# Write a Program to Find The Year is Leap Year Or Not

Find The Year is Leap Year Or Not
Introduction
In the following  C program, we have to find that a year which is given by the user is a leap year or not. We know that if a year contains 366 days or 29 days in February then the year is called a leap year.
Roman general Julius Caesar introduced the first leap years over 2000 years ago. But the Julian Calendar had only one rule: any year evenly divisible by four would be a leap year.
This formula produced way too many leap years but was not corrected until the introduction of the Gregorian calendar more than 1500 years later.

Algorithm:
Step 1: START
Step 3: IF ((YEAR%400==0) OR (YEAR%100!=0)
AND (YEAR%4==0))  THEN  go to Step 4
ELSE    go to Step 5
Step 4: PRINT THE YEAR IS LEAP YEAR
Step 5: PRINT THE YEAR IS NOT LEAP YEAR
Step 6: STOP

Program to find the year is leap year or not:
#include<stdio.h>
#include<conio.h>
main()
{
int year;
printf("\n Enter the Year:");
scanf("%d",&year);
if((year%400==0)||(year%100!=0)&&(year%4==0))
{
printf("\n The Year is Leap Year");
}
else
{
printf("\n The Year is Not Leap Year");
}
getch();
}

Input:
We Have To Check the Program using the following input-
⦁1st  case by  giving  input  as  1900

⦁2nd  case by  giving  input  as  2008

Output:
1st Case:

2nd Case:

Discussion:
If efficiency is not a concern computing a year is a leap year or not through the algorithm is successively done.

We check the number with 3 different conditions to confirm that the year is a leap year or not. If the number is divisible by 4 and 400 then it is a leap year and if the number is divisible by 4 and not divisible by 100 then it is also a leap year but otherwise, the year is not a leap year.