Question

how to make a lip year finder in python???​

Answer 1

Answer:

See this example:

• year = int(input("Enter a year: "))
• if (year % 4) == 0:
• if (year % 100) == 0:
• if (year % 400) == 0:
• print("{0} is a leap year". format(year))

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have an nice day dear!!! :-)

Answer 2

year = 2000

# To get year (integer input) from the user

# year = int(input("Enter a year: "))

if (year % 4) == 0:

  if (year % 100) == 0:

      if (year % 400) == 0:

          print("{0} is a leap year".format(year))

      else:

          print("{0} is not a leap year".format(year))

  else:

      print("{0} is a leap year".format(year))

else:

  print("{0} is not a leap year".format(year))

Output

2000 is a leap year