looking for a good free calendar program download?

Answer 1

Try Mozilla Sunbird/Lightining
http://www.mozilla.org/projects/calendar/

From the makers of the Firefox browser, it runs on Windows, Mac OS X, Linux etc. Sunbird is standalone, while Lightning is an extension for the Mozilla Thunderbird mail and newsgroups client.

You can also try Chandler:
http://chandler.osafoundation.org/

Answer 2

i know where a good free calender is but it requires no download, www.bravenet.com, sighn up and then it gives you a list of free web tools. click calender follow the steps and your on your way.

Answer 3

try Google Desktop

Answer 4

If u've Turo C++ IDE, then u can run my program, it's good-
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
#include // for printf(), scanf(), getchar()
#define lpyr(year) ((!(year%4) && (year%100)) || (!(year%400) && (year%1000)))

int isdatevalid(int day, int month, int year);
int weekday(int day, int month, int year);
char week[7][10] ={"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};
main()
{
int month, day, year;
clrscr();
printf("Return the day of the week given the date.");
printf("Enter date in the form dd/mm/yyyy : ");
scanf("%d/%d/%d",&day,&month,&year);
if (isdatevalid(day,month,year))
{
printf("The day of the week for this date is %s",week[weekday(day,month,year)]);
}
else
printf("%d/%d/%d not a valid date!",day,month,year);
getch();
}

// return 1 if date is valid, 0 otherwise.
int isdatevalid(int day, int month, int year)
{
if (day <= 0) return 0 ;
switch( month )
{
case 1 :
case 3 :
case 5 :
case 7 :
case 8 :
case 10 :
case 12 : if (day > 31) return 0 ; else return 1 ;
case 4 :
case 6 :
case 9 :
case 11 : if (day > 30) return 0 ; else return 1 ;
case 2 : if ( day > 29 ) return 0 ;
if ( day < 29 ) return 1 ;
if (lpyr(year)) return 1 ; // leap year
else return 0 ;
}
return 0 ;
}
//given day, month, year, returns day of week, eg. Monday = 0 etc.
// tested for 1901 to 2099 (seems to work from 1800 on too)
int weekday(int day, int month, int year)
{
int ix, tx, vx;
switch (month) {
case 2 :
case 6 : vx = 0; break;
case 8 : vx = 4; break;
case 10 : vx = 8; break;
case 9 :
case 12 : vx = 12; break;
case 3 :
case 11 : vx = 16; break;
case 1 :
case 5 : vx = 20; break;
case 4 :
case 7 : vx = 24; break;
}
if (year > 1900) // 1900 was not a leap year
year = year-1900;
ix = ((year - 21) % 28) + vx + (month > 2); // take care of February
tx = (ix + (ix / 4)) % 7 + day; // take care of leap year
return (tx % 7);
}

Answer 5

This not exactly what you asked for, but it might suit your needs.

Hope it helps.

http://www.timeanddate.com/calendar/