Unit 2 · Number Systems / 進位制

Start Here / 開始做題

1. Identify the base first. 中文：先找出進位制的底數。
2. Read every digit from left to right when converting to decimal. 中文：轉十進位時由左至右計算。
3. For decimal to another base, record remainders from right to left. 中文：除法餘數要倒序讀取。
4. Check your answer by converting it back once. 中文：可反向轉換檢查答案。

Unit Dashboard / 單元地圖

Core Idea / 核心概念

Fast Conversion / 快速轉換

C++ Syntax / C++ 語法

int remainder = n % base;  // last digit in the target base
n /= base;                 // remove the last digit
value = value * base + digit;  // read one digit into decimal value

Yellow Syntax Guide / 黃色語法重點

Common Errors / 常見錯誤

Examples / 例題

Example 2-1 · Binary to Decimal

Sample Input / 範例輸入

101101

Sample Output / 範例輸出

45

C++ focus / 語法重點：For each digit: value = value * 2 + digit.

Reference Answer / 參考答案

#include <iostream>
#include <string>
using namespace std;

int main() {
    string b;
    cin >> b;
    int value = 0;
    for (char ch : b) value = value * 2 + (ch - '0');
    cout << value << endl;
    return 0;
}

Example 2-2 · Decimal to Binary

Sample Input / 範例輸入

45

Sample Output / 範例輸出

101101

C++ focus / 語法重點：Repeatedly take n % 2, then reverse the digits.

Reference Answer / 參考答案

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main() {
    int n;
    cin >> n;
    if (n == 0) { cout << 0 << endl; return 0; }
    string result;
    while (n > 0) { result += char('0' + n % 2); n /= 2; }
    reverse(result.begin(), result.end());
    cout << result << endl;
    return 0;
}

Example 2-3 · Octal to Decimal

Sample Input / 範例輸入

157

Sample Output / 範例輸出

111

C++ focus / 語法重點：Change the base from 2 to 8: value = value * 8 + digit.

Reference Answer / 參考答案

#include <iostream>
#include <string>
using namespace std;

int main() {
    string o;
    cin >> o;
    int value = 0;
    for (char ch : o) value = value * 8 + (ch - '0');
    cout << value << endl;
    return 0;
}

Example 2-4 · Decimal to Octal

Sample Input / 範例輸入

111

Sample Output / 範例輸出

157

C++ focus / 語法重點：Use n % 8 for the last octal digit and n /= 8 to continue.

Reference Answer / 參考答案

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main() {
    int n;
    cin >> n;
    if (n == 0) { cout << 0 << endl; return 0; }
    string result;
    while (n > 0) { result += char('0' + n % 8); n /= 8; }
    reverse(result.begin(), result.end());
    cout << result << endl;
    return 0;
}

Example 2-5 · Hex Digit Value

Sample Input / 範例輸入

B

Sample Output / 範例輸出

11

C++ focus / 語法重點：Digits A to F represent 10 to 15.

Reference Answer / 參考答案

#include <iostream>
using namespace std;

int main() {
    char ch;
    cin >> ch;
    if (ch >= '0' && ch <= '9') cout << ch - '0' << endl;
    else cout << ch - 'A' + 10 << endl;
    return 0;
}

Example 2-6 · Hexadecimal to Decimal

Sample Input / 範例輸入

2F

Sample Output / 範例輸出

47

C++ focus / 語法重點：For every digit, multiply the current answer by 16 first.

Reference Answer / 參考答案

#include <iostream>
#include <string>
using namespace std;

int main() {
    string h;
    cin >> h;
    int value = 0;
    for (char ch : h) {
        int digit = (ch <= '9') ? ch - '0' : ch - 'A' + 10;
        value = value * 16 + digit;
    }
    cout << value << endl;
    return 0;
}

Example 2-7 · Decimal to Hexadecimal

Sample Input / 範例輸入

47

Sample Output / 範例輸出

2F

C++ focus / 語法重點：Use n % 16; values 10–15 become A–F.

Reference Answer / 參考答案

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main() {
    int n;
    cin >> n;
    if (n == 0) { cout << 0 << endl; return 0; }
    string result, digits = "0123456789ABCDEF";
    while (n > 0) { result += digits[n % 16]; n /= 16; }
    reverse(result.begin(), result.end());
    cout << result << endl;
    return 0;
}

Example 2-8 · Binary Addition

Sample Input / 範例輸入

101 11

Sample Output / 範例輸出

8

C++ focus / 語法重點：Convert each binary number, then add the two decimal values.

Reference Answer / 參考答案

#include <iostream>
#include <string>
using namespace std;

int valueOf(string s) {
    int value = 0;
    for (char ch : s) value = value * 2 + (ch - '0');
    return value;
}

int main() {
    string a, b;
    cin >> a >> b;
    cout << valueOf(a) + valueOf(b) << endl;
    return 0;
}

Example 2-9 · Four Binary Bits to Hex

Sample Input / 範例輸入

1101

Sample Output / 範例輸出

D

C++ focus / 語法重點：Four bits have values 8, 4, 2, 1; use a hex digit string to look up the result.

Reference Answer / 參考答案

#include <iostream>
#include <string>
using namespace std;

int main() {
    string b, digits = "0123456789ABCDEF";
    cin >> b;
    int value = 0;
    for (char ch : b) value = value * 2 + (ch - '0');
    cout << digits[value] << endl;
    return 0;
}

Example 2-10 · Display in Three Bases

Sample Input / 範例輸入

26

Sample Output / 範例輸出

11010
32
1A

C++ focus / 語法重點：The stream manipulators oct and hex change output base; uppercase makes A–F uppercase.

Reference Answer / 參考答案

#include <iostream>
#include <iomanip>
using namespace std;

int main() {
    int n;
    cin >> n;
    int bit = 1;
    while (bit <= n / 2) bit *= 2;
    if (n == 0) cout << 0 << endl;
    else {
        for (; bit > 0; bit /= 2) cout << (n / bit) % 2;
        cout << endl;
    }
    cout << oct << n << endl;
    cout << uppercase << hex << n << endl;
    return 0;
}

ClassWork / 課後練習

Complete each task independently. Submit your C++ source file and one screenshot of successful output. 中文：獨立完成，提交程式碼及成功輸出截圖。

ClassWork 2-1 · Decimal to Binary

Sample Input / 範例輸入

64

Sample Output / 範例輸出

1000000

C++ focus / 語法重點：Use repeated division by 2. Remember the special case N = 0.

ClassWork 2-2 · Decimal to Octal

Sample Input / 範例輸入

83

Sample Output / 範例輸出

123

C++ focus / 語法重點：Repeatedly use % 8 and /= 8, then reverse the collected digits.

ClassWork 2-3 · Decimal to Hex

Sample Input / 範例輸入

255

Sample Output / 範例輸出

FF

C++ focus / 語法重點：Prepare 0123456789ABCDEF for digit lookup.

ClassWork 2-4 · Binary to Decimal

Sample Input / 範例輸入

100000

Sample Output / 範例輸出

32

C++ focus / 語法重點：Read from left to right: answer = answer × 2 + current digit.

ClassWork 2-5 · Octal to Decimal

Sample Input / 範例輸入

70

Sample Output / 範例輸出

56

C++ focus / 語法重點：The base is 8, so multiply the running value by 8.

ClassWork 2-6 · Hex to Decimal

Sample Input / 範例輸入

1A

Sample Output / 範例輸出

26

C++ focus / 語法重點：Convert A–F into values 10–15 before adding.

ClassWork 2-7 · Count One Bits

Sample Input / 範例輸入

13

Sample Output / 範例輸出

3

C++ focus / 語法重點：When n % 2 == 1, increase the counter; then use n /= 2.

ClassWork 2-8 · Last Binary Digit

Sample Input / 範例輸入

18

Sample Output / 範例輸出

0

C++ focus / 語法重點：The last binary digit is the same as N % 2.

ClassWork 2-9 · Hex Digit to Binary

Sample Input / 範例輸入

A

Sample Output / 範例輸出

1010

C++ focus / 語法重點：First convert the digit to 0–15, then print bit values 8, 4, 2, 1.

ClassWork 2-10 · Same Value Check

Sample Input / 範例輸入

10
1010

Sample Output / 範例輸出

Same

C++ focus / 語法重點：Convert B to decimal first, then compare with ==.

ClassWork 2-11 · Bytes in Hex

Sample Input / 範例輸入

10

Sample Output / 範例輸出

0A

C++ focus / 語法重點：The first digit is N / 16; the second is N % 16.

ClassWork 2-12 · Binary Place Value

Sample Input / 範例輸入

10000001

Sample Output / 範例輸出

128 1

C++ focus / 語法重點：The leftmost bit has place value 128; the rightmost bit has place value 1.

Next Step / 下一步

You should now be able to convert one value between decimal, binary, octal and hexadecimal. Continue to Unit 3 when you can explain why remainders are read in reverse order. 中文：能解釋為何餘數要倒序讀取後，再進入 Unit 3。