蓝桥杯单片机第十三届模拟题

Posted 2022-11-30 一心向月

全部文件：提取码：azxy
这个模拟题主要容易出错的地方是通道1和3的切换时只会读取通道1的值，只要把IIC_SendAck(0)改成IIC_SendAck(1)就可以了。

#include "STC15F2K60S2.h"
#include "iic.h"
typedef unsigned char u8;
typedef unsigned int u16;
sbit S4=P3^3;
sbit S5=P3^2;
sbit S6=P3^1;
sbit S7=P3^0;

//					0	 1	  2	   3    4    5    6    7    8    9  
u8 code duan[12]=0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90,0xff,0xbf;
u8 model=1;//界面标志
u16 pl=0,pl_hc=0;//频率和频率缓存
u16 tt=0;//周期
u16 uu=0,uu_hc=0;//通道电压和通道3电压缓存
u8 td=1;//通道
u16 count=0;//定时器0脉冲
u8 count1=0;//1S标志
u16 count2=0;//长按判断
u8 led_close=0;//LED关闭标志

void delay(u16 i)

	while(i--);


void Delay5ms()	

	unsigned char i, j;

	i = 59;
	j = 90;
	do
	
		while (--j);
	 while (--i);


void TimerInit(void)		

	AUXR |= 0x40;		
	TMOD = 0x16;		
	TL1 = 0xA0;		//5ms 
	TH1 = 0x15;		
	TL0 = 0xff;
	TH0 = 0xff;
	TR1 = 1;	
	TR0 = 1;	
	ET0 = 1;
	ET1 = 1;
	EA = 1;


void Service_T0() interrupt 1

	count++;


void Service_T1() interrupt 3

	count1++;
	if(count1==200)
	
		pl=count;
		count=0;
		count1=0;
	


void Chan_hc573(chose)

	switch(chose)
	
		case 4:
			P2 = (P2 & 0x1f) | 0x80; break;
		case 5:
			P2 = (P2 & 0x1f) | 0xa0; break;
		case 6:
			P2 = (P2 & 0x1f) | 0xc0; break;
		case 7:
			P2 = (P2 & 0x1f) | 0xe0; break;
	


void Init_system()

	Chan_hc573(5);
	P0=0x00;
	Chan_hc573(4);
	P0=0xff;


void Smg_display(u8 com,u8 dat)

	Chan_hc573(7);
	P0=0xff;
	Chan_hc573(6);
	P0=0x01<<com;
	Chan_hc573(7);
	P0=dat;


void Close_smg()

	Chan_hc573(6);
	P0=0x00;
	Chan_hc573(7);
	P0=0xff;


void Display_f_n_u()

	if(model==1)
	
		Smg_display(0,0x8e); delay(1000);
		Smg_display(1,duan[10]); delay(1000);
		Smg_display(2,duan[10]); delay(1000);
		if(pl>9999) Smg_display(3,duan[pl/10000]); delay(1000);
		if(pl>999) Smg_display(4,duan[(pl/1000)%10]); delay(1000);
		if(pl>99) Smg_display(5,duan[(pl/100)%10]); delay(1000);
		if(pl>9) Smg_display(6,duan[(pl/10)%10]); delay(1000);
		Smg_display(7,duan[pl%10]); delay(1000);
		Close_smg();
	
	else if(model==2)
	
		tt=1000000/pl;
		Smg_display(0,0xc8); delay(1000);
		Smg_display(1,duan[10]); delay(1000);
		Smg_display(2,duan[10]); delay(1000);
		if(tt>9999) Smg_display(3,duan[tt/10000]); delay(1000);
		if(tt>999) Smg_display(4,duan[(tt/1000)%10]); delay(1000);
		if(tt>99) Smg_display(5,duan[(tt/100)%10]); delay(1000);
		if(tt>9) Smg_display(6,duan[(tt/10)%10]); delay(1000);
		Smg_display(7,duan[tt%10]); delay(1000);
		Close_smg();
	
	else if(model==3)
	
		Smg_display(0,0xc1); delay(1000);
		Smg_display(1,duan[11]); delay(1000);
		Smg_display(2,duan[td]); delay(1000);
		Smg_display(3,duan[10]); delay(1000);
		Smg_display(4,duan[10]); delay(1000);
		Smg_display(5,duan[(uu/100)%10]+0x80); delay(1000);
		Smg_display(6,duan[(uu/10)%10]); delay(1000);
		Smg_display(7,duan[uu%10]); delay(1000);
		Close_smg();
	


void Key_down()

	if(S4==0)
	
		delay(100);
		if(S4==0)
		
			while(S4==0) Display_f_n_u();
			model++;
			if(model==4) model=1;
		
	
	if(S5==0)
	
		delay(100);
		if(S5==0)
		
			while(S5==0) Display_f_n_u();
			if(model==3)
			
				if(td==1) td=3;
				else if(td==3) td=1;
			
		
	
	if(S6==0)
	
		delay(100);
		if(S6==0)
		
			while(S6==0) Display_f_n_u();
			if(td==3) uu_hc=uu;
		
	
	if(S7==0)
	
		delay(100);
		if(S7==0)
		
			while(S7==0) Display_f_n_u();count2++;Delay5ms();
			if(count2<=200) count2=0;pl_hc=pl;
			else if(count2>200) count2=0;led_close=~led_close;
		
	


void Read_td()

	if(td==1) uu=read_pcf8591(1)/0.51;
	else if(td==3) uu=read_pcf8591(3)/0.51;


void Led()

	Chan_hc573(4);
	if(led_close==0)
	
		if(uu>uu_hc) P00=0;
		if(pl>pl_hc) P01=0;
		if(model==1) P02=0;
		else if(model==2) P03=0;
		else if(model==3) P04=0;
	
	else P0=0xff;


void main()

	Init_system();
	TimerInit() ;
	while(1)
	
		Led();
		Read_td();
		Key_down();
		Display_f_n_u();