Sieve of Eratosthenes algorithm

Question

I am currently reading \"Programming: Principles and Practice Using C++\", in Chapter 4 there is an exercise in which:

I need to mak

Answer 1

I think im late to this party but im reading the same book as you, this is the solution in came up with! Feel free to make suggestions (you or any!), for what im seeing here a couple of us extracted the operation to know if a number is multiple of another to a function.

#include "../../std_lib_facilities.h"

bool numIsMultipleOf(int n, int m) {
  return n%m == 0;
}

int main() {
  vector<int> rawCollection = {};
  vector<int> numsToCheck = {2,3,5,7};

  // Prepare raw collection
  for (int i=2;i<=100;++i) {
    rawCollection.push_back(i);
  }

  // Check multiples
  for (int m: numsToCheck) {
    vector<int> _temp = {};

    for (int n: rawCollection) {
      if (!numIsMultipleOf(n,m)||n==m) _temp.push_back(n);
    }
    rawCollection = _temp;
  }

  for (int p: rawCollection) {
    cout<<"N("<<p<<")"<<" is prime.\n";
  }

  return 0;
}

Answer 2

I am following the same book now. I have come up with the following implementation of the algorithm.

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
inline void keep_window_open() { char ch; cin>>ch; }

int main ()
{
    int max_no = 100;

    vector <int> numbers (max_no - 1);
    iota(numbers.begin(), numbers.end(), 2);

    for (unsigned int ind = 0; ind < numbers.size(); ++ind)
    {
        for (unsigned int index = ind+1; index < numbers.size(); ++index)
        {
            if (numbers[index] % numbers[ind] == 0)
            {
                numbers.erase(numbers.begin() + index);
            }
        }
    }
    cout << "The primes are\n";
    for (int primes: numbers)
    {
        cout << primes << '\n';
    }
}