题解 | #学英语#

对输入字符型数字，从右往左提取每三位进行处理，以字符串长度整除3的值作为循环次数判断

d = {
    '1': 'one','2': 'two','3':'three', '4':'four', '5':'five', 
    '6':'six', '7':'seven', '8':'eight', '9':'nine', '10':'ten', 
    '11':'eleven', '12':'twelve', '13':'thirteen', '14':'fourteen', 
    '15':'fifteen', '16':'sixteen', '17':'seventeen', '18':'eighteen',
    '19':'nineteen', '20':'twenty', '30':'thirty', '40':'forty', '50':'fifty', 
    '60':'sixty', '70':'seventy', '80':'eighty', '90':'ninety', '100': 'hundred',
    '1000': 'thousand', '1000000': 'million', '1000000000': 'billion', 
}
while True:
    try:
        num = input()
        res = []
        i = 0
        while i <= len(num)//3:  # 字符串数字，从右到左每三位进行处理
            # 获取分段三位(注意num[-3:0]无效)
            seg = num[-3:] if i==0 else num[-3*(i+1):-3*i]  
            if seg == '':  # 分段为空推出循环
                break
            if i >= 1:  # 分段从i=1开始处理thousand、million等
                seg_digit = '1' + '000'*i
                res.append(d[seg_digit])
            # 分段三位不满三位向左填充0
            seg = '{:0>3}'.format(seg)
            ge, shi, bai = seg[2], seg[1], seg[0]  # 个位、十位、百位
            if ge != '0' and shi != '1':  # 个位不为0且十位不为1，nineteen等特殊处理
                res.append(d[ge])
            if shi != '0':
                if shi == '1': # 十位为1，用seg[1:]十位+个位获取英文
                    res.append(d[seg[1:]])
                else:
                    res.append(d[f'{shi}0'])
            if bai != '0':
                if shi!='0' or ge!='0':  # 百位后无十位或个位则不需要添加and
                    res.append('and')
                res.append(f'{d[bai]} hundred')
            i += 1
        print(' '.join(res[::-1]))
    except:
        break