如何在C++中实现类二进制计数的十进制数值及英文转写算法？

Hey Joseph, this is a fun problem—combining binary counting logic with decimal representation and English number words! Let's break this down into two manageable parts: generating that binary-like decimal sequence, and converting each number to its English cardinal equivalent. Here's how to pull this off in C++:

First, let's clarify what this sequence actually is: each number is the binary representation of an integer, interpreted as a decimal number. For example:

• Binary 1 → decimal 1
• Binary 10 → decimal 10
• Binary 11 → decimal 11
• Binary 100 → decimal 100, and so on.

The simplest way to generate this is to simulate binary increment on a string, then convert that string to a decimal integer. Here's a straightforward implementation:

#include <iostream>
#include <string>
#include <algorithm>

// Helper to increment a binary string (e.g., "11" → "100")
std::string increment_binary_string(std::string s) {
    int i = s.size() - 1;
    // Flip 1s to 0s until we hit a 0 or the start of the string
    while (i >= 0 && s[i] == '1') {
        s[i] = '0';
        i--;
    }
    // If we reached the start, prepend a 1
    if (i < 0) {
        s.insert(s.begin(), '1');
    } else {
        s[i] = '1';
    }
    return s;
}

int main() {
    std::string binary_str = "1";
    // Generate the first 7 numbers in the sequence
    for (int i = 0; i < 7; i++) {
        long long decimal_num = stoll(binary_str); // Convert binary string to decimal
        std::cout << decimal_num << " ";
        binary_str = increment_binary_string(binary_str);
    }
    // Output: 1 10 11 100 101 110 111
    return 0;
}

Alternatively, you can generate the sequence by taking integers starting from 1, converting each to a binary string, then parsing that string as a decimal number. The string method above is just more direct for this specific use case.

Next, we need a function to turn those decimal numbers into words like "one", "ten", "one hundred and eleven". This requires handling different number ranges (ones, teens, tens, hundreds, thousands, etc.). Here's a robust implementation that works for numbers up to billions:

#include <iostream>
#include <vector>
#include <string>

// Handle numbers 0-19
std::string convert_less_than_20(int num) {
    const std::vector<std::string> words = {
        "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine",
        "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen",
        "seventeen", "eighteen", "nineteen"
    };
    return words[num];
}

// Handle multiples of 10 (20, 30, ..., 90)
std::string convert_tens(int num) {
    const std::vector<std::string> words = {
        "", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"
    };
    std::string result = words[num / 10];
    if (num % 10 != 0) {
        result += "-" + convert_less_than_20(num % 10);
    }
    return result;
}

// Handle numbers 0-999
std::string convert_less_than_1000(int num) {
    std::string result;
    if (num >= 100) {
        result += convert_less_than_20(num / 100) + " hundred";
        num %= 100;
        if (num != 0) {
            result += " and "; // Remove this line for American English
        }
    }
    if (num != 0) {
        result += (num < 20) ? convert_less_than_20(num) : convert_tens(num);
    }
    return result;
}

// Handle larger numbers by breaking into thousands/millions/billions
std::string number_to_english(long long num) {
    if (num == 0) return "zero";
    
    const std::vector<std::string> scales = {"", "thousand", "million", "billion"};
    std::string result;
    int scale_idx = 0;
    
    while (num > 0) {
        int part = num % 1000;
        if (part != 0) {
            std::string part_str = convert_less_than_1000(part);
            if (!result.empty()) {
                result = part_str + " " + scales[scale_idx] + ", " + result;
            } else {
                result = part_str + " " + scales[scale_idx];
            }
        }
        num /= 1000;
        scale_idx++;
    }
    
    // Clean up trailing spaces/commas
    if (!result.empty() && result.back() == ' ') result.pop_back();
    if (!result.empty() && result.back() == ',') result.pop_back();
    
    return result;
}

Now just combine the sequence generator with the word converter. Here's the full example:

// Include all helper functions from above here

int main() {
    std::string binary_str = "1";
    for (int i = 0; i < 7; i++) {
        long long decimal_num = stoll(binary_str);
        std::cout << decimal_num << " → " << number_to_english(decimal_num) << std::endl;
        binary_str = increment_binary_string(binary_str);
    }
    
    /* Output:
    1 → one
    10 → ten
    11 → eleven
    100 → one hundred
    101 → one hundred and one
    110 → one hundred and ten
    111 → one hundred and eleven
    */
    return 0;
}

• Overflow Handling: Use long long instead of int to handle larger numbers in the sequence (since binary strings get long quickly, e.g., binary 1000000 is decimal 1000000, which fits in a 32-bit int, but longer strings will need 64-bit integers).
• English Variations: The "and" in "one hundred and one" is standard in British English. For American English, just remove the line adding " and " in convert_less_than_1000.
• Scalability: The word converter works up to billions, but you can extend the scales vector to handle trillions and beyond if needed.

内容的提问来源于stack exchange，提问作者Joseph