Hacked By AnonymousFox
#! /usr/bin/python2.7
# Calculate your unbirthday count (see Alice in Wonderland).
# This is defined as the number of days from your birth until today
# that weren't your birthday. (The day you were born is not counted).
# Leap years make it interesting.
import sys
import time
import calendar
def main():
if sys.argv[1:]:
year = int(sys.argv[1])
else:
year = int(raw_input('In which year were you born? '))
if 0 <= year < 100:
print "I'll assume that by", year,
year = year + 1900
print 'you mean', year, 'and not the early Christian era'
elif not (1850 <= year <= time.localtime()[0]):
print "It's hard to believe you were born in", year
return
if sys.argv[2:]:
month = int(sys.argv[2])
else:
month = int(raw_input('And in which month? (1-12) '))
if not (1 <= month <= 12):
print 'There is no month numbered', month
return
if sys.argv[3:]:
day = int(sys.argv[3])
else:
day = int(raw_input('And on what day of that month? (1-31) '))
if month == 2 and calendar.isleap(year):
maxday = 29
else:
maxday = calendar.mdays[month]
if not (1 <= day <= maxday):
print 'There are no', day, 'days in that month!'
return
bdaytuple = (year, month, day)
bdaydate = mkdate(bdaytuple)
print 'You were born on', format(bdaytuple)
todaytuple = time.localtime()[:3]
todaydate = mkdate(todaytuple)
print 'Today is', format(todaytuple)
if bdaytuple > todaytuple:
print 'You are a time traveler. Go back to the future!'
return
if bdaytuple == todaytuple:
print 'You were born today. Have a nice life!'
return
days = todaydate - bdaydate
print 'You have lived', days, 'days'
age = 0
for y in range(year, todaytuple[0] + 1):
if bdaytuple < (y, month, day) <= todaytuple:
age = age + 1
print 'You are', age, 'years old'
if todaytuple[1:] == bdaytuple[1:]:
print 'Congratulations! Today is your', nth(age), 'birthday'
print 'Yesterday was your',
else:
print 'Today is your',
print nth(days - age), 'unbirthday'
def format((year, month, day)):
return '%d %s %d' % (day, calendar.month_name[month], year)
def nth(n):
if n == 1: return '1st'
if n == 2: return '2nd'
if n == 3: return '3rd'
return '%dth' % n
def mkdate((year, month, day)):
# January 1st, in 0 A.D. is arbitrarily defined to be day 1,
# even though that day never actually existed and the calendar
# was different then...
days = year*365 # years, roughly
days = days + (year+3)//4 # plus leap years, roughly
days = days - (year+99)//100 # minus non-leap years every century
days = days + (year+399)//400 # plus leap years every 4 centirues
for i in range(1, month):
if i == 2 and calendar.isleap(year):
days = days + 29
else:
days = days + calendar.mdays[i]
days = days + day
return days
if __name__ == "__main__":
main()
Hacked By AnonymousFox1.0, Coded By AnonymousFox