已知C源程序如下：/*Input today’s date,output tomor

题目：

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);

画出程序中所有函数的控制流程图