#=
Created on 20 Sep 2021

@author: David Doblas Jiménez
@email: daviddoji@pm.me

Solution for Problem 50 of Project Euler
https://projecteuler.net/problem=50 =#

using BenchmarkTools
using Primes

function Problem50()
    #=
    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? =#

    ans = 0
    result = 0
    prime_list = primes(1_000_000)

    for i in 1:length(prime_list)
        sum = 0
        count = 0
        for j in prime_list[i:end]
            sum += j
            count += 1
            if isprime(sum) && count > result
                result = count
                ans = sum
            end
            if sum > 1_000_000
                break
            end
        end
    end
    return ans
end


println("Time to evaluate Problem $(lpad(50, 3, "0")):")
@btime Problem50()
println("")
println("Result for Problem $(lpad(50, 3, "0")): ", Problem50())