Calculation of elapsed years and elapsed months including leap years

It's not a problem in competitive programming because it depends on the library, but it's a memo.

--Supplement (2020/1/13)

dd = relativedelta(end, begin)
delta_years =  dd.years
delta_months = dd.months

It seems to be okay.

--The following is the previous article

01.01.2000
01.01.2016
From the input
16 years, total 5844 days

I want to get the output. First, it's very easy for days, just subtract datetime.datetime. However, the timedelta obtained by subtracting datetime has days, but not month. In addition, dateimte itself can add days, but cannot add or subtract on a monthly basis.

Solution 1:

Solution 2

Monthly addition and subtraction can be performed by using dateutil.relativedelta. Since this result is saved in datetime, it can be implemented as follows, for example.

import sys
import math
from dateutil.relativedelta import relativedelta

#Input / output processing and datetime
import datetime
begin = input()
end = input()
bd, bm, by = map(int, begin.split("."))
b = datetime.datetime(year=by, month=bm, day=bd)
ed, em, ey = map(int, end.split("."))
e = datetime.datetime(year=ey, month=em, day=ed)

#delta days are easy to get
dt = e - b
resd = dt.days

#Use relativedelta to get the "month difference"
deltamonth = 0
while True:
    b = b + relativedelta(months=1)
    if b <= e:
        deltamonth += 1
        continue
    break

#The year is always December, so divide it into months and years.
resy = deltamonth // 12
resm = deltamonth % 12


#This is the process of distinguishing between the plural and the singular from the problem
res = ""
if resy > 0:
    if resy == 1:
        res += "1 year, "
    else:
        res += "{0} years, ".format(resy)

if resm > 0:
    if resm == 1:
        res += "1 month, "
    else:
        res += "{0} months, ".format(resm)

if resd == 1 or resd == 0:
    res += "total {0} days".format(resd)
else:
    res += "total {0} days".format(resd)

print(res)