The following code does the job but, as noted above, Python provides a module, datetime, with many methods for handling dates and times and it is better to use this.
# Set us_date_style to True for dates written M/D/Y, otherwise assume D/M/Y
us_date_style = False
# Number of days per month indexed from zero (Jan=0, Feb=1, ..., Dec=11)
days_per_month = [31,28,31,30,31,30,31,31,30,31,30,31]
# Consider the given date as "today"
today = '4/3/2013'
# Get day, month and year from the today string as integers
day, month, year = today.split('/')
if us_date_style:
day, month = month, day
day, month, year = int(day), int(month), int(year)
# establish if today's year is a leap year
if not year % 400:
is_leap_year = True
elif not year % 100:
is_leap_year = False
elif not year % 4:
is_leap_year = True
else:
is_leap_year = False
# Form tomorrow's date by adding one to today's day...
tday, tmonth, tyear = day + 1, month, year
# ... but check if we've fallen off the end of the month
if tday > days_per_month[month-1]:
if month == 2 and is_leap_year and tday == 29:
# corner case of today being 28th of February of a leap year
pass
else:
tday, tmonth = 1, month + 1
# Have we fallen off the end of the year?
if tmonth == 13:
# Yup: it must be 1 January of the next year
tday, tmonth, tyear = 1, 1, year+1
date_tuple = (tmonth, tday, tyear) if us_date_style else (tday, tmonth, tyear)
tomorrow = '{:d}/{:d}/{:d}'.format(*date_tuple)
print('The day after {:s} is {:s}'.format(today, tomorrow))