Leap year calculation

Question

The length of a year is (more or less) 365.242196 days. So we have to subtract, more or less, a quarter of a day to make it fit :

365.242196 - 0.25 = 364.992196 (by adding 1 day in 4 years) : but oops, now it's too small!! lets add a hundreth of a day (by not adding that day once in a hundred year :-))

364.992196 + 0,01 = 365.002196 (oops, a bit too big, let's add that day anyway one time in about 400 years)

365.002196 - 1/400 = 364.999696

Almost there now, just play with leapseconds now and then, and you're set.

(Note : the reason no more corrections are applied after this step is because a year also CHANGES IN LENGTH!!, that's why leapseconds are the most flexible solution, see for examlple here)

That's why i guess

There's an algorithm on wikipedia to determine leap years:

function isLeapYear (year):
    if ((year modulo 4 is 0) and (year modulo 100 is not 0))
    or (year modulo 400 is 0)
        then true
    else false

There's a lot of information about this topic on the wikipedia page about leap years, inclusive information about different calendars.

In general terms the algorithm for calculating a leap year is as follows...

A year will be a leap year if it is divisible by 4 but not by 100. If a year is divisible by 4 and by 100, it is not a leap year unless it is also divisible by 400.

Thus years such as 1996, 1992, 1988 and so on are leap years because they are divisible by 4 but not by 100. For century years, the 400 rule is important. Thus, century years 1900, 1800 and 1700 while all still divisible by 4 are also exactly divisible by 100. As they are not further divisible by 400, they are not leap years

this is enough to check if a year is a leap year.

if( (year%400==0 || year%100!=0) &&(year%4==0))
    cout<<"It is a leap year";
else
    cout<<"It is not a leap year";