How to find leap year programatically in C

Question

I made a program using C to find whether the entered year is a leap year or not. But unfortunately its not working well. It says a year is leap and the preceding year is not leap.

#include<stdio.h>
#include<conio.h>
int yearr(int year);
void main(void)
{
    int year;
    printf("Enter a year:");
    scanf("%d",&year);
    if(!yearr(year))
    {
        printf("It is a leap year.");
    }
    else
    {
    printf("It is not a leap year");
    }


getch();
}
int yearr(int year)
{
    if((year%4==0)&&(year/4!=0))
    return 1;
    else
    return 0;
}

After reading the comments i edited my coding as:

#include<stdio.h>
#include<conio.h>
int yearr(int year);
void main(void)
{
    int year;
    printf("Enter a year:");
    scanf("%d",&year);
    if(!yearr(year))
    {
        printf("It is a leap year.");
    }
    else
    {
    printf("It is not a leap year");
    }


getch();
}
int yearr(int year)
{
    if((year%4==0)
    {
    if(year%400==0)
    return 1;
    if(year%100==0)
    return 0; 
    }
    else
    return 0;
}

Answer 1

From Wikipedia article on Leap year:

if (year modulo 4 is 0) and (year modulo 100 is not 0) or (year modulo 400 is 0)
   then is_leap_year
else
   not_leap_year

Answer 2

Although the logic that divides by 400 first is impeccable, it is not as computationally efficient as dividing by 4 first. You can do that with the logic:

#define LEAPYEAR(y) (((y) % 4) == 0 && (((y) % 100) != 0 || ((y) % 400) == 0))

This divides by 4 for every value, but for 3/4 of them, the testing terminates there. For the 1/4 that pass the first test, it then divides by 100, eliminating 24/25 values; for the remaining 1 out of a 100, it divides by 400 too, coming up with a final answer. Granted, this is not a huge saving.

Answer 3

The problem with your code is that you are returning a non-zero value from yearr if you think that the year is a leap year. So you don't need the ! in your if statement.