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  1. <?php
  2. $year=2036;
  3. if($year%400==0 || ($year%100!=0 && $year%4==0))
  4. echo $year.' is Leap Year';
  5. else echo $year.' is Not a Leap Year';
  6. ?>
  7.  
Success #stdin #stdout 0.02s 26044KB
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2036 is Leap Year
https://ideone.com/TiXVEM
language:
PHP (php 7.3.5)
created:
6 hours ago
visibility:
public

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