Calculate the day when Date is given as input using the Doomsday algorithm
Your code is hard to understand, even after watching the video.


Make some functions. years_doomsday would help move some of the hard to understand information to be self-contained.
If you need to floor divide and get the remainder use divmod.
You can check the leap year using calendar.isleap.
You have a bug, if you ever enter a leap year then the non-leap years will return incorrect values for January and February.
You should make a function that calls years_doomsday, and returns the weekday.
You should make centdoom a global constant.
By rotating centdoom once you can remove the need for the -3.


import calendar

WEEKDAYS = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"]
MDOOMSDAY = [3, 28, 14, 4, 9, 6, 11, 8, 5, 10, 7, 12]
MDOOMSDAY_LEAP = list(MDOOMSDAY)
MDOOMSDAY_LEAP[:2] = [4, 29]
CDOOMSDAY = [2, 0, 5, 3]


def years_doomsday(year):
    a, b = divmod(year, 100)
    c, d = divmod(b, 12)
    e = d // 4
    return (CDOOMSDAY[a % 4] + c + d + e) % 7


def doomsday(year, month, day):
    y_doomsday = years_doomsday(year)
    mdoomsday = MDOOMSDAY_LEAP if calendar.isleap(year) else MDOOMSDAY
    return WEEKDAYS[(day - mdoomsday[month-1] + y_doomsday) % 7]


if __name__ == '__main__':
    tests = [
        (2305, 7, 13, 'Thursday'),
        (1776, 7, 4, 'Thursday'),
        (1969, 7, 20, 'Sunday'),
        (1984, 1, 6, 'Friday'),
        (1902, 10, 19, 'Sunday'),
    ]
    for test in tests:
        if doomsday(*test[:3]) != test[-1]:
            print('Broken for', test)


If you don't want any fun, you can replace your code with datetime.date.isoweekday.

import datetime

for test in tests:
    if WEEKDAYS[datetime.date(*test[:3]).isoweekday()%7] != test[-1]:
        print('Broken for', test)