Leap year
Seymore4Head
Seymore4Head at Hotmail.invalid
Fri Sep 26 23:49:43 EDT 2014
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Fri Sep 26 23:49:43 EDT 2014
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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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