Python:實現doomsday末日算法
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()
