问答题

已知C源程序如下：
/*Input today’s date,output tomorrow’s date*/
/*version 2*/
#include＜stdio.h＞
struct ydate
{ int day;int month;int year;};
int leap(struct ydate d)
{ if((d.year%4==0&&d.year%100!=0)||(d.year%400==0))
　　　　return 1;
　　 else
　　　　return 0;
}
int numdays(struct ydate d)
{ int day;
　　 static int daytab[]=
　　　　{31,28,31,30,31,30,31,31,30,31,30,31);
　　 if(leap(d)&&d.month==2)
　　　　day=29;
　　 else
　　　　day=daytab[d.month-1];
　　 return day;
}
int main(void)
{ struct ydate today,tomorrow;
　　 printf("format of date is:year,month,day 输入的年、月、日之间应用逗号隔开\n")；
　　 printf("today is:");
　　 scanf("%d,%d,%d",&today.year,&today.month,&today.day);
　　 while(0＞=today.year
|| today.year＞65535 || 0＞=today.month || today.month＞12) ||
0＞=today.day || today.day＞numdays(today))
　　 {　　printf("input date error!reenter the day!\n");
printf("today is：");
scanf("%d,%d,%d",&today.year,&today.month,&today.day);
}
if(today.day!=numdays(today))
{ tomorrow.year=today.year;
tomorrow.month=today.month;
tomorrow.day=today.day+1;
}
else if(today.month==12)
{ tomorrow.year=today.year+1;
tomorrow.month=1;
tomorrow.day=1;
}
else
{ tomorrow.year=today.year;
tomorrow.month=today.month+1;
tomorrow.day=1;
}
printf("tomorrow is:%d,%d,%d\n\",
tomorrow.year,tomorrow.month,tomorrow.day);
}

设计一组测试用例，使该程序所有函数的语句覆盖率和分支覆盖率均能达到100%。如果认为该程序的语句或分支覆盖率无法达到100%．需说明为什么。