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  1.  
  2. <?php
  3. $year=2036;
  4. if($year%400==0 || ($year%100!=0 && $year%4==0))
  5. echo $year.' is Leap Year';
  6. else echo $year.' is Not a Leap Year';
  7. ?>
  8.  
  9.  
Success #stdin #stdout 0.03s 26004KB
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2036 is Leap Year
https://ideone.com/2DTsJf
language:
PHP (php 7.3.5)
created:
14 days ago
visibility:
public

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