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第三题

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一弦三思

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s1=s.split('.')[0]
s2=s.split('.')[1]
return reduce(lambda x,y:x*10+y,map(char2num,s1+s2))/10**len(s2)

def char2num(r): return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[r]


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