Python/project_euler/problem_050/sol1.py
Christian Clauss 24d3cf8244
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Python

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