"leap year" Code Answer's

leap year algorithm

By Marton on Feb 18, 2021

def isLeapYear (year):
    if ((year % 4 == 0) and (year % 100 != 0)) or (year % 400 == 0):
		true
    else:
    	false

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find leap year javascript

By Jealous Jaguar on Jul 01, 2020

function isLeap(year) {
   if (year % 4 === 0) {
      if (year % 100 === 0){
         if (year % 400 == 0){
            return ("Leap year.");
         } else {
            return ("Not leap year.");
         }
      } else {
         return ("Leap year.");
      }
   } else{
      return ("Not leap year.");
   }
}

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leap year

By Outrageous Ocelot on Feb 27, 2021

I'll simplfy this. Humans cut every year 6 hours short (1/4 of a day[it is
actually less than 1/4 but see below for the rest]).
This means that every 4 years, 24 hours have been lost and to make up for
that, we just add the 24 hours to a much shorter month (like february - 28d)
giving us the 'leap year' and making the year 1 day longer.


This is to keep up with the astronomical years which the western calender
is based off of, in which the terrarial 'lap' of the sun is the equivilant
of 265.24 full rotations of the earth (day/night cycles). The reason we don't
measure individual years in 365.24 earth rotations is because we would end up
having a super long day at the end of the year and then days would end up
being dark or light depending on the year and it would all get in a 
big muddle. Instead, we keep the calendar normal and just remove the .24 of 
a day off every 3 years, only to add all the lost time together and make
another day every four years (or every leap year). You may have noticed
that 4x0.24 is actually not 100 but 0.96, so surely after 100 years,
we would miss a day? Well actually, in the year 2000, we did still have a leap
year because it is a multiple of 400(a leap century if you will),
but every other 100 years, we do not have an extra day because of that 0.24.

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Check Leap Year

By BreadCode on May 03, 2021

#include <iostream>
using namespace std;

int main() {
    int year;
    cout << "Enter a year: ";
    cin >> year;
    if (year % 4 == 0) {
        if (year % 100 == 0) {
            (year % 400 == 0) ?
            cout << year << " is a leap year." :
            cout << year << " is not a leap year.";
        }
        else
            cout << year << " is a leap year.";
    }
    else
        cout << year << " is not a leap year.";
    return 0;
}

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leap year

By Weary Weevil on May 06, 2021

import calendar

def is_leap(year):
    return calendar.isleap(year)

Source: www.hackerrank.com

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