使用C语言，写出3个可操作的STC89C52RC的流水灯程序

以下是3个可操作的STC89C52RC的流水灯程序：

程序1：

#include <reg52.h>

sbit LED1=P2^0;
sbit LED2=P2^1;
sbit LED3=P2^2;
sbit LED4=P2^3;
sbit LED5=P2^4;
sbit LED6=P2^5;
sbit LED7=P2^6;
sbit LED8=P2^7;

void delay(unsigned int i)
{
while(i--);
}

void main()
{
while(1)
{
LED1=0;
delay(50000);
LED1=1;
LED2=0;
delay(50000);
LED2=1;
LED3=0;
delay(50000);
LED3=1;
LED4=0;
delay(50000);
LED4=1;
LED5=0;
delay(50000);
LED5=1;
LED6=0;
delay(50000);
LED6=1;
LED7=0;
delay(50000);
LED7=1;
LED8=0;
delay(50000);
LED8=1;
}
}

程序2：

#include <reg52.h>

sbit LED1=P2^0;
sbit LED2=P2^1;
sbit LED3=P2^2;
sbit LED4=P2^3;
sbit LED5=P2^4;
sbit LED6=P2^5;
sbit LED7=P2^6;
sbit LED8=P2^7;

void delay(unsigned int i)
{
while(i--);
}

void main()
{
while(1)
{
LED1=0;
delay(50000);
LED1=1;
LED2=0;
delay(50000);
LED2=1;
LED3=0;
delay(50000);
LED3=1;
LED4=0;
delay(50000);
LED4=1;
LED5=0;
delay(50000);
LED5=1;
LED6=0;
delay(50000);
LED6=1;
LED7=0;
delay(50000);
LED7=1;
LED8=0;
delay(50000);
LED8=1;
LED7=0;
delay(50000);
LED7=1;
LED6=0;
delay(50000);
LED6=1;
LED5=0;
delay(50000);
LED5=1;
LED4=0;
delay(50000);
LED4=1;
LED3=0;
delay(50000);
LED3=1;
LED2=0;
delay(50000);
LED2=1;
}
}

程序3：

#include <reg52.h>

sbit LED1=P2^0;
sbit LED2=P2^1;
sbit LED3=P2^2;
sbit LED4=P2^3;
sbit LED5=P2^4;
sbit LED6=P2^5;
sbit LED7=P2^6;
sbit LED8=P2^7;

void delay(unsigned int i)
{
while(i--);
}

void main()
{
while(1)
{
LED1=0;
delay(50000);
LED2=0;
delay(50000);
LED3=0;
delay(50000);
LED4=0;
delay(50000);
LED5=0;
delay(50000);
LED6=0;
delay(50000);
LED7=0;
delay(50000);
LED8=0;
delay(50000);
LED1=1;
LED2=1;
LED3=1;
LED4=1;
LED5=1;
LED6=1;
LED7=1;
LED8=1;
LED7=0;
delay(50000);
LED8=0;
delay(50000);
LED6=0;
delay(50000);
LED5=0;
delay(50000);
LED3=0;
delay(50000);
LED2=0;
delay(50000);
LED1=0;
delay(50000);
LED4=0;
delay(50000);
}
}