求一份由32*16的led点阵屏 AT89C51控制的protues仿真图及程序 谢谢

Answer 1

#include<reg51.h>

unsigned  char  code  P2H[]={  0x10,0x20,0x40,0x80 }; 

unsigned  char  code  P2L[]={  0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07 }; 

int a,b,n,i,j,k,t;

unsigned  char  code  ziku[]={ //字符码的存放格式为:从左到右共16列，并按列的上下部分依次存放

//一组即为一列

/* EL */

0x08,0x20,  0xF8,0x3F,  0x88,0x20,  0x88,0x20,  0xE8,0x23,  0x08,0x20,  0x10,0x18,  0x00,0x00,  

0x08,0x20,  0xF8,0x3F,  0x08,0x20,  0x00,0x20, 0x00,0x20, 0x00,0x20,  0x00,0x30, 0x00,0x00,

/* Va */

0x08,0x00,  0x78,0x00,  0x88,0x07,  0x00,0x38,  0x00,0x0E,  0xC8,0x01,  0x38,0x00,  0x08,0x00,

0x00,0x00,  0x00,0x18,  0x80,0x24,  0x80,0x22, 0x80,0x22, 0x80,0x22,  0x00,0x3F, 0x00,0x20,

/* 秦 */  

0x20,0x82,  0x20,0x82,  0x2A,0x45,  0x2A,0x45,  0xAA,0x25,  0x6A,0x15,  0x3E,0x0D,  0x2B,0xFF,

0xAA,0x04,  0xAA,0x0C,  0xEA,0x14,  0xAA,0x24,  0x2A,0x65,  0x22,0xC2,  0x20,0x42,  0x00,0x00,

/* 波 */  

0x10,0x04,  0x60,0x04,  0x01,0xFF,  0x16,0x40,  0x00,0x30,  0xF8,0x8F,  0x88,0x80,  0x88,0x41,

0x88,0x26,  0xFF,0x18,  0x88,0x18,  0x88,0x26,  0x88,0x61,  0x18,0xC0,  0x08,0x40,  0x00,0x00,

/* , */

0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x80,  

0x00,0xB0,  0x00,0x70,  0x00,0x00,  0x00,0x00, 0x00,0x00, 0x00,0x00,  0x00,0x00, 0x00,0x00,

/* 自 */

0x00,0x00,  0x00,0x00,  0x00,0x00,  0xF8,0x7F,  0x48,0x24,  0x4C,0x24,  0x4A,0x24,  0x48,0x24, 

0x48,0x24,  0x48,0x24,  0x48,0x24,  0xF8,0x7F,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,

/* 动 */

0x00,0x00,  0x40,0x08,  0x48,0x1C,  0x48,0x0A,  0xC8,0x09,  0x48,0x08,  0x48,0x4E,  0x40,0x28,  

0x20,0x10,  0x20,0x0C,  0xFE,0x03,  0x20,0x20,  0x20,0x40,  0xF0,0x3F,  0x20,0x00,  0x00,0x00,

/* 化 */

0x00,0x00,  0x00,0x01,  0x80,0x00,  0x40,0x00,  0xF0,0x7F,  0x0E,0x00,  0x00,0x04,  0x00,0x04,

0x00,0x02,  0xFE,0x3F,  0x80,0x40,  0x40,0x40,  0x20,0x40,  0x10,0x40,  0x00,0x78,  0x00,0x00,

/* 04 */

0x00,0x00,  0xE0,0x0f,  0x10,0x10,  0x08,0x20,  0x08,0x20,  0x10,0x10,  0xE0,0x0F,  0x00,0x00,

0x00,0x00,  0x00,0x07,  0xC0,0x04,  0x20,0x24,  0x10,0x24,  0xF8,0x3F,  0x00,0x24,  0x00,0x00,

/* 1 */

0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x10,0x20,  0x10,0x20,  0xF8,0x3F,  

0x00,0x20,  0x00,0x20,  0x00,0x00,  0x00,0x00, 0x00,0x00, 0x00,0x00,  0x00,0x00, 0x00,0x00,

/* 班 */

0x42,0x10,  0x42,0x10,  0xFE,0x0F,  0x42,0x88,  0x42,0x48,  0xE0,0x20,  0x00,0x18,  0xFF,0x07,  

0x00,0x20,  0x42,0x20,  0x42,0x20,  0xFE,0x3F, 0x42,0x20, 0x42,0x20,  0x42,0x20, 0x00,0x00,

/* . */

0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  0x00,0x00,  

0x00,0x30,  0x00,0x30,  0x00,0x00,  0x00,0x00, 0x00,0x00, 0x00,0x00,  0x00,0x00, 0x00,0x00,

};

void   delay(int t)

{    

     for(t;t>0;t--);

}

void  show()

{

for(n=0;n<384;n+=2) //共六个字

{

   for(k=0;k<10;k++)    //每个字显示的时间

   {

          for(i=0;i<4;i++)        //4个字符区

        {  

           for(j=0;j<8;j++)    //8列

           {

   a=j+i*8+n;

b=j+i*8+n+32;

           P2=P2H[i]+P2L[j];   //位选端

if(a>383) 

    P0=ziku[a-384];

   else   

    P0=ziku[a];

if(b>383) 

    P1=ziku[b-384];  

   else   

P1=ziku[b];

             delay(10); //80～110

           }

       }

   }

}

}

void main()

{

while(1)

{

   show();

}

}

Answer 2

我可以做的

我做单片机的