新人求助，郭天祥的TX—c1板子。刚学中断。关于中断的

用定时器0的方式1，让P0^1以200MS间隔闪烁，用定时器1的方式1实现数码管前两位59S循环计时~ 我都把自己写的改的面目全非啦，间隔时间就是短压根达不到1S一次计时，调试的时候曾经把num2==70,才能达到1S计时60S循环。
以下书上的程序，，看看他对吗?郭天祥的板子 TX_C1 晶振11.0592.

#include <reg52.h>
#define uchar unsigned char
#define uint unsigned int
sbit led1=P1^0;
sbit dula=P2^6;
sbit wela=P2^7;
uchar code table[]={
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71};
uchar num,num1,num2,shi,ge;
void delayms(uint);
void display(uchar,uchar);
void main()
{
TMOD=0x11;
TH0=(65536-45872)/256;//装初值11.0592晶振定时器定时50ms数为45872
TL0=(65536-45872)%256;
TH1=(65536-45872)/256;
TH1=(65536-45872)%256;
EA=1;
ET0=1;
ET1=1;
TR0=1;
TR1=1;
while(1)
{
display(shi,ge);

}
}
void display(uchar shi,uchar ge)
{
dula=1;
P0=table[shi];
dula=0;
P0=0Xff;
wela=1;
P0=0xfe;
wela=0;
delayms(5);


dula=1;
P0=table[ge];
dula=0;
P0=0xff;
wela=1;
P0=0xfd;
delayms(5);
wela=0;

}
void delayms(uint ms)
{
uint i,j;
for(i=ms;i>0;i--)
for(j=110;j>0;j--);
}
void T0_time() interrupt 1
{
TH0=(65536-45872)/256;
TL0=(65536-45872)%256;
num1++;
if(num1==4)
{
num1=0;
led1=~led1;
}
}
void T1_time() interrupt 3
{
TH1=(65536-45872)/256;
TH1=(65536-45872)%256;
num2++;
if(num2==70)
{
num2=0;
num++;
if(num==60)
num=0;
shi=num/10;
ge=num%10;
}
}

#include <reg52.h>
#define uchar unsigned char
#define uint unsigned int
sbit led1=P1^0;
sbit dula=P2^6;
sbit wela=P2^7;
uchar code table[]={
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71};
uchar num,num1,num2,shi,ge;
void delayms(uint);
void display(uchar,uchar);
void main()
{
TMOD=0x11;
TH0=(65536-45872)/256;//装初值11.0592晶振定时器定时50ms数为45872
TL0=(65536-45872)%256;
TH1=(65536-45872)/256;
TH1=(65536-45872)%256;
EA=1;
ET0=1;
ET1=1;
TR0=1;
TR1=1;
while(1)
{
display(shi,ge);

}
}
void display(uchar shi,uchar ge)
{
dula=1;
P0=table[shi];
dula=0;
P0=0Xff;
wela=1;
P0=0xfe;
wela=0;
delayms(5);


dula=1;
P0=table[ge];
dula=0;
P0=0xff;
wela=1;
P0=0xfd;
delayms(5);
wela=0;

}
void delayms(uint ms)
{
uint i,j;
for(i=ms;i>0;i--)
for(j=110;j>0;j--);
}
void T0_time() interrupt 1
{
TH0=(65536-45872)/256;
TL0=(65536-45872)%256;
num1++;
if(num1==4)
{
num1=0;
led1=~led1;
}
}
void T1_time() interrupt 3
{
TH1=(65536-45872)/256;
TH1=(65536-45872)%256;
num2++;
if(num2==20)
{
num2=0;
num++;
if(num==60)
num=0;
shi=num/10;
ge=num%10;
}
}

按道理说if（num2==20）此时应该就是1S吧?

20个50ms   就是1s      按道理说   是这样的