Python/project_euler/problem_191/sol1.py
Joyce bcfca67faa
[mypy] fix type annotations for all Project Euler problems ()
* [mypy] fix type annotations for problem003/sol1 and problem003/sol3

* [mypy] fix type annotations for project euler problem007/sol2

* [mypy] fix type annotations for project euler problem008/sol2

* [mypy] fix type annotations for project euler problem009/sol1

* [mypy] fix type annotations for project euler problem014/sol1

* [mypy] fix type annotations for project euler problem 025/sol2

* [mypy] fix type annotations for project euler problem026/sol1.py

* [mypy] fix type annotations for project euler problem037/sol1

* [mypy] fix type annotations for project euler problem044/sol1

* [mypy] fix type annotations for project euler problem046/sol1

* [mypy] fix type annotations for project euler problem051/sol1

* [mypy] fix type annotations for project euler problem074/sol2

* [mypy] fix type annotations for project euler problem080/sol1

* [mypy] fix type annotations for project euler problem099/sol1

* [mypy] fix type annotations for project euler problem101/sol1

* [mypy] fix type annotations for project euler problem188/sol1

* [mypy] fix type annotations for project euler problem191/sol1

* [mypy] fix type annotations for project euler problem207/sol1

* [mypy] fix type annotations for project euler problem551/sol1
2021-10-12 00:33:44 +08:00

106 lines
3.2 KiB
Python

"""
Prize Strings
Problem 191
A particular school offers cash rewards to children with good attendance and
punctuality. If they are absent for three consecutive days or late on more
than one occasion then they forfeit their prize.
During an n-day period a trinary string is formed for each child consisting
of L's (late), O's (on time), and A's (absent).
Although there are eighty-one trinary strings for a 4-day period that can be
formed, exactly forty-three strings would lead to a prize:
OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA
OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO
AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL
AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA
LAOO LAOA LAAO
How many "prize" strings exist over a 30-day period?
References:
- The original Project Euler project page:
https://projecteuler.net/problem=191
"""
cache: dict[tuple[int, int, int], int] = {}
def _calculate(days: int, absent: int, late: int) -> int:
"""
A small helper function for the recursion, mainly to have
a clean interface for the solution() function below.
It should get called with the number of days (corresponding
to the desired length of the 'prize strings'), and the
initial values for the number of consecutive absent days and
number of total late days.
>>> _calculate(days=4, absent=0, late=0)
43
>>> _calculate(days=30, absent=2, late=0)
0
>>> _calculate(days=30, absent=1, late=0)
98950096
"""
# if we are absent twice, or late 3 consecutive days,
# no further prize strings are possible
if late == 3 or absent == 2:
return 0
# if we have no days left, and have not failed any other rules,
# we have a prize string
if days == 0:
return 1
# No easy solution, so now we need to do the recursive calculation
# First, check if the combination is already in the cache, and
# if yes, return the stored value from there since we already
# know the number of possible prize strings from this point on
key = (days, absent, late)
if key in cache:
return cache[key]
# now we calculate the three possible ways that can unfold from
# this point on, depending on our attendance today
# 1) if we are late (but not absent), the "absent" counter stays as
# it is, but the "late" counter increases by one
state_late = _calculate(days - 1, absent, late + 1)
# 2) if we are absent, the "absent" counter increases by 1, and the
# "late" counter resets to 0
state_absent = _calculate(days - 1, absent + 1, 0)
# 3) if we are on time, this resets the "late" counter and keeps the
# absent counter
state_ontime = _calculate(days - 1, absent, 0)
prizestrings = state_late + state_absent + state_ontime
cache[key] = prizestrings
return prizestrings
def solution(days: int = 30) -> int:
"""
Returns the number of possible prize strings for a particular number
of days, using a simple recursive function with caching to speed it up.
>>> solution()
1918080160
>>> solution(4)
43
"""
return _calculate(days, absent=0, late=0)
if __name__ == "__main__":
print(solution())