Leap year
Seymore4Head
Seymore4Head at Hotmail.invalid
Sat Sep 27 05:49:43 CEST 2014
Still practicing. Since this is listed as a Pseudocode, I assume this
is a good way to explain something. That means I can also assume my
logic is fading with age.
http://en.wikipedia.org/wiki/Leap_year#Algorithm
Me trying to look at the algorithm, it would lead me to try something
like:
if year % 4 !=0:
return False
elif year % 100 !=0:
return True
elif year % 400 !=0:
return False
**** Since it is a practice problem I have the answer:
def is_leap_year(year):
return ((year % 4) == 0 and ((year % 100) != 0 or (year % 400) == 0))
I didn't have any problem when I did this:
if year % 400 == 0:
print ("Not leap year")
elif year % 100 == 0:
print ("Leap year")
elif year % 4 == 0:
print ("Leap year")
else:
print ("Not leap year")
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