题解 | #学英语#

# 2024年11月8日   周五   下午16:07

# long类型是一种整数数据类型,通常用于存储较大范围的整数值
# 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。
#  1,652,510:  one million six hundred and fifty two thousand five hundred and ten

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


def f100(n):
    if n > 0:  # 这个n > 0一定要写
        if n < 20:
            word.append(num1[n])
        elif 20 <= n < 100:  # 39//30

            word.append(num2[n // 10])
            if n % 10 != 0:
                word.append(num1[n % 10])


def f1000(n):
    if n >= 100:  # 789

        word.append(num1[n // 100])
        word.append("hundred")
        if n % 100 != 0:
            word.append("and")
    f100(n % 100)  # 注意这里的缩进


n = int(input())
word = []
# 处理n
a = n % 1000
b = (n // 1000) % 1000  # thousand
c = (n // 1000000) % 1000  # million
d = (n // 1000000000) % 1000  # billion
if d > 0:
    f1000(d)
    word.append("billion")
if c > 0:
    f1000(c)
    word.append("million")

if b > 0:
    f1000(b)
    word.append("thousand")
if a > 0:
    f1000(a)
print(*word)