题解 | #学英语#

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

string change(long long n);

int main() {
    long long  n;
    while (cin >> n) { // 注意 while 处理多个 case
        cout << change(n) << endl;
    }
}

string change(long long n) {
    vector<string> shi = {"", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
    vector<string> other = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
    
    if (n >= 0 && n <= 19) {
        return other[n];
    }
    else if (n >= 20 && n <= 99) {
        if (n % 10 == 0) {
            return shi[n / 10];
        } else {
            return shi[n / 10] + " " + change(n % 10);
        }
    }
    else if (n >= 100 && n <= 999) {
        if (n % 100 == 0) {
            return other[n / 100] + " hundred";
        } else {
            return other[n / 100] + " hundred and " + change(n % 100);
        }
    }
    else if (n >= 1000 && n <= 999999) {
        if (n % 1000 == 0) {
            return other[n / 1000] + " thousand";
        } else {
            return change(n / 1000) + " thousand " + change(n % 1000);
        }
    }
    else if (n >= 1000000 && n <= 99999999) {
        if (n % 1000000 == 0) {
            return other[n / 1000000] + " million";
        } else {
            return change(n / 1000000) + " million " + change(n % 1000000);
        }
    }
    else {
        if (n % 1000000000 == 0) {
            return other[n / 1000000000] + " billion";
        } else {
            return change(n / 1000000000) + " billion " + change(n % 1000000000);
        }
    }
}

// 64 位输出请用 printf("%lld")