题解 | #学英语#

这题我先从后往前，3个为1组，生成数组。 然后针对一个数组，比如123这个数字，进行输出 最后有几个3位一组就分别对应百万，千，百

``` python []
def numtostr(f):
    gw=["one","two","three","four","five","six","seven","eight","nine"]
    xe=["ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"]
    sw=["twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"]
    s=''#用于计数，我是把他理解成字符串来做的
    if len(f)==1:#转换成3位
        f='00'+f
    if len(f)==2:#转换成3位
        f="0"+f
    if len(f)==3:
        if int(f[-2:])==0:#考虑100，200，300这种情况
            s=s+" "+gw[int(f[0])-1]+" hundred"
        else:
            if int(f[0])!=0:
                s=s+gw[int(f[0])-1]+" hundred and"
                if 0<int(f[-2:])<=9:#解决01,02这种情况
                    s=s+" "+gw[int(f[-2:])-1]
                elif 10<=int(f[-2:])<=19:#解决11，12，13~19这种情况
                    s=s+" "+xe[int(f[-2:])-10]
                elif int(f[2])==0:#考虑个位数为0的这种情况，比如10，20，30等
                    s=s+" "+sw[int(f[1])-2]
                else:
                    s=s+" "+sw[int(f[1])-2]+" "+gw[int(f[2])-1]
            else:
                if 0<int(f[-2:])<=9:#解决01,02这种情况
                    s=s+" "+gw[int(f[-2:])-1]
                elif 10<=int(f[-2:])<=19:#解决11，12，13~19这种情况
                    s=s+" "+xe[int(f[-2:])-10]
                elif int(f[2])==0:#考虑个位数为0的这种情况，比如10，20，30等
                    s=s+" "+sw[int(f[1])-2]
                else:
                    s=s+" "+sw[int(f[1])-2]+" "+gw[int(f[2])-1]
    return str(s).lstrip(" ")#最后还有把前面的空格去除
#numtostr("209")#测试用，测试一些特殊情况是否符合            
while True:
    try:
        n=str(input())
        #tmpnum=[]#每三位为一组
        tmpnum=n[::-1]#先倒叙
        tmp=[]
        for j in range(0,len(tmpnum),3):
            tmp.append("".join(tmpnum[j:j+3][::-1]))#然后负负得正
        #3个一组得到的结果大概是这样的以1652510举例，得到结果为['510', '652', '1']
        #print(tmp)
        #numtostr("123")
        if len(tmp)==1:#如果只有3位
            print(numtostr(tmp[0]))#就直接打印,还得取首页空格
        if len(tmp)==2:#如果超过3位就加thousand
            print(numtostr(tmp[1])+" thousand "+numtostr(tmp[0]))
        if len(tmp)==3:#如果超过6位就加million
            print(numtostr(tmp[2])+" million "+numtostr(tmp[1])+" thousand "+numtostr(tmp[0]))
        if len(tmp)==4:#
            print(numtostr(tmp[2])+" billion "+numtostr(tmp[2])+" million "+numtostr(tmp[1])+" thousand "+numtostr(tmp[0]))
    except:
        break