一、题目

🔸题目描述

Jessi初学英语，为了快速读出一串数字，编写程序将数字转换成英文：
具体规则如下:
1.在英语读法中三位数字看成一整体，后面再加一个计数单位。从最右边往左数，三位一单位，例如12,345 等
2.每三位数后记得带上计数单位分别是thousand, million, billion.
3.公式：百万以下千以上的数 X thousand X, 10亿以下百万以上的数：X million X thousand X, 10 亿以上的数：X billion X million X thousand X. 每个X分别代表三位数或两位数或一位数。
4.在英式英语中百位数和十位数之间要加and，美式英语中则会省略，我们这个题目采用加上and，百分位为零的话，这道题目我们省略and

下面再看几个数字例句：
22: twenty two
100: one hundred
145: one hundred and forty five
1,234: one thousand two hundred and thirty four
8,088: eight thousand (and) eighty eight (注:这个and可加可不加，这个题目我们选择不加)
486,669: four hundred and eighty six thousand six hundred and sixty nine
1,652,510: one million six hundred and fifty two thousand five hundred and ten

说明：
数字为正整数，不考虑小数，转化结果为英文小写；
保证输入的数据合法
关键字提示：and，billion，million，thousand，hundred

🔸输入输出

输入
输入一个long型整数

输出
输出相应的英文写法

🔸样例1

输入
22


输出
twenty two

二、代码参考


/*
小于20的直接读数，如果个位和十位是0，则不需要读数（正百的100，例：one hundred）；
如果有百位读取百位，
如果个位和十位是0，则不需要加and，否则需要加and，加hubdred，加百位 2、大于20的，依次添加个位，十位，百位；
添加百位的时候，先加and，加hundred，加百位
*/
import java.util.*;
public class Main {
     public static void main(String[] args){
         Scanner in = new Scanner(System.in);
         String[] NUMS = {"zero", "one", "two", "three", "four", "five", "six", "seven",
                         "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen",
                         "fifteen", "sixteen", "seventeen", "eighteen", "nineteen", "twenty"};
         String[] NUMSSHI = {"zero", "ten", "twenty", "thirty", "forty", "fifty",
                            "sixty", "seventy", "eighty", "ninety"};
         String[] POWER = {"", "hundred", "thousand", "million", "billion"};
         while(in.hasNext()){
             String line = in.nextLine();
             StringBuilder sb = new StringBuilder();
             ArrayList<String> lists = new ArrayList<>();
             if(!line.matches("\\d+")){//如果匹配的不是数字
                 System.out.println("error");
             }
             int linenum = Integer.parseInt(line);
             int power = 1;//单位
             while(linenum != 0){
                 if(power != 1){
                     lists.add(POWER[power]);//添加单位
                 }
                 int t = linenum % 1000;//取低三位
                 //注意小于20，直接读
                 if(t % 100 <= 20){
                     if(t % 100 != 0){//十位和个位是零的话就不需要读数了
                         lists.add(NUMS[t % 100]);
                     }
                     if(t / 100 != 0){//有百位
                         if(t % 100 != 0){//十位和个位是零的话就不需要添加and了
                             lists.add("and");
                         }
                         lists.add("hundred");
                         lists.add(NUMS[t / 100]);
                     }
                 }else{//大于20
                     // 有个位
                     if(t % 10 != 0){
                         lists.add(NUMS[t % 10]);
                     }
                     t /= 10;
                     // 有十位
                     if(t % 10 != 0){
                         lists.add(NUMSSHI[t % 10]);
                     }
                     t /= 10;
                     // 有百位
                     if(t % 10 != 0){
                         lists.add("and");
                         lists.add("hundred");
                         lists.add(NUMS[t % 10]);
                     }
                 }
                 linenum /= 1000;//每次缩小1000倍
                 power++;//单位*1000
             }
             //添加的时候，先添加低位，读数的时候先读高位，倒着读
             for(int i = lists.size() - 1; i >= 0; i--){
                 if(i!=0){
                     sb.append(lists.get(i) + " ");
                 }else{
                     sb.append(lists.get(i));//最后一个不加空格
                 }
             }
             System.out.println(sb.toString());
         }
     }   
}



--------------------------------------------------------


num1 = ['zero','one','two','three','four','five','six',
       'seven','eight','nine','ten','eleven','twelve',
       'thirteen','fourteen','fifteen','sixteen',
       'seventeen','eighteen','nineteen']
num2 = [0,0,'twenty','thirty','forty','fifty','sixty',
       'seventy','eighty','ninety']

# 100以内转英文
def n2w(n):
    if n > 0:
        if n < 20:
            word.append(num1[n])
        else:
            word.append(num2[n//10])
            if n%10 != 0:
                word.append(num1[n%10])

# 1000以内转英文
def hun(n):
    if n >= 100:
        word.append(num1[n//100])
        word.append('hundred')
        if n % 100 != 0:
            word.append('and')
    n2w(n%100)

while True:
    try:
        n = int(input())
    except:
        break
    else:
        word = []
        a = n % 1000  # 个十百位
        b = (n//1000) % 1000  # 个十百千
        c = (n//1000000) % 1000  #个十百m
        d = n // 1000000000    # 个十百b
        
        if d > 0:
            hun(d)
            word.append('billion')
        if c > 0 :
            hun(c)
            word.append('million')
        if b > 0:
            hun(b)
            word.append('thousand')
        if a > 0 :
            hun(a)
        print(' '.join(word))



--------------------------------------------------------------


#include<stdio.h>


int main(){
    int in;
    char s2[10][10] = {"zero", "one", "two", "three", "four","five", "six", "seven", "eight", "nine"};
    char s1[10][10] = {"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",
                                      "sixteen", "seventeen", "eighteen", "nineteen"};
    char s0[10][10] = {"ten", "twenty", "thirty", "forty", "fifty",
                                    "sixty", "seventy", "eighty", "ninety"};
    scanf("%d",&in);
    int m[9];
    int k=8;
    while(k>=0){
        m[k] = in%10;
        in = in/10;
        k--;
    }
    
    int f_and = 0;
    for(int i=0;i<3;i++){
        int sum = 0;
        for(int j=0;j<3;j++){
            int idx = i*3+j;
            sum += m[idx];
            if(m[idx]!=0){
                if(j==0){
                    printf("%s hundred ",s2[m[idx]]);
                    f_and = 1;
                }
                else if(j==1){
                     if(f_and==1){
                         printf("and ");
                         f_and = 0;
                     }
                    if(m[idx]>1) //输出>=20的整10数
                         printf("%s ",s0[m[idx]-1]);
                    else if(m[idx]==1 && m[idx+1]==0){  // ten
                        printf("ten ");
                    }
                }
                else{
                    if(f_and==1){
                         printf("and ");
                         f_and = 0;
                     }
                    if(m[idx-1]==1){
                        printf("%s ",s1[m[idx]]);
                    }
                    else{
                        printf("%s ",s2[m[idx]]);
                    }
                }
                
            }
            
        }
        if(sum!=0){
            if(i==0)
                printf("million ");
            else if(i==1)
                printf("thousand ");
        }
    }
}