Project Euler/10

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Project Euler Problem 10 is a problem related to prime numbers. The task is to find the sum of all prime numbers below 2000000.

Implementations

C++

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main() {
    const int LIMIT = 2000000;
    vector<bool> is_prime(LIMIT, true);
    is_prime[0] = is_prime[1] = false;

    int sqrt_limit = static_cast<int>(sqrt(LIMIT));
    for (int i = 2; i <= sqrt_limit; ++i) {
        if (is_prime[i]) {
            for (int j = i * i; j < LIMIT; j += i) {
                is_prime[j] = false;
            }
        }
    }

    long long sum = 0;
    for (int i = 2; i < LIMIT; ++i) {
        if (is_prime[i]) {
            sum += i;
        }
    }

    cout << sum << endl;
    return 0;
}

C♯

using System;

class Program
{
    static void Main()
    {
        const int Limit = 2000000;
        bool[] isPrime = new bool[Limit];
        for (int i = 0; i < Limit; i++) isPrime[i] = true;
        isPrime[0] = isPrime[1] = false;

        int sqrtLimit = (int)Math.Sqrt(Limit);
        for (int i = 2; i <= sqrtLimit; i++)
        {
            if (isPrime[i])
            {
                for (int j = i * i; j < Limit; j += i)
                {
                    isPrime[j] = false;
                }
            }
        }

        long sum = 0;
        for (int i = 2; i < Limit; i++)
        {
            if (isPrime[i]) sum += i;
        }

        Console.WriteLine(sum);
    }
}

JavaScript (Node.js)

function sumPrimesBelow(limit) {
    const isPrime = new Uint8Array(limit);
    isPrime.fill(1);
    isPrime[0] = isPrime[1] = 0;

    const sqrtLimit = Math.floor(Math.sqrt(limit));
    for (let i = 2; i <= sqrtLimit; i++) {
        if (isPrime[i]) {
            for (let j = i * i; j < limit; j += i) {
                isPrime[j] = 0;
            }
        }
    }

    let sum = 0n;
    for (let i = 2; i < limit; i++) {
        if (isPrime[i]) {
            sum += BigInt(i);
        }
    }
    return sum;
}

const result = sumPrimesBelow(2000000);
console.log(result.toString());

Python

primes=[]
sieve=[0]*2000005
for i in range(2,2000000):
    if not sieve[i]:
        primes.append(i)
    for j in primes:
        if j*i>=2000000:
            break
        sieve[j*i]=1
        if not i%j:
            break
print(sum(primes))

Polynomix

Works in theory, but very ineffective.

1=(2 2*6.)#\l l@{l/0#\s l#&.(%s=0-) s}|.

This does not work on the answer checking in Project Euler Mirror due to a bug in that site's source that makes it literally impossible to get a correct answer.