单片机计数器实现减一答:sbit k3=P1^7;//加1 uchar seg[]={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90};uchar counter;void delay(uint m){ uint i,j;for(i=0;i<m;i++)for(j=0;j<10;j++);} void display(void){ P3=0x01;P2=seg[counter/10];//显示十位 delay(20);P3=0x02;P2...