2、解题步骤 (1)分离数据 A3B3A2B2AB1ADBOC3C2C Co=1011 0110 0111 A3A241A0-=1101,B3B2B1B=0110, ccC, C=01l1 (2)A3A2A1A与B3B2B1B0相比 A3A2A1A0>B3B2B1B0,FAB=A=1,则地址A=1 TY2Y3Y4Y=B3B2B1B0=01102､解题步骤 (1)分离数据 (2) A3A2A1A0与B3B2B1B0相比 A3A2A1A0 >B3B2B1B0 , FA>B=A= 1,则地址A=1 1Y 2Y 3Y 4Y= B3B2B1B0 =0110 A3B3 A2B2 A1B1 A0B0C3C2C1C0=“1011 0110 0111 A3A2A1A0=1101 , B3B2B1B0=0110 , C3C2C1C0=0111