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第三题

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digits = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
index=s.index('.')

def char2num(s):
    return digits[s]

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

pass

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