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  1. #!/usr/bin/python
  2. import sys
  3.  
  4. # this example assumes proper dates are entered (there are no safety checks performed)
  5.  
  6. # split string and convert to integers
  7. month, day, year = map( lambda x: int(x), raw_input().split('/') )
  8.  
  9. # gregorian?
  10. is_gregorian = True
  11. if (year < 0):
  12. 	is_gregorian = False
  13. 	year = year - 1 # account for there being no year zero
  14. else:
  15. 	from datetime import date
  16. 	julian_calendar_fail = date(1582, 10, 15) # julian calendar goes away
  17. 	is_gregorian = date(year, month, day) > julian_calendar_fail
  18.  
  19. # leap year calculation:
  20. if is_gregorian and year % 4 == 0 and not ( year % 100 == 0 and year % 400 != 0 ):
  21. 	is_leap_year = True
  22. elif not is_gregorian and year % 4 == 0: # julian
  23. 	is_leap_year = True
  24. else:
  25. 	is_leap_year = False
  26.  
  27. if (month == 2):
  28. 	days_in_month = 29 if is_leap_year else 28
  29. else:
  30. 	from calendar import monthrange
  31. 	days_in_month = monthrange(2000, month) # year irrelevant for non-February months
  32.  
  33. # safety:
  34. century = int( str(year)[:-2] ) if str(year)[:-2] != '' else 0
  35. year    = int( str(year)[-2:] ) if str(year)[-2:] != '' else 0
  36.  
  37. # the doomsday calculation: http://e...content-available-to-author-only...a.org/wiki/Doomsday_rule
  38. doomsdays = [ 4 if is_leap_year else 3, 29 if is_leap_year else 28, 0, 4, 9, 6, 11, 8, 5, 10, 7, 12 ]
  39. # weekdays, starting with Monday == 0
  40. days = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']
  41.  
  42. a = year / 12
  43. b = year % 12
  44. c = b / 4
  45. d = ( a + b + c )
  46.  
  47. # "Finding a year's Doomsday" section @ http://e...content-available-to-author-only...a.org/wiki/Doomsday_rule:
  48. if is_gregorian:
  49. 	d = d % 7
  50.  
  51. 	anchor_days = [ 1, 6, 4, 2 ] # from wikipedia link above
  52. 	anchor = anchor_days[ century % 4 ]
  53. 	offset = day - doomsdays[ month - 1 ]
  54. 	day_of_week = (d + anchor + offset)
  55. else:
  56. 	anchor = 6 + d - century
  57. 	anchor += 7 if anchor < 0 else -7 # normalize between -7 and 7
  58. 	offset = day - doomsdays[ month - 1 ]
  59. 	day_of_week = (anchor + offset)
  60.  
  61. # normalize day_of_week to be between 0 and 6
  62. day_of_week = day_of_week % 7
  63. print days[ day_of_week ]
Success #stdin #stdout 0.09s 10960KB
comments ()
stdin 
2/15/2000
stdout 
Tuesday
https://ideone.com/HIyf7
language:
Python (cpython 2.7.16)
created:
12 years ago
visibility:
public

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