Printing prime numbers from 1 through 100

Question

This c++ code prints out the following prime numbers: 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97.

But I don\'t think tha

Answer 1

I check if a number is prime or not with the following code( of course using sqrt ):

bool IsPrime(const unsigned int x)
{
  const unsigned int TOP
  = static_cast<int>(
      std::sqrt( static_cast<double>( x ) )
    ) + 1;

  for ( int i=2; i != TOP; ++i )
  {
    if (x % i == 0) return false;
  }
  return true;
}

I use this method to determine the primes:

#include <iostream>

using std::cout;
using std::cin;
using std::endl;

#include <cmath>

void initialize( unsigned int *, const unsigned int );
void show_list( const unsigned int *, const unsigned int );
void criba( unsigned int *, const unsigned int );
void setItem ( unsigned int *, const unsigned int, const unsigned int );

bool IsPrime(const unsigned int x)
{
  const unsigned int TOP
  = static_cast<int>(
      std::sqrt( static_cast<double>( x ) )
    ) + 1;

  for ( int i=2; i != TOP; ++i )
  {
    if (x % i == 0) return false;
  }
  return true;
}

int main()
{

    unsigned int *l;
    unsigned int n;

    cout << "Ingrese tope de criba" << endl;
    cin >> n;

    l = new unsigned int[n];

    initialize( l, n );

    cout << "Esta es la lista" << endl;
    show_list( l, n );

    criba( l, n );  

    cout << "Estos son los primos" << endl;
    show_list( l, n );
}

void initialize( unsigned int *l, const unsigned int n)
{
    for( int i = 0; i < n - 1; i++ )
        *( l + i ) = i + 2;
}

void show_list( const unsigned int *l, const unsigned int n)
{
    for( int i = 0; i < n - 1; i++ )
    {
        if( *( l + i ) != 0)
            cout << l[i] << " - ";
    }
    cout << endl;
}

void setItem( unsigned int *l, const unsigned int n, const unsigned int p)
{
    unsigned int i = 2;
    while( p * i <= n)
    {
        *( l + (i * p - 2) ) = 0;
        i++;
    }
}

void criba( unsigned int *l, const unsigned int n)
{
    for( int i = 0;  i * i <= n ; i++ )
     if( IsPrime ( *( l + i) ) )
        setItem( l, n, *(l + i) );      
}

Answer 2

this is my approach from a simple blog:

//Prime Numbers generation in C++
//Using for loops and conditional structures
#include <iostream>
using namespace std;

int main()
{
int a = 2;       //start from 2
long long int b = 1000;     //ends at 1000

for (int i = a; i <= b; i++)
{

 for (int j = 2; j <= i; j++)
 {
    if (!(i%j)&&(i!=j))    //Condition for not prime
        {
            break;
        }

    if (j==i)             //condition for Prime Numbers
        {
              cout << i << endl;

        }
 }
}
}

- See more at: http://www.programmingtunes.com/generation-of-prime-numbers-c/#sthash.YoWHqYcm.dpuf

Answer 3

A simple program to print "N" prime numbers. You can use N value as 100.

    #include  <iostream >
    using  namespace  std;

    int  main()
    {
        int  N;
        cin  >>  N;
        for (int  i =  2;  N > 0;  ++i)
        {
            bool  isPrime  =  true ;
            for (int  j =  2;  j < i;  ++j)
            {
                if (i  % j ==  0)
                {
                    isPrime  =  false ;
                    break ;
                }
            }
            if (isPrime)
            {
                --N;
                cout  <<  i  <<  "\n";
            }
        }
        return  0;
    }

Answer 4

actually the better solution is to use "A prime sieve or prime number sieve" which "is a fast type of algorithm for finding primes" .. wikipedia

The simple (but not faster) algorithm is called "sieve of eratosthenes" and can be done in the following steps(from wikipedia again):

1. Create a list of consecutive integers from 2 to n: (2, 3, 4, ..., n).
2. Initially, let p equal 2, the first prime number.
3. Starting from p, count up in increments of p and mark each of these numbers greater than p itself in the list. These numbers will be 2p, 3p, 4p, etc.; note that some of them may have already been marked.
4. Find the first number greater than p in the list that is not marked. If there was no such number, stop. Otherwise, let p now equal this number (which is the next prime), and repeat from step 3.