P1126 - 【提高】英文翻译 -

难度：8+

输入格式

一个自然数n，0< =n< =2^31-1。

输出格式

输出这个数的英文，最后不要有多余的空格。

输入数据 1

1111111111

输出数据 1

one billion one hundred and eleven million one hundred and eleven thousand one hundred and eleven

 代码：

#include <iostream>
#include <string>
#include <vector>
#include <bits/stdc++.h>
using namespace std;

// 定义数字到英文的映射
const string units[] = {"", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
const string teens[] = {"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
const string tens[] = {"", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
const string thousands[] = {"", "thousand", "million", "billion"};

// 将三位数转换为英文
string numberToWords(int num) {
    if (num == 0) return "zero";
    if (num < 10) return units[num];
    if (num < 20) return teens[num - 10];
    if (num < 100) return tens[num / 10] + (num % 10 != 0 ? " " + units[num % 10] : "");
    if (num < 1000) return units[num / 100] + " hundred" + (num % 100 != 0 ? " and " + numberToWords(num % 100) : "");
    return "";
}

// 将大数字转换为英文
string convertToEnglish(long long num) {
    if (num == 0) return "zero";
    vector<string> parts;
    int unit = 0;
    while (num > 0) {
        if (num % 1000 != 0) {
            parts.push_back(numberToWords(num % 1000) + " " + thousands[unit]);
        }
        unit++;
        num /= 1000;
    }
    reverse(parts.begin(), parts.end());
    return parts[0] + (parts.size() > 1 ? " " + parts[1] : "");
}
int main() {
    long long n;
    cin >> n;
    cout << convertToEnglish(n) << endl;
    return 0;
}