SRM144 DIV1 300

一般的编程题，注意考虑边界情况和各种约束的完备性，详情见测试用例

class BinaryCode:

    def decode(self, message):
        enc = [int(ch) for ch in message]
        return (self._decodeBy(enc,0), self._decodeBy(enc,1))

    def _conact(self, nlist):
        s = ''
        for i in nlist:
            s = s + str(i)
        return s

    def _get(self, a, i):
        if 0 <= i < len(a):
            return a[i]
        else:
            return 0

    def _decodeBy(self, enc, first):
        ori = [first] 
        for i in range(1, len(enc)):
            x = enc[i-1] - self._get(ori,i-1) - self._get(ori,i-2)
            if not x in [0, 1]:
                return 'NONE'
            ori.append(x)

        if enc[-1] != self._get(ori, len(ori)-1) + self._get(ori, len(ori)-2):
            return 'NONE'

        return self._conact(ori)

# test
o = BinaryCode()

# test: len(message) = 1
assert(('0', 'NONE') == o.decode("0"))

# test: 最后一个数的推导受到2个条件的约束, 是否都进行了验证
assert(('NONE', 'NONE') == o.decode("112"))

# test: 是否每个数都进行了01验证
assert(('NONE', 'NONE') == o.decode("932"))

# test case
assert(( "011100011",  "NONE" ) == o.decode("123210122"))
assert(( "01",  "10" ) == o.decode("11"))
assert(("NONE",  "NONE") == o.decode("123210120"))
assert(("NONE",  "NONE") == o.decode("3"))
assert(("01101001101101001101001001001101001",
  "10110010110110010110010010010110010") == o.decode("12221112222221112221111111112221111"))