Sieve of Eratosthenes with bitset finding prime numbers - C++ STL

C++ examples for STL:bitset

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Description

Sieve of Eratosthenes with bitset finding prime numbers

Demo Code

#include <iostream> 
#include <iomanip> 
#include <cmath> 
#include <bitset> // bitset class definition 
using namespace std; 

int main() /*w ww .j  a  v  a 2 s.  c o m*/
{ 
    const int SIZE = 1024; 
    int value; 
    bitset< SIZE > sieve; // create bitset of 1024 bits 
    sieve.flip(); // flip all bits in bitset sieve 
    sieve.reset( 0 ); // reset first bit (number 0) 
    sieve.reset( 1 ); // reset second bit (number 1) 

    // perform Sieve of Eratosthenes 
    int finalBit = sqrt( static_cast< double >( sieve.size() ) ) + 1; 

    // determine all prime numbers from 2 to 1024 
    for ( int i = 2; i < finalBit; ++i ) 
    { 
       if ( sieve.test( i ) ) // bit i is on 
       { 
           for ( int j = 2 * i; j < SIZE; j += i ) 
               sieve.reset( j ); // set bit j off 
       }
    }

    cout << "The prime numbers in the range 2 to 1023 are:\n" ; 

    // display prime numbers in range 2-1023 
    for ( int k = 2, counter = 1; k < SIZE; ++k ) 
    { 
       if ( sieve.test( k ) ) // bit k is on 
       { 
          cout << setw( 5 ) << k; 

          if ( counter++ % 12 == 0 ) // counter is a multiple of 12 
              cout << '\n'; 
       }
   }

   cout << endl; 

   // get value from user to determine whether value is prime 
   cout << "\nEnter a value from 2 to 1023 (-1 to end): "; 
   cin >> value; 

   // determine whether user input is prime 
   while ( value != -1 ) 
   { 
       if ( sieve[ value ] ) // prime number 
          cout << value << " is a prime number\n"; 
       else // not a prime number 
          cout << value << " is not a prime number\n"; 

       cout << "\nEnter a value from 2 to 1023 (-1 to end): "; 
       cin >> value; 
   }
}

Result


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