54 lines
1.2 KiB
Julia
54 lines
1.2 KiB
Julia
#=
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Created on 20 Sep 2021
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for Problem 50 of Project Euler
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https://projecteuler.net/problem=50
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=#
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using BenchmarkTools
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using Primes
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function Problem50()
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#=
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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 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 consecutive primes?
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=#
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ans = 0
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result = 0
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prime_list = primes(1_000_000)
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for i in 1:length(prime_list)
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sum = 0
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count = 0
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for j in prime_list[i:end]
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sum += j
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count += 1
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if isprime(sum) && count > result
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result = count
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ans = sum
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end
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if sum > 1_000_000
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break
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end
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end
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end
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return ans
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end
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println("Time to evaluate Problem 50:")
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@btime Problem50()
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println("")
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println("Result for Problem 50: ", Problem50()) |