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* Pyupgrade to Python 3.9
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86 lines
1.9 KiB
Python
86 lines
1.9 KiB
Python
"""
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Project Euler Problem 50: https://projecteuler.net/problem=50
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Consecutive prime sum
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The prime 41, can be written as the sum of six consecutive primes:
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41 = 2 + 3 + 5 + 7 + 11 + 13
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This is the longest sum of consecutive primes that adds to a prime below
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one-hundred.
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The longest sum of consecutive primes below one-thousand that adds to a prime,
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contains 21 terms, and is equal to 953.
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Which prime, below one-million, can be written as the sum of the most
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consecutive primes?
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"""
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from __future__ import annotations
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def prime_sieve(limit: int) -> list[int]:
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"""
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Sieve of Erotosthenes
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Function to return all the prime numbers up to a number 'limit'
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https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
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>>> prime_sieve(3)
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[2]
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>>> prime_sieve(50)
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[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
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"""
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is_prime = [True] * limit
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is_prime[0] = False
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is_prime[1] = False
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is_prime[2] = True
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for i in range(3, int(limit ** 0.5 + 1), 2):
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index = i * 2
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while index < limit:
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is_prime[index] = False
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index = index + i
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primes = [2]
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for i in range(3, limit, 2):
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if is_prime[i]:
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primes.append(i)
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return primes
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def solution(ceiling: int = 1_000_000) -> int:
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"""
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Returns the biggest prime, below the celing, that can be written as the sum
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of consecutive the most consecutive primes.
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>>> solution(500)
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499
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>>> solution(1_000)
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953
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>>> solution(10_000)
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9521
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"""
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primes = prime_sieve(ceiling)
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length = 0
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largest = 0
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for i in range(len(primes)):
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for j in range(i + length, len(primes)):
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sol = sum(primes[i:j])
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if sol >= ceiling:
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break
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if sol in primes:
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length = j - i
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largest = sol
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return largest
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if __name__ == "__main__":
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print(f"{solution() = }")
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