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Sieve of Eratosthenes c++

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luco answered on November 9, 2020 Popularity 8/10 Helpfulness 5/10

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Sieve of Eratosthenes c++

Comment

    #include <cmath>
    #include <vector>
	using namespace std;
	int n;
    vector<int> ans;
    vector<bool> is_prime(n+1, true);
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i <= sqrt(n); i++) {
        if (is_prime[i]) {
            for (int j = i * i; j <= n; j += i)
                is_prime[j] = false;
        }
    }
    for(int i=2;i<is_prime.size();i++){
        if(is_prime[i]){
            ans.push_back(i);
        }
    }

Popularity 8/10 Helpfulness 5/10 Language cpp

Source: Grepper

Tags: c++ sieve-of-eratosthenes

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Contributed on Sep 20 2022

luco

0 Answers Avg Quality 2/10

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int n;
vector<bool> is_prime(n+1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i <= n; i++) {
    if (is_prime[i] && (long long)i * i <= n) {
        for (int j = i * i; j <= n; j += i)
            is_prime[j] = false;
    }
}

Popularity 10/10 Helpfulness 7/10 Language cpp

Source: cp-algorithms.com

Tags: c++

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Contributed on Nov 09 2020

tough life

0 Answers Avg Quality 2/10

c++ sieve of eratosthenes

Comment

#include <iostream>
#include <vector>
#include <algorithm>
#include <bitset>

#define N 1000000	//N is the Range (0..N)

bitset < N+1 > numbers;
vector < int > primes;

void sieve(){
    numbers.set();
    numbers[1] = 0;

    for (int i = 2; i < N; i++){
        if (numbers[i] == 1){
            cout<<i<<endl;
            primes.push_back(i);
            for (int j = i*i; j<=N; j+=i)
                numbers[j] = 0;
        }
    }
}

Popularity 10/10 Helpfulness 4/10 Language cpp

Source: Grepper

Tags: algorithm algo

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Contributed on Jul 11 2022

ALeonidou

0 Answers Avg Quality 2/10

Sieve of Eratosthenes in C++

Comment

#include <vector>

std::vector<bool> sieveOfEratosthenes(int limit) {
    std::vector<bool> isPrime(limit + 1, true);
    isPrime[0] = isPrime[1] = false;

    for (int p = 2; p * p <= limit; ++p) {
        if (isPrime[p]) {
            for (int i = p * p; i <= limit; i += p)
                isPrime[i] = false;
        }
    }

    return isPrime;
}

bool isPrime(int number) {
    if (number <= 1)
        return false;

    std::vector<bool> isPrimeList = sieveOfEratosthenes(number);
    return isPrimeList[number];
}

Popularity 7/10 Helpfulness 3/10 Language cpp

Source: chat.openai.com

Tags: c++ sieve-of-eratosthenes sieve-of-e

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Contributed on Jun 18 2023

Darshan Trivedi

0 Answers Avg Quality 2/10

sieve of eratosthenes c++

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int n;
vector<char> is_prime(n+1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i <= n; i++) {
    if (is_prime[i] && (long long)i * i <= n) {
        for (int j = i * i; j <= n; j += i)
            is_prime[j] = false;
    }
}

Popularity 9/10 Helpfulness 2/10 Language cpp

Source: cp-algorithms.com

Tags: c++

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Contributed on Nov 09 2020

tough life

0 Answers Avg Quality 2/10

Sieve of Eratosthenes c++

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