Python/project_euler/problem_009/sol1.py
Joyce bcfca67faa
[mypy] fix type annotations for all Project Euler problems (#4747)
* [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

81 lines
1.7 KiB
Python

"""
Project Euler Problem 9: https://projecteuler.net/problem=9
Special Pythagorean triplet
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product a*b*c.
References:
- https://en.wikipedia.org/wiki/Pythagorean_triple
"""
def solution() -> int:
"""
Returns the product of a,b,c which are Pythagorean Triplet that satisfies
the following:
1. a < b < c
2. a**2 + b**2 = c**2
3. a + b + c = 1000
>>> solution()
31875000
"""
for a in range(300):
for b in range(a + 1, 400):
for c in range(b + 1, 500):
if (a + b + c) == 1000:
if (a ** 2) + (b ** 2) == (c ** 2):
return a * b * c
return -1
def solution_fast() -> int:
"""
Returns the product of a,b,c which are Pythagorean Triplet that satisfies
the following:
1. a < b < c
2. a**2 + b**2 = c**2
3. a + b + c = 1000
>>> solution_fast()
31875000
"""
for a in range(300):
for b in range(400):
c = 1000 - a - b
if a < b < c and (a ** 2) + (b ** 2) == (c ** 2):
return a * b * c
return -1
def benchmark() -> None:
"""
Benchmark code comparing two different version function.
"""
import timeit
print(
timeit.timeit("solution()", setup="from __main__ import solution", number=1000)
)
print(
timeit.timeit(
"solution_fast()", setup="from __main__ import solution_fast", number=1000
)
)
if __name__ == "__main__":
print(f"{solution() = }")