forked from TheAlgorithms/Python
-
Notifications
You must be signed in to change notification settings - Fork 0
/
sol1.py
54 lines (39 loc) · 1.57 KB
/
sol1.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
"""
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 sum 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)
222
>>> solution(1000,10)
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 <= n_limit)
if __name__ == "__main__":
print(f"{solution() = }")