Calculate the day when Date is given as input using the Doomsday algorithm Was too bored in holidays...here's a code which finds the day of the week using the algorithm in this video. Please give a logical date in the correct format cause I didn't account for improper inputs cause I'm too lazy. My code is short and sweet.

day=["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"]    #some database numbers and strings
mdoom=[3,28,14,4,9,6,11,8,5,10,7,12]
while True:
    entry=input('\nEnter a date in DD/MM/YYYY format:')         #Input date in a specified format
    date,month,year = map(int,entry.split('/'))                 #Get the date,month and year
    centdoom=[3,2,0,5]
    a=centdoom[((year//100)%4)-3]
    b=(year%100)//12
    c=(year%100)%12
    d=c//4                                                      #Random calculations for doomsday :)
    e=a+b+c+d

    if (year%4==0) and ((year%100!=0) or (year%400==0)):        #Checks for leap year
        mdoom[0],mdoom[1]=4,29

    ddoom=e%7                                                   #finds a date in the 1st week which has the same day as the doomsday    
    r=mdoom[month-1]                                            #Gives the doomsdate in the input month 
    r=r%7
    r=ddoom-r                                                   
    date=date%7                                                 #finds a date in the 1st week which has the same day as the input date
    print('\n',day[(r+date)%7])                                 #prints the output
    k=input('\nTry again?(y/n)')
    if k=='y':
        continue
    else:
        break


And any suggestions or improvements are welcome.
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)