mirror of
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bcfca67faa
* [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
81 lines
1.7 KiB
Python
81 lines
1.7 KiB
Python
"""
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Project Euler Problem 9: https://projecteuler.net/problem=9
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Special Pythagorean triplet
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A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
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a^2 + b^2 = c^2
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For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
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There exists exactly one Pythagorean triplet for which a + b + c = 1000.
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Find the product a*b*c.
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References:
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- https://en.wikipedia.org/wiki/Pythagorean_triple
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"""
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def solution() -> int:
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"""
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Returns the product of a,b,c which are Pythagorean Triplet that satisfies
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the following:
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1. a < b < c
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2. a**2 + b**2 = c**2
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3. a + b + c = 1000
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>>> solution()
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31875000
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"""
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for a in range(300):
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for b in range(a + 1, 400):
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for c in range(b + 1, 500):
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if (a + b + c) == 1000:
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if (a ** 2) + (b ** 2) == (c ** 2):
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return a * b * c
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return -1
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def solution_fast() -> int:
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"""
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Returns the product of a,b,c which are Pythagorean Triplet that satisfies
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the following:
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1. a < b < c
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2. a**2 + b**2 = c**2
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3. a + b + c = 1000
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>>> solution_fast()
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31875000
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"""
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for a in range(300):
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for b in range(400):
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c = 1000 - a - b
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if a < b < c and (a ** 2) + (b ** 2) == (c ** 2):
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return a * b * c
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return -1
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def benchmark() -> None:
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"""
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Benchmark code comparing two different version function.
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"""
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import timeit
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print(
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timeit.timeit("solution()", setup="from __main__ import solution", number=1000)
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)
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print(
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timeit.timeit(
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"solution_fast()", setup="from __main__ import solution_fast", number=1000
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)
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)
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if __name__ == "__main__":
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print(f"{solution() = }")
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