将非负整数num转换为其对应的英文表示

示例1：
输入：num=123
输出：one hundred twenty three

示例2：
输入：num=12345
输出：twelve thousand three hundred forty five

示例3：
输入：num=1234567
输出：one million two hundred thirty four thousand five hundred sixty seven

示例4：
输入：1234567891
输出：one million two hundred thirty four million five hundred sixty seven thousand eight hundred ninety one

提示：

• 0<=num<=2^31-1

实现代码

num=input("请输入一个整数:\n")
def less_Hundred(num):
    L1 = ["zero","one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"]
    L2 = ["eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen",
          "twenty"]
    L3 = ["thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]
    if int(num)<=10:
        return L1[int(num)]
    elif 10<int(num)<=20:
        return L2[int(num)%10-1]
    elif 20<int(num)<30:
        return L2[-1]+" "+L1[int(num)%10]
    elif 30<=int(num)<100:
        return L3[int(num)//10-3]+" "+L1[int(num)%10]
def less_thousand(num):
    if int(num)<100:
        return less_Hundred(num)
    if 100<=int(num)<1000:
        return less_Hundred(num[0])+" hundred "+less_Hundred(num[1:])
def less_million(num):
    s=""
    if int(num)<1000:
        s+=less_thousand(num)
    elif 1000<=int(num)<1000000:
        s+=less_thousand(num[:-3])+" thousand "
        if num[-3:]!="000":
            s+=less_thousand(num[-3:])
    elif 1000000000>int(num)>=1000000:
        s+=less_thousand(num[:-6])+" million "
        if num[-6:-3]!="000":
            s += less_thousand(num[-6:-3]) + " thousand "
        if num[-3:] != "000":
            s += less_thousand(num[-3:])
    elif int(num)>=1000000000:
        s+=less_thousand(num[:-9])+" billion "
        if num[-9:-6]!="000":
            s+=less_thousand(num[-9:-6])+" million "
        if num[-6:-3]!="000":
            s+=less_thousand(num[-6:-3]) + " thousand "
        if num[-3:]!="000":
            s+=less_thousand(num[-3:])
    return s


print(less_million(num))