1010 Example 2-3 judging leap year

The title Description enters an arbitrary year between 2000-2,500 years (including 2000 and 2500) and determines whether it is a leap year. enter An integer year that represents the years. Enter the guaranteed 2000≤year≤2500. output If the year entered is a leap years, output "leap yearly", otherwise output "not leap year".
Note that output quotation marks are not required and line endings are output wrapped. Sample Input

2000

Sample output

Leap year

#include <stdio.h>
int main ()
{
int year,leap;
scanf ("%d", &year);
if (year%4==0)
{
if (year%100==0)
{
if (year%400==0)
leap=1;
else leap=0;
}
else leap=1;
}
else
leap=0;
if (LEAP)
printf ("Leap year\n");
else printf ("Not leap year\n");
return 0;
}