小明编程|
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Code# By Websten from forums # # Given your birthday and the current date, calculate your age in days. # Compensate for leap days. # Assume that the birthday and current date are correct dates (and no time travel). # Simply put, if you were born 1 Jan 2012 and todays date is 2 Jan 2012 # you are 1 day old. # # Hint # A whole year is 365 days, 366 if a leap year. def isLeap(year): if year%400==0: return True if year%4==0 and year%100!=0: return True return False def nextDay(year, month, day): """Simple version: assume every month has 30 days""" if isLeap(year): years=[31,29,31,30,31,30,31,31,30,31,30,31] else: years=[31,28,31,30,31,30,31,31,30,31,30,31] if day < years[month-1]: return year, month, day + 1 else: if month == 12: return year + 1, 1, 1 else: return year, month + 1, 1 def dateIsAfter(year1, month1, day1, year2, month2, day2): """Returns True if year1-month1-day1 is after year2-month2-day2. Otherwise, returns False.""" if year1 > year2: return True if year1 == year2: if month1 > month2: return True if month1 == month2: return day1 > day2 return False def daysBetweenDates(year1, month1, day1, year2, month2, day2): """Returns the number of days between year1/month1/day1 and year2/month2/day2. Assumes inputs are valid dates in Gregorian calendar.""" # program defensively! Add an assertion if the input is not valid! if dateIsAfter(year2, month2, day2, year1, month1, day1)==False: return "AssertionError" days = 0 while dateIsAfter(year2, month2, day2, year1, month1, day1): days += 1 (year1, month1, day1) = nextDay(year1, month1, day1) return days def test(): test_cases = [((2012,1,1,2012,2,28), 58), ((2012,1,1,2012,3,1), 60), ((2011,6,30,2012,6,30), 366), ((2011,1,1,2012,8,8), 585 ), ((1900,1,1,1999,12,31), 36523)] for (args, answer) in test_cases: result = daysBetweenDates(*args) if result != answer: print "Test with data:", args, "failed" else: print "Test case passed!" test() |
