Printing 100 prime numbers

This is a program I wrote to print out 100 prime numbers. How can I improve it?

```
// Calculating primes
#include
#include
#include
using namespace std;

int main()
{
const int MAX(100); // Number of primes to be identified
long primes[MAX] = {};
long trial(5); // Hold a candidate prime
int count(0); // Count primes found
bool found(false); // Indicates when a prime is found

vector nonPrimes; //Vector to hold non prime numbers
vector disqualify; //Vector to hold prime numbers that disqualify the non prime numbers as prime numbers.

do
{
trial += 2; // Produce next candidate value
found = false; // Reset indicator - assume it is not prime

for (int i = 0; i < count; i++) // Try division by existing primes
{
found = (trial % primes[i]) == 0; // True if no remainder
if (found) // No remainder means number is not a prime
break;
// Terminate the division loop
}
// The above loop will exit either due to a break or after trying all
// existing primes as a divisor. found was initialized to false. If
// found is false after exiting the loop, a divisor was not found.
if (!found) // We have a new prime: not false = true
primes[count++] = trial; // Save candidate in next array position
} while (count < MAX);

// Main loop has completed - we have found MAX prime numbers.
// Display the prime numbers, presenting five numbers on one line.
cout << "Prime numbers found during the program execution:" << endl;
for (int i = 0; i < MAX; i++)
{
if (i % 5 == 0) // For a new line on first line of output
cout << endl; // and on every fifth line that follows
cout << setw(10) << primes[i]; // Provide space between numbers
}

cout << endl; // All primes displayed - for a n

• STOP. Do not use system("pause"). This is not portable. If you

haven't already, read
this

• Stop using using namespace std. Read

this.
I mentioned this in your last review as well.

• You don't need to use std::endl for every new line of your output

table. Doing so flushes the output stream needlessly.

• You don't need two vectors. Just two for loops and one vector will

suffice. Take a look at the example below.

• For beginner code, I think you have good documentation(comments).

Although for some people, it might err on the side of excessive.

-
You could have initialized the array with the first three primes.
That would have cut down the number of iterations.

long primes[MAX] = {2,3,5};
 int count(0);             // Count primes found

Example:

#include  
#include 
#include 

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

int  main()
{
    int counter = 0;
    std::vector primes;

    int N, output;
    cin >> N;
    output = N;
    for (int i=2; N > 0; ++i)
    {
        bool  isPrime = true;
        for (int j=0; j < primes.size(); ++j)  //First pass(i==2), will not enter second loop because size() == 0
        {
            if (i % primes[j] ==  0)
            {
                isPrime = false;
                break;
            }
        }

        if (isPrime)
        {
            primes.push_back(i);
            --N;
            cout  <<  i  <<  "\n";
        }
    }

    for(int i=0; i < output; i++)
    {
        if (i % 5 == 0)      // For a new line on first line of output
            cout << "\n";      // and on every fifth line that follows

        cout << setw(10) << primes[i];  // Provide space between numbers
    }
    cout << endl;

    return  0;
}

StackExchange Code Review Q#66468, answer score: 12