Project Euler > Problem 50 > Consecutive prime sum (Java Solution)

Problem:

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?

Solution:

997651

Code:
The solution may include methods that will be found here: Library.java .

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print?

1	public interface EulerSolution{

2

3	public String run();

4

5

}

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print?

01

/*

02	* Solution to Project Euler problem 50

03	* By Nayuki Minase

04

*

05	* http://nayuki.eigenstate.org/page/project-euler-solutions

06	* https://github.com/nayuki/Project-Euler-solutions

07

*/

08

09

10	public final class p050 implements EulerSolution {

11

12	public static void main(String[] args) {

13	System.out.println(new p050().run());

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}

15

16

17	private static final int LIMIT = Library.pow(10, 6);

18

19

20	public String run() {

21	boolean[] isPrime = Library.listPrimality(LIMIT);

22	int[] primes = Library.listPrimes(LIMIT);

23

24	long maxSum = 0;

25	int maxRun = -1;

26	for (int i = 0; i < primes.length; i++) {  // For each index of a starting prime number

27	int sum = 0;

28	for (int j = i; j < primes.length; j++) {  // For each end index (inclusive)

29	sum += primes[j];

30	if (sum > LIMIT)

31

break;

32	else if (j - i > maxRun && sum > maxSum && isPrime[sum]) {

33	maxSum = sum;

34	maxRun = j - i;

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}

36

}

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}

38	return Long.toString(maxSum);

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}

40

41

}