📚 Year 11 Cambridge Computer Science: Formula & Theorem Quick Reference | 剑桥计算机：公式定理速查手册
This quick reference guide compiles the essential formulas, theorems, and conversion rules needed for the Year 11 Cambridge Computer Science course. Use it to revise data representation, logic, storage, and file calculations efficiently.
本速查手册汇总了 Year 11 剑桥计算机科学课程所需的核心公式、定理和转换规则，帮助你高效复习数据表示、逻辑、存储和文件大小计算。
1. Number Systems & Conversions | 数制与转换
. Sum the results. For example, binary 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8+0+2+1 = 11₁₀.
二进制转十进制：将每个二进制位（0或1）乘以 2 的位置次幂，从最右位0开始，求和。例如二进制 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8+0+2+1 = 11₁₀。
verse order. Example: 13₁₀ → 13÷2=6 r1, 6÷2=3 r0, 3÷2=1 r1, 1÷2=0 r1 → read upward: 1101₂.
十进制转二进制：不断将十进制数除以 2，记录余数，当商为0时停止，然后倒序读取余数。例如 13₁₀ → 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：1101₂。
Hexadecimal to binary: each hex digit corresponds to a 4‑ 0011 1010₂. Binary to hex: group bits in fours from the right, pad with leading zeros if necessary, then convert each group.
十六进制转二进制：每个十六进制数字对应 4 位二进制组。例如 3A₁₆ = 0011 1010₂。二进制转十六进制：从右起每4位分组，必要时在高位补0，然后转换每组。
32+15 = 47₁₀). Decimal to hex: repeatedly divide by 16 and read remainders upward, using A–F for 10–15.
十六进制转十进制：每位乘以16对应次幂（如 2F₁₆ = 2×16¹ + 15×16⁰ = 32+15 = 47₁₀）。十进制转十六进制：反复除以16，倒序取余，10–15 用 A–F 表示。
2. Binary Addition & Overflow | 二进制加法与溢出
arry of 1 to the next column. Add column by column from right to left, carrying as needed.
二进制加法遵循四项规则：0+0=0, 0+1=1, 1+0=1, 1+1=0 并进位1。从右向左逐位相加，需要时进位。
Overflow detection in fixed‑width two's complement arithmetic: overflow occurs when adding two numbers with the same sign produces a result with a different sign. This indicates the result cannot be represented within the given number of bits. For 8‑bit numbers, if the carry into the sign bit differs from the carry out of the sign bit, overflow has happened.
固定宽度二进制补码溢出检测：当两个同符号数相加得到的结果符号不同，即发生溢出，表示结果无法在给定位数内表示。对于8位数字，如果进入符号位的进位与出符号位的进位不同，则产生溢出。
3. Two’s Complement Representation | 二进制补码表示
For an n‑bit two's complement integer, the leftmost bit is the sign bit (0 for positive, 1 for negative). The decimal value is calculated as:
Decimal = –bₙ₋₁ × 2ⁿ⁻¹ + bₙ₋₂ × 2ⁿ⁻² + … + b₁ × 2¹ + b₀ × 2⁰
对于 n 位二进制补码整数，最左位是符号位（0正1负）。十进制值公式：
十进制 = –bₙ₋₁ × 2ⁿ⁻¹ + bₙ₋₂ × 2ⁿ⁻² + … + b₁ × 2¹ + b₀ × 2⁰
To convert a negative decimal to n‑bit two' nt: write the positive magnitude in binary (n bits), invert all bits, then add 1. Example: –5 in 8 bits → +5 = 00000101₂, invert = 11111010₂, add 1 = 11111011₂.
将负数十进制转换为 n 位补码：写出其正数的 n 位二进制形式，全部取反，再加1。例：–5 用8位 → +5 = 00000101₂，取反 = 11111010₂，加1 = 11111011₂。
To manually check a negative two's complement value: keep all bits from the right up to and including the first '1', then invert all bits to the left of that '1'.
手工快速求负补码的值：从右向左直到第一个‘1’保留沿袭，该‘1’左侧所有位取反。
4. Logic Gates – Truth Tables | 逻辑门真值表
Basic logic gates and their symbols: NOT (¬), AND (·), OR (+), NAND, NOR, XOR (⊕). The following tables summarise their behaviour.
基本逻辑门及其符号：非门 (¬)、与门 (·)、或门 (+)、与非门、或非门、异或门 (⊕)。下表总结其行为。
NOT gate: Y = ¬A
| A | Y |
|---|---|
| 0 | 1 |
| 1 | 0 |
AND gate: Y = A·B
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
OR gate: Y = A+B
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
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