P1.1键为数字加1键 P1.0为数字减1键

           使LED显示数据从000到999之间一次加一或减一

解决方案：（C语言编程）

#include      
#include
#define uint unsigned int
#define uchar unsigned char
uchar bai,shi,ge,num;
uint shu;
sbit uk= P1^1;
sbit dk= P1^0;
sbit le=P3^6;
void delay(uint a);
void display(uchar ba,uchar sh,uchar g);
uchar code tabledu[]={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,
0xf8,0x80,0x90,0x88,0x83,0xc6,0xa1,0x86,0x8e,0x89,0x86,0xc7,0xc7,0xc0,0xff};
uchar code tablewe[]={0xff,0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};
void init()
{
 shu=0;
 bai=0;
 shi=0;
 ge=0;
}
void main()
{
 init();
 while(1){  
  if(uk==0){
   delay(5);
   if(uk==0){
    if(shu==999)
     shu=0;
    shu++;
    bai=shu/100;
             shi=shu%100/10;
    ge=shu%10;
    }
   while(!uk) ;
   }
     if(dk==0){
   delay(5);
   if(dk==0){
    if(shu==0)
     shu=999;
    bai=shu/100;
             shi=shu%100/10;
    ge=shu%10;
    shu--;
    }
   while(!dk) ;
   }
  display(bai,shi,ge); 
 }
}
void display(uchar ba,uchar sh,uchar g)
{
 le=1;
 P2=tablewe[1];
 le=0;
 le=1;
 P0=tabledu[ba];
 le=0;
 delay(5);
 le=1;
 P2=tablewe[2];
 le=0;
 le=1;
 P0=tabledu[sh];
 le=0;
 delay(5);
 le=1;
 P2=tablewe[3];
 le=0;
 le=1;
 P0=tabledu[g];
 le=0;
 delay(5); 
}
void delay(uint a) {
 uint i,j;
 for(j=a;j>0;j--)
     for(i=250;i>0;i--);
}