/
xddxdd
/
Python
Обзор
Документация
Войти
/
xddxdd
/
Python
Код
Запросы
0
Задачи
Вики
Пакеты
0
Релизы
0
Аналитика
Безопасность
master
other/doomsday.py
59 строк
2 KB
Nguyen Thi Thanh Minh
Doomsday Algorithm: Fix leap year check (#12396)
18 янв 2025, 06:07
Не верифицирован
18 янв 2025, 06:07
533767f
Код
Авторство
О чём код?
#!/bin/python3 # Doomsday algorithm info: https://en.wikipedia.org/wiki/Doomsday_rule DOOMSDAY_LEAP = [4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5] DOOMSDAY_NOT_LEAP = [3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5] WEEK_DAY_NAMES = { 0: "Sunday", 1: "Monday", 2: "Tuesday", 3: "Wednesday", 4: "Thursday", 5: "Friday", 6: "Saturday", } def get_week_day(year: int, month: int, day: int) -> str: """Returns the week-day name out of a given date. >>> get_week_day(2020, 10, 24) 'Saturday' >>> get_week_day(2017, 10, 24) 'Tuesday' >>> get_week_day(2019, 5, 3) 'Friday' >>> get_week_day(1970, 9, 16) 'Wednesday' >>> get_week_day(1870, 8, 13) 'Saturday' >>> get_week_day(2040, 3, 14) 'Wednesday' """ # minimal input check: assert len(str(year)) > 2, "year should be in YYYY format" assert 1 <= month <= 12, "month should be between 1 to 12" assert 1 <= day <= 31, "day should be between 1 to 31" # Doomsday algorithm: century = year // 100 century_anchor = (5 * (century % 4) + 2) % 7 centurian = year % 100 centurian_m = centurian % 12 dooms_day = ( (centurian // 12) + centurian_m + (centurian_m // 4) + century_anchor ) % 7 day_anchor = ( DOOMSDAY_NOT_LEAP[month - 1] if year % 4 != 0 or (centurian == 0 and year % 400 != 0) else DOOMSDAY_LEAP[month - 1] ) week_day = (dooms_day + day - day_anchor) % 7 return WEEK_DAY_NAMES[week_day] if __name__ == "__main__": import doctest doctest.testmod()