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leap year condition

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Visitor answered on August 21, 2021 Popularity 10/10 Helpfulness 4/10

answer leap year condition

related Check if a year is leap year

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related def is_leap(year): leap = False

leap year condition

Comment

function isLeapYear(year){

  if(year % 4 === 0 && year % 100 !== 0 || year % 400 === 0){
    console.log('Leap Year')
  }
  else{
    console.log('Not Leap Year')
  } 
}
isLeapYear(1988)

Popularity 10/10 Helpfulness 4/10 Language javascript

Source: Grepper

Tags: javascript leap-year

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Contributed on Mar 30 2023

Visitor

0 Answers Avg Quality 2/10

Closely Related Answers

Check if a year is leap year

Comment

-1

#include<iostream>
using namespace std;
int main() {
   int year = 2016;
   if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))
   cout<<year<<" is a leap year";
   else
   cout<<year<<" is not a leap year";
   return 0;
}

Popularity 10/10 Helpfulness 3/10 Language whatever

Source: www.tutorialspoint.com

Tags: leap-year leap

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Contributed on Aug 21 2021

Awful Alligator

0 Answers Avg Quality 2/10

leap year

Comment

// works in C, C++, C#, Java, JavaScript, and other C-like languages

// optimal technique
if (((year & 3) == 0 && ((year % 25) != 0) || (year & 15) == 0)) {
  // leap year
}

// classic technique
if (year % 4 == 0 && (year % 100 != 0 || year % 400 == 0)) {
	// leap year
}

Popularity 10/10 Helpfulness 3/10 Language c

Source: stackoverflow.com

Tags: c

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Contributed on Aug 22 2022

ambl

0 Answers Avg Quality 2/10

leap year formula

Comments(1)

$day = [ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
$year = 2021;

if (($year % 4 == 0 && $year % 100 != 0) || ($year % 400 == 0)) {
  $day = [ 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
  echo "This is leap year - get 366 day";
  
} else {
   echo "This is NOT leap year - get ONLY 365 day";
}

Popularity 10/10 Helpfulness 3/10 Language php

Source: Grepper

Tags: php

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Contributed on Dec 17 2021

Zidane (Vi Ly - VietNam)

0 Answers Avg Quality 2/10

leap year

Comment

return (year % 4 == 0 && (year % 100 != 0 || year % 400 == 0));

Popularity 9/10 Helpfulness 3/10 Language c

Source: Grepper

Tags: c leap-year leap-yea

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Contributed on Aug 30 2023

ycomb1nator

0 Answers Avg Quality 2/10

find leap year

Comment

year = int(input("Enter a year: "))

if year % 400 == 0 or (year % 4 == 0 and year % 100 != 0):
    print(year, "is a leap year")
else:
    print(year, "is not a leap year")

Popularity 9/10 Helpfulness 2/10 Language python

Source: chat.openai.com

Tags: find

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Contributed on Jun 28 2023

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Check Year is Leap Year?

Comment

import java.util.Scanner;

public class LeapYearChecker {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter a year: ");

        int year = Integer.parseInt(scanner.nextLine());

        String result = (year % 4 == 0 && (year % 100 != 0 || year % 400 == 0)) ? "is a leap year." : "is not a leap year.";
        System.out.println(year + " " + result);

        scanner.close();
    }
}

Popularity 9/10 Helpfulness 2/10 Language java

Source: chat.openai.com

Tags: java

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Contributed on Aug 02 2023

Dark Dogfish

0 Answers Avg Quality 2/10

def is_leap(year): leap = False

Comment

-1

def is_leap(year):
    leap = False
    if(year%4==0):
        #century year check
        if(year%100==0):
            #century leap year check
            if(year%400==0):
                leap=True
        else:
            leap=True
    return leap

Popularity 9/10 Helpfulness 1/10 Language python

Source: stackoverflow.com

Tags: python py

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Contributed on May 07 2022

Alex Sylvain Luenga

0 Answers Avg Quality 2/10

leap year condition

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leap year condition

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Check if a year is leap year

leap year

leap year formula

leap year

find leap year

Check Year is Leap Year?

def is_leap(year): leap = False

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