Leap Year

A year Y will be passed as input.  The program must find if the given year is a leap year or not.  If it is a leap year, print yes else no.

Example 1:

Input:
2000
Output:
yes

Example 2:

Input:
1980
Output:
yes

Example 3:

Input:
400
Output:
yes

Example 4:

Input:
2019
Output:
no

Note: A year is a leap year if it is divisible by 4. If it is a century then it should be divisible by 400.The pseudocode is as given below:

If year is divisible by 400:
    then is_leap_year
else if year is divisible by 100
    then not_leap_year
else if year is divisible by 4
    then is_leap_year
else
    not_leap_year

year = int(input())
if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0):
    print("yes")
else:
    print("no")

#include <iostream>
using namespace std;

int main() {
    int year;
    cin >> year;
    if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
        cout << "yes";
    else
        cout << "no";
    return 0;
}

using System;

class Program {
    static void Main(string[] args) {
        int year = Convert.ToInt32(Console.ReadLine());
        if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
            Console.WriteLine("yes");
        else
            Console.WriteLine("no");
    }
}

#include<stdio.h>
#include <stdlib.h>

int main() {
    int year;
    scanf(“ % d”, & year);
    if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
        printf(“yes”);
    else
        printf(“no”);
}

import java.util.*;
public class Hello {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        if ((n % 400 == 0) || (n % 4 == 0) && (n % 100 != 0))
            System.out.print("yes");
        else
            System.out.print("no");
    }
}