C ++：Eratosthenes筛子不总是有效

Question

I am making a C++ program to find prime numbers using the Sieve of Eratosthenes

Currently I have the following code:

C++

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

class is_multiple_of {
    int m_div;
public:
    is_multiple_of(int div) : m_div(div) {}
    bool operator()(int n) { return ((n > m_div) && (0 == n % m_div)); }
};

int main ()
{
    vector<int> v;
    for (int i=2; i<=100; i++) v.push_back(i);

    v.erase( remove_if(v.begin(), v.end(), is_multiple_of(2)), v.end() );
    v.erase( remove_if(v.begin(), v.end(), is_multiple_of(3)), v.end() ); 
    v.erase( remove_if(v.begin(), v.end(), is_multiple_of(5)), v.end() ); 
    v.erase( remove_if(v.begin(), v.end(), is_multiple_of(7)), v.end() ); 

    for (unsigned i=0; i<v.size(); ++i)
        cout << v[i] << " ";


    cout << endl << v.size();
    return 0;
}

which works fine ie it says there are 25 primes between 1 and 100 (which is correct ). However, if i want to know the first 500 primes for example, it will say there are 118 where in reality there are 95. To correct this issue, I have to add more multiples to remove ie v.erase( remove_if(v.begin(), v.end(), is_multiple_of(11)), v.end() ); and then add more is_multiple_of() 's.

Is there a way to make this more efficient rather than just having it remove more multiples of previously found primes?

Answer 1

Since you want to implement your version of finding the primes, then the following should work (we are using 500)

 int stopInt = static_cast<int>(sqrt(500));
 int j = 0;
 for (int i = 0; i < stopInt; ++i, ++j)
     v.erase(remove_if(v.begin(), v.end(), is_multiple_of(v[j])), v.end());

Here is an alternative that "marks" the items by just moving them to the end of the vector:

     vector<int>::iterator sMarked = v.end();
     int stopInt = static_cast<int>(sqrt(500));
     int j = 0;
     for (int i = 0; i < stopInt; ++i, ++j)
         sMarked = remove_if(v.begin(), sMarked, is_multiple_of(v[j]));
     v.erase(sMarked, v.end());

So we're constantly marking the elements by keeping track of the return value of remove_if . Then at the end, we do one single vector::erase to remove the multiples. This should be more efficient than calling the v_erase constantly in the loop.