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[]=
{31，28，31，30，31，30，3l，31，30，31，30，31}；
if(1eap(d)＆＆d. month==2)
day=29；
else
day=daytabEd. 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．%"，&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\n"，
tomorrow. year，tomorrow. month，tomorrow. day)；
}
(1) 画出程序中所有函数的控制流程图；
(2) 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。