sol1.py

"""
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__ importannotations


defprime_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

foriinrange(3, int(limit**0.5+1), 2):
index=i*2
whileindex<limit:
is_prime[index] =False
index=index+i

primes= [2]

foriinrange(3, limit, 2):
ifis_prime[i]:
primes.append(i)

returnprimes


defsolution(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

foriinrange(len(primes)):
forjinrange(i+length, len(primes)):
sol=sum(primes[i:j])
ifsol>=ceiling:
break

ifsolinprimes:
length=j-i
largest=sol

returnlargest


if__name__=="__main__":
print(f"{solution() =}")