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\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%，则说明为什么。
