第二题
return reduce(lambda x, y : x * y, L)




第三题
    pos = s.find('.')
str = s[0:pos] + s[pos+1:]
print(str)
def fn(x, y):
    return x * 10 + y
def char2num(s):
    return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
def pown(base,n = 2):
    result = 1
    while n :
        n = n - 1
        result *= base
    return result
integer = reduce(fn, map(char2num, str))
return integer / pown(10,pos)