有关日历计算,

题目

有关日历计算,
要求输入n,输出从2000.1.1开始往后n天的那一天的年月日星期.
Given the number of days that have elapsed since January 1,2000 A.D,your mission is to find the date and the day of the week.
Input
The input consists of lines each containing a positive integer,which is the number of days that have elapsed since January 1,2000 A.D.The last line contains an integer -1,which should not be processed.You may assume that the resulting date won't be after the year 9999.
Output
For each test case,output one line containing the date and the day of the week in the format of "YYYY-MM-DD DayOfWeek",where "DayOfWeek" must be one of "Sunday","Monday","Tuesday","Wednesday","Thursday","Friday" and "Saturday".
Sample Input
1730
1740
1750
1751
-1
Sample Output
2004-09-26 Sunday
2004-10-06 Wednesday
2004-10-16 Saturday
2004-10-17 Sunday
要c的

答案

以下是用C++写的
#include
struct InNum
{
int num;
InNum *next;
};
void main()
{
int year=2000,day=1,mouth=1,dayTemp=0,week0=6,week;
InNum *head,*last,*p;
head=new InNum;
head->next=NULL;
last=head;
p=new InNum;
cin>>p->num;
p->next=NULL;
last->next=p;
last=last->next;
while(p->num>0)
{
p=new InNum;
cin>>p->num;
p->next=NULL;
last->next=p;
last=last->next;
}
p=head->next;
while(p->num>=0)
{
p->num--;
while(p->num>=365)
{
year++;
p->num-=365;
}
dayTemp=p->num;
while(p!=NULL && p->num>28)
{
if(2==mouth && 0==year%4 && p->num>=29)
{
mouth++;
p->num-=29;
}
else if(2==mouth && p->num>=28)
{
mouth++;
p->num-=28;
}
else if(mouthnum>mouth%2+30)
{
p->num-=(mouth%2+30);
mouth++;
}
else if(mouth>=9 && p->num>(mouth+1)%2+30)
{
p->num-=((mouth+1)%2+30);
mouth++;
}
else
break;
}
day = p->num;
week = (dayTemp + year-2001+1)%7;
cout