1. C语言编程，怎么编写 时 分 秒 的程序

#include<stdio.h>

int main(){

int hour,minute,second

printf("请输入时间：")

scanf("%d:%d:%d",&hour,&minute,&second)

printf("Time:%02d:%02d:%02d\n",hour,minute,second)

return 0

}

#include<reg51.h>

#include<intrins.h>

unsigned char code table[]={0x3F,0x06,0x5b,0x4F,0x66,0x6d,0x7d,0x07,0x7f,0x6F}

char hour=10

char minute=59

char second=50

int number=1000

sbit P3_0=P3^0

sbit P3_1=P3^1

sbit P3_2=P3^2

sbit P3_3=P3^3

sbit P3_4=P3^4

sbit P3_5=P3^5

sbit P3_6=P3^6

sbit P3_7=P3^7

void delay2ms(void)   //误差 0us

{

  unsigned char a,b

  for(b=4b>0b--)

      for(a=248a>0a--)

  _nop_() //if Keil,require use intrins.h

}

void DisPlay(char hour,char minute,char second)

{

P0=table[hour/10]

P3_0=0//共阴低电平0为亮

delay2ms()

P3_0=1

P0=table[hour%10]

P3_1=0

delay2ms()

P3_1=1

P0=0x40

P3_2=0

delay2ms()

P3_2=1

P0=table[minute/10]

P3_3=0

delay2ms()

P3_3=1

P0=table[minute%10]

P3_4=0

delay2ms()

P3_4=1

P0=0x40

P3_5=0

delay2ms()

P3_5=1

P0=table[second/10]

P3_6=0

delay2ms()

P3_6=1

P0=table[second%10]

P3_7=0

delay2ms()

P3_7=1

}

void main()

{

TMOD=0x01

TH0=(65536-1000)/256

TL0=(65536-1000)%256

EA=1

ET0=1

TR0=1

while(1)

{

DisPlay(hour,minute,second)

}

}

void timeerr(void) interrupt 1 using 1

{

TR0=0

TH0=(65536-1000)/256

TL0=(65536-1000)%256

if(number==0)

{

if(second==59)

{

second=0

if(minute==59)

{

minute=0

if(hour==23)

{

hour=0

second=0

minute=0

}

else

{

hour++

}

}

else

{

minute++

}

}

else

{

second++

}

number=1000

}

else

{

number--

}

TR0=1

}

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