Solution to Project Euler Problem 10: Summation of primes - The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.

• Java Code

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

public class SikademyEulerSolution {
	
	private static final int LIMIT = 2000000;


        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 boolean isPrime(int x) {
		if (x < 0)
			throw new IllegalArgumentException("Negative number");
		if (x == 0 || x == 1)
			return false;
		else if (x == 2)
			return true;
		else {
			if (x % 2 == 0)
				return false;
			for (int i = 3, end = sqrt(x); i <= end; i += 2) {
				if (x % i == 0)
					return false;
			}
			return true;
		}
	}

        // Returns a Boolean array 'isPrime' where isPrime[i] indicates whether i is prime, for 0 <= i <= n.
	// For a large batch of queries, this is faster than calling isPrime() for each integer.
	// For example: listPrimality(100) = {false, false, true, true, false, true, false, true,
	// false, false, false, true, false, true, false, false, false, true, ...} (array length 101).
	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;
		for (int i = 3, end = sqrt(n); i <= end; i += 2) {
			if (result[i]) {
				for (int j = i * i, inc = i * 2; j <= n; j += inc)
					result[j] = false;
			}
		}
		return result;
	}
	
	
	// Returns all the prime numbers less than or equal to n, in ascending order.
	// For example: listPrimes(97) = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ..., 83, 89, 97}.
	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 String run() {
		long sum = 0;
		for (int p : Library.listPrimes(LIMIT - 1))
			sum += p;
		return Long.toString(sum);
	}

	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 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

# Returns a list of True and False indicating whether each number is prime.
# For 0 <= i <= n, result[i] is True if i is a prime number, False otherwise.
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

# Returns all the prime numbers less than or equal to n, in ascending order.
# For example: listPrimes(97) = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ..., 83, 89, 97].
def list_primes(n):
	return [i for (i, isprime) in enumerate(list_primality(n)) if isprime]

def compute():
	ans = sum(list_primes(1999999))
	return str(ans)

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