【算法题】数字转英文

灰太狼 2022-05-23 08:44 197阅读 0赞

import java.util.*;
    public class Main {
        public static String[] _0_19 = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen","seventeen", "eighteen", "nineteen" };
        public static String[] _20_90 = { "", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" };
        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            while (sc.hasNext()) {
                String s = sc.nextLine();
                long n = Long.parseLong(s);
                if (n < 0) {
                    System.out.println("error");
                    continue;
                }
                String res = "";
                if (n > 999999999L) {
                    res = res + threeDigit(n / 1000000000L) + " billion ";
                    n %= 1000000000L;
                }
                if (n > 999999L) {
                    res = res + threeDigit(n / 1000000L) + " million ";
                    n %= 1000000L;
                }
                if (n > 999L) {
                    res = res + threeDigit(n / 1000L) + " thousand ";
                    n %= 1000L;
                }
                if (n <= 999L) {
                    res = res + threeDigit(n);
                }
                System.out.println(res.trim());
            }
            sc.close();
        }
        
        public static String threeDigit(long ln) {
            int n = (int)ln;
            String res = "";
            if (n >= 100) {
                res = res + _0_19[n / 100] + " hundred";
                n = n % 100;
                if (n > 0) {
                    res += " and ";
                } else if (n == 0) {
                    return res;
                }
            }
            if (n >= 20) {//>=20
                res = res + _20_90[n / 10];
                n = n % 10;
                if (n > 0) {//0-9
                    res = res + " " + _0_19[n];//注意此处的空格
                }
            } else {//0-19
                res += _0_19[n];
            }
            return res;
        }
    }