Optimizing prime numbers code?

Question

I wrote this code to show the primes between 1 and 100. The only condition is to do not use functions, whole code should be inline. I would ask if I can improve (optimize) it mu

Answer 1

If you just want primes below 100, there's no need to write code to compute them. It's perhaps a stupid answer, but it solves your problem as stated efficiently and concisely.

int main() {
    cout << "Prime numbers are:" << endl << "2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97" << endl;
    return 0;
}

Answer 2

/*** Return an array of primes from 2 to n. ***/
int[] function nPrimes(int n) {
   int primes[];  //memory allocation may be necessary depending upon language.
   int p = 0;
   bool prime;
   for (int i = 2; i <= n; i++) {
       prime = true;
       //use (j <= (int)sqrt(i)) instead of (j < i) for cost savings.
       for (int j = 2; j <= (int)sqrt(i); j++)  {
           if (i % j == 0) {
               prime = false;
               break;
           }
       }
       if (prime) {
           primes[p++] = i;
       }    
   }
   return primes;
}

Answer 3

Avoid using the square root function, and increment your divisor by 2. Also some tricky things in the i loop to increment your possible prime by 2. Inner loop doesn't even need to check divisibility by 2 since no even numbers will even be tested.

int i,j,sq;
int min;
for(sq = 2; sq <= 10; sq++)
{
  min = (sq-1)*(sq-1);
  min = min + (min+1)%2; //skip if it's even, so we always start on odd
  for(i = min; i < sq*sq; i+=2)
  {
    for(j = 3; j <= sq; j+=2)
    {
      if (i%j == 0)
        bad;
    }
  }
}

note that the sq loop doesn't add time because it shrinks the inner loops proportionately.