**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 2: READ YEAR

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

**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.

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