HJ42 题解 | #学英语#

#思路：
#把需要用到的词汇存起来
#定义一个函数f处理1-1000的，函数中需要处理各种特殊情况
#使用函数f，分别处理1000以内级、1000级、百万级

a=["one","two","three","four","five","six","seven","eight","nine","ten","elven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen",
"nineteen","twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety","hundred","thousand","million"]
b=["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17"
,"18","19","20","30","40","50","60","70","80","90","100","1000","1000000"]

s=input()
s1=int(s)
ans=""
#对于1-1000的定义一个函数，因为后面大量使用
def f(s):
    #如果在b中直接返回s，这样解决整十还要计算个位的问题
    if s in b:
        if int(s)<100:
            res=a[b.index(s)]
        else:
            res="one "+a[b.index(s)]
        return res
    ge=s[-1]
    if int(s)<10:
        if ge=="0":
            res=""
        else:
            res=a[b.index(ge)]
    elif int(s)<100:#小于100
        shi=s[-2]+"0"
        if ge=="0":
            res=a[b.index(shi)]
        else:
            res=a[b.index(shi)]+" "+a[b.index(ge)]#拼接结果

    else:#大于100，小于1000
        shi=s[-2]+"0"
        bai=s[-3]
        if s[-2:] in b:#类似919这种，后两位是直接存在的
            res=a[b.index(bai)]+" hundred and "+a[b.index(s[-2:])]
            return res
        if ge=="0":#类似400或430这样的
            if shi=="00":#特殊情况，403这样的，十位为0
                res=a[b.index(bai)]+" hundred"
            else:
                res=a[b.index(bai)]+" hundred and "+a[b.index(shi)]
        else:
            if shi=="00":#特殊情况，403这样的，十位为0
                res=a[b.index(bai)]+" hundred and "+a[b.index(ge)]
            else:
                res=a[b.index(bai)]+" hundred and "+a[b.index(shi)]+" "+a[b.index(ge)]
    return res

#按规则实现
#小于1000
if s1<1000:
    ans=f(s)
#大于1000，小于百万
elif s1<1000000:
    ans=f(s[:-3])+" thousand "+f(s[-3:])
#大于百万
else:
    ans=f(s[:-6])+" million "+f(s[-6:-3])+" thousand "+f(s[-3:])

print(ans)