mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-24 05:21:09 +00:00
b96e6c7075
* Added solution for Project Euler problem 174. * Fixed import order and removed executable permission from sol1.py * Update docstrings, doctests, and annotations. Reference: #3256 * Update docstring * Update sol1.py Co-authored-by: Dhruv <dhruvmanila@gmail.com>
53 lines
1.5 KiB
Python
53 lines
1.5 KiB
Python
"""
|
||
Project Euler Problem 174: https://projecteuler.net/problem=174
|
||
|
||
We shall define a square lamina to be a square outline with a square "hole" so that
|
||
the shape possesses vertical and horizontal symmetry.
|
||
|
||
Given eight tiles it is possible to form a lamina in only one way: 3x3 square with a
|
||
1x1 hole in the middle. However, using thirty-two tiles it is possible to form two
|
||
distinct laminae.
|
||
|
||
If t represents the number of tiles used, we shall say that t = 8 is type L(1) and
|
||
t = 32 is type L(2).
|
||
|
||
Let N(n) be the number of t ≤ 1000000 such that t is type L(n); for example,
|
||
N(15) = 832.
|
||
|
||
What is ∑ N(n) for 1 ≤ n ≤ 10?
|
||
"""
|
||
|
||
from collections import defaultdict
|
||
from math import ceil, sqrt
|
||
|
||
|
||
def solution(t_limit: int = 1000000, n_limit: int = 10) -> int:
|
||
"""
|
||
Return the sum of N(n) for 1 <= n <= n_limit.
|
||
|
||
>>> solution(1000,5)
|
||
249
|
||
>>> solution(10000,10)
|
||
2383
|
||
"""
|
||
count: defaultdict = defaultdict(int)
|
||
|
||
for outer_width in range(3, (t_limit // 4) + 2):
|
||
if outer_width * outer_width > t_limit:
|
||
hole_width_lower_bound = max(
|
||
ceil(sqrt(outer_width * outer_width - t_limit)), 1
|
||
)
|
||
else:
|
||
hole_width_lower_bound = 1
|
||
|
||
hole_width_lower_bound += (outer_width - hole_width_lower_bound) % 2
|
||
|
||
for hole_width in range(hole_width_lower_bound, outer_width - 1, 2):
|
||
count[outer_width * outer_width - hole_width * hole_width] += 1
|
||
|
||
return sum(1 for n in count.values() if 1 <= n <= 10)
|
||
|
||
|
||
if __name__ == "__main__":
|
||
print(f"{solution() = }")
|