我正在尝试对串行协议进行逆向工程，除了校验和字节外，一切似乎都很简单。似乎是一些简单的算法 - 当所有字节的总和相同时，校验和/crc 保持不变。但是还是想不出确切的算法：

最后一个字节是校验和。

80,6F,A3,01,02,B0,08,18
80,6F,A3,03,00,80,00,40
80,6F,A3,02,00,80,00,41
80,6F,A3 ,04,00,80,00,43
80,6F,A3,01,01,81,08,48
80,6F,A3,01,02,80,08,48
80,6F,A3,01,00, 81,08,49
80,6F,A3,01,04,80,08,4A
80,6F,A3,01,00,80,08,4E
80,6F,A3,01,02,A0,08,68
80,6F,A3,01,01,A2,08,6B
80,6F,A3,01,02,88,08,70
80,6F,A3,01,06,80,08,74
80,6F,A3 ,01,00,88,08,76
80,6F,A3,01,02,90,08,78
80,6F,A3,01,12,80,08,78

回应：

10,03,A3,00,01,00,38,06,73,9C,9B,94,00,12,12,02 10,03
,A3,00,01,00,38,0A,74,A0 ,8E,94,00,12,12,06 10,03
,A3,00,01,00,40,06,72,9E,99,94,00,12,12,0B 10,03
,A3,00 ,01,00,30,07,72,9B,9A,8E,00,12,12,12 10,03
,A3,00,01,00,38,04,74,77,78,CC,01, 12,12,12 10,03
,A3,00,01,00,30,07,71,9B,99,90,00,12,12,12 10,03
,A3,00,01,00,30, 07,6F,9F,92,94,00,12,12,13 10,03
,A3,00,01,00,30,07,6E,A1,8C,94,00,12,12,14
10, 03,A3,00,01,00,30,07,6E,A0,8E,94,00,12,12,17 10,03
,A3,00,01,00,30,07,72,9B,9A ,90,00,12,12,1C 10,03
,A3,00,01,00,30,06,72,A1,91,94,00,12,12,1C 10,03
,A3,00,01 ,00,30,07,72,9B,99,90,00,12,12,1D 10,03
,A3,00,01,00,30,07,70,9C,98,94,00,12, 12,1F 10,03
,A3,00,01,00,38,05,74,76,77,00,00,12,12,2C
10,03,A3,00,01,00,B4,03,6C,3E,3A,00,00,12,00,31 10,03
,A3,00,01,00,30,06,72,72 ,71,00,00,12,12,33

第一个字节保持不变，不确定我是否应该将它们计入校验和。

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