Solution to Project Euler 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?

• Java Code

package sikademy;
/**
 *
 * @author Archangel Macsika
 * Copyright (c) Sikademy. All rights reserved
 */

public class SikademyEulerSolution {
	
	private static final int LIMIT = pow(10, 6);
	
	
	public String run() {
		boolean[] isPrime = listPrimality(LIMIT);
		int[] primes = listPrimes(LIMIT);
		
		long maxSum = 0;
		int maxRun = -1;
		for (int i = 0; i < primes.length; i++) {  // For each index of a starting prime number
			int sum = 0;
			for (int j = i; j < primes.length; j++) {  // For each end index (inclusive)
				sum += primes[j];
				if (sum > LIMIT)
					break;
				else if (j - i > maxRun && sum > maxSum && isPrime[sum]) {
					maxSum = sum;
					maxRun = j - i;
				}
			}
		}
		return Long.toString(maxSum);
	}

        public static boolean[] listPrimality(int n) {
		if (n < 0)
			throw new IllegalArgumentException("Negative array size");
		boolean[] result = new boolean[n + 1];
		if (n >= 2)
			result[2] = true;
		for (int i = 3; i <= n; i += 2)
			result[i] = true;
		// Sieve of Eratosthenes
		for (int i = 3, end = sqrt(n); i <= end; i += 2) {
			if (result[i]) {
				// Note: i * i does not overflow
				for (int j = i * i, inc = i * 2; j <= n; j += inc)
					result[j] = false;
			}
		}
		return result;
	}
	
	public static int[] listPrimes(int n) {
		boolean[] isPrime = listPrimality(n);
		int count = 0;
		for (boolean b : isPrime) {
			if (b)
				count++;
		}
		
		int[] result = new int[count];
		for (int i = 0, j = 0; i < isPrime.length; i++) {
			if (isPrime[i]) {
				result[j] = i;
				j++;
			}
		}
		return result;
	}

        public static int pow(int x, int y) {
		if (x < 0)
			throw new IllegalArgumentException("Negative base not supported");
		if (y < 0)
			throw new IllegalArgumentException("Negative exponent");
		int z = 1;
		for (int i = 0; i < y; i++) {
			if (Integer.MAX_VALUE / z < x)
				throw new ArithmeticException("Overflow");
			z *= x;
		}
		return z;
	}

        public static int sqrt(int x) {
		if (x < 0)
			throw new IllegalArgumentException("Square root of negative number");
		int y = 0;
		for (int i = 1 << 15; i != 0; i >>>= 1) {
			y |= i;
			if (y > 46340 || y * y > x)
				y ^= i;
		}
		return y;
	}

	public static void main(String[] args) {
                SikademyEulerSolution solution = new SikademyEulerSolution();
		System.out.println(solution.run());
	}	
}

• Python Code

# 
# @author Archangel Macsika
# Copyright (c) Sikademy. All rights reserved.
# 

def compute():
	ans = 0
	isprime = list_primality(999999)
	primes = list_primes(999999)
	consecutive = 0
	for i in range(len(primes)):
		sum = primes[i]
		consec = 1
		for j in range(i + 1, len(primes)):
			sum += primes[j]
			consec += 1
			if sum >= len(isprime):
				break
			if isprime[sum] and consec > consecutive:
				ans = sum
				consecutive = consec
	return str(ans)

def sqrt(x):
	assert x >= 0
	i = 1
	while i * i <= x:
		i *= 2
	y = 0
	while i > 0:
		if (y + i)**2 <= x:
			y += i
		i //= 2
	return y

def list_primality(n):
	result = [True] * (n + 1)
	result[0] = result[1] = False
	for i in range(sqrt(n) + 1):
		if result[i]:
			for j in range(i * i, len(result), i):
				result[j] = False
	return result


def list_primes(n):
	return [i for (i, isprime) in enumerate(list_primality(n)) if isprime]

if __name__ == "__main__":
	print(compute())