C program to find the year is leap year or not?

Answer 1

Algorithm:

If the year is not divisible by 4 then it is not a leap year.

Else if the year is not divisible by 100 then it is a leap year.

Else if the year is not divisible by 400 then is not a leap year.

Else it is a leap year.

Implementation:

bool is_leap_year (unsigned year) {

if (year%4) return false;

if (year%100) return true;

if (year%400) return false;

return true;

}

Answer 2

#include <stdio.h>

#include<conio.h>

void main()

{

int year;

clrscr();

printf("Enter the year which is to be checked for being leap or not");

scanf("%d",&year);

if (year%4==0 && (year%100!=0 year%400==0))

{

printf("leap year");

}

else

{

printf("not a leap year");

}

getch();

}

Answer 3

Use the following function to determine if a given year is a leap year or not:bool is_leap_year(int year) {

if (year%4) return false;

if (year%100) return true;

if (year%400) return false;

return true;

}

This works by a process of elimination, dealing with the most common cases first:

Years that are not divisible by 4 are not leap years.

Years that are divisible by 4 but not by 100 are leap years.

Years that are divisible by 100 but not by 400 are not leap years.

Years that are divisible by 400 are leap years.

Answer 4

The simplest code for leap year in Java is something like this:
public static boolean isLeapYear(int year) {

return ((year % 4 0);

}

But recommend to use the standard Java library to test for a leap year:

new GregorianCalendar().isLeapYear(year);