| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 8985 | Accepted: 3366 |
Description
According to the Gregorian calendar, which is the civil calendar in use today, years evenly divisible by 4 are leap years, with the exception of centurial years that are not evenly divisible by 400. Therefore, the years 1700, 1800, 1900 and 2100 are not leap years, but 1600, 2000, and 2400 are leap years.
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
You may assume that the resulting date won’t be after the year 9999.
Output
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
#include
int isRunnian(int a)
{
if(a%400==0||(a%4==0&&a%100))
return 1;
return 0;
}
int main()
{ int num=0;
int i;
int t,k,m;
struct
{
int year;
int mounth;
int day;
}date;
char str[7][10]={"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};
int b[2][12]={{31,28,31,30,31,30,31,31,30,31,30,31},{31,29,31,30,31,30,31,31,30,31,30,31}};//0平年1闰年
int n;
while(scanf("%d",&n)!=EOF&&n!=-1)
{
date.year=2000;date.mounth=1;date.day=1;
m=(n+6)%7;
n++;
while(n>(t=isRunnian(date.year)?366:365))
{
n-=t;
date.year++;
}
k=isRunnian(date.year);
i=0;
while(n>b[k][i])
{
n-=b[k][i];
++i;
date.mounth++;
}
date.day=n;
printf("%d-%02d-%02d %s\n",date.year,date.mounth,date.day,str[m]);
}
return 0;
}