我在互联网上搜索过，找不到它，请帮助解决这个
我找到了答案
Number of inputs = 16(A) + 16(B) + 1(Cin) = 33 address bits
Number of outputs = 16(sum/diff) + 1(Cout) = 17
Thus, this would require a 2^33 x 17-bit ROM.
这是参考资料点击这里下载David Money Harris的解决方案第181页练习5.57 a(