Leap Year Checker...even more to it

[Colin Klipsch]

In article <4404 at catfish11.UUCP> yedinak at motcid.UUCP (Mark A. Yedinak) writes:
>johnb at srchtec.UUCP (John Baldwin) writes:
>: (Stan Brown, Oak Road Systems) writes:
>
>:|In article <24700010 at sunc1>, mccaugh at sunc1.cs.uiuc.edu writes:
>:|> Ordinarily, a leap-year is a multiple of four, so that--given leap-year y--
>:|> (y%4 == 0) ought to indicate if y designates a leap-year.
>
>:|Bzzzzzt!  Nope, but thanks for playing.  Vanna has lovely gifts for you.

To be nitpicky to the point of nausea, I will point out that the rule
for deciding leap years has FOUR conditions, not three.

A correct algorithm is:

#define FALSE 0
#define TRUE  1

int is_leap_year(int year)
{
   if (year%4000 == 0) return FALSE;
   else if (year%400 == 0) return TRUE;
   else if (year%100 == 0) return FALSE;
   else if (year%4 == 0) return TRUE;
   else return FALSE;
}

With the addition of the divisible-by-4000 condition, the Gregorian
calendar is now accurate to one day in 20,000 years.  (Assuming its
term of use lasts long enough for the rule to be practiced, which I
doubt!)

Source: _Exploration of the Universe_, G. Abell

--------------------------------------------------------------------------
"May the forces of evil become confused on the way to your house."
                                                          -- George Carlin
Bemusedly,                   | Disclaimers:
  Colin Klipsch              |   Not guaranteed to fulfill any purpose,
  Property of UVa Ast. Dept. |   express or implied.  Contents may have
  Charlottesville, Virginia  |   settled during shipping.  Not rated.  May
  cak3g at virginia.edu         |   cause drowsiness.  Use before 29-Feb.
____________________________/ \___________________________________________