1
1 + 1 + 1 = 1，进位1
An overflow occurs when the result of an addition exceeds the maximum value that can be stored with the available number of bits. For example, with 8 bits the maximum is 255 (11111111). Adding two numbers that produce a 9-bit result triggers an overflow flag.
当加法结果超出可用位数所能存储的最大值时，便发生溢出。例如，8位时最大值为255 (11111111)。两个数相加得到9位结果就会触发溢出标志。
6. Logic Gates & Truth Tables | 逻辑门与真值表
Logic gates are the building blocks of digital circuits. Each gate has a defined truth table.
逻辑门是数字电路的构建模块，每个门都有确定的真值表。
The NOT gate inverts the input.
非门将输入取反。
| A | NOT A |
|---|---|
| 0 | 1 |
| 1 | 0 |
The AND gate outputs 1 only if all inputs are 1.
与门仅当所有输入均为1时输出1。
| A | B | A AND B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
The OR gate outputs 1 if at least one input is 1.
或门只要至少一个输入为1就输出1。
| A | B | A OR B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
The NAND gate is an AND followed by a NOT.
与非门是与门后接非门。
| A | B | A NAND B |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
The NOR gate is an OR followed by a NOT.
或非门是或门后接非门。
| A | B | A NOR B |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
The XOR gate outputs 1 when the inputs are different.
异或门在输入不同时输出1。
| A | B | A XOR B |
|---|---|---|
| 0 | 0 | 0 |
