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第三题

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    n = s.find('.')

    def char2num(s):

        DIGITS = {'0':0, '1':1, '2':2, '3':3, '4':4, '5':5, '6':6, '7':7, '8':8, '9':9,}

        return DIGITS[s]

    return reduce(lambda x, y: x * 10 + y,map(char2num,s[:n])) + reduce(lambda x, y: x * 10 + y,map(char2num,s[n+1:]))/10**n


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