Check whether an integer is a prime number in C

A prime number is an integer, which is divided only by 1 and itself.

Algorithm 1: dividers between 2 and N-1 will be tested

Code:

---

/************************** \
* * xyz_abc1.c
/ * Algorithm: testing all dividers * /
 # include <stdio.h>
 int main (void)
 (
    int i, nb, count, test;
    test = count = 0;
    printf ( "Enter integer");
    if (scanf ( "% d", & nb)! = 1)
      return -1;
   
for (i = 2; i <nb i + + count + +)
  if (nb% i == 0)
  test = 1;
if (! test)
  printf ( "% d prime number, many iterations =% d \ n", nb, count);
else
  printf ( "% d is not the first number, many iterations =% d \ n", nb, count);
return 0;
)

---

Algorithm 2: Dividing peers will not be tested, research is limited to odd dividers

Code:

---

/************************** \

  * * xyz_abc1.c
  \ **************************/

  / * Algorithm: exclude even numbers and

  * Test all dividers * /
  # include <stdio.h>


  int main (void)
  (
    int i, nb, count, test;
    test = count = 0;
    printf ( "Enter integer");
    if (scanf ( "% d", & nb)! = 1)
      return -1;

    if (nb% 2 == 0)
            test = 1;
    else (
        for (i = 3; i <nb i + = 2, count + +)
          if (nb% i == 0)
            test = 1;
    )
    if (! test)
            printf ( "% d prime number, many iterations =% d \ n",
                            nb, count);
  else
            printf ( "% d is not the first number, many iterations =% d \ n", nb, count);
    return 0;
  )

---

Algorithm 3: Dividing up the odd square root of N will be tested

Dividing up the odd square root of N will be tested

Code:

---

/************************** \

  * * xyz_abc3.c
  \ **************************/

  / * Algorithm: exclude even numbers and

  * Test all dividers to the square root * /
  # include <stdio.h>
  # include <math.h>

  int main (void)
  (
    int i, nb, count, test limits;
    test = count = 0;
    printf ( "Enter integer");
    if (scanf ( "% d", & nb)! = 1)
      return -1;
    limit = sqrt (nb) + 1;

    if (nb% 2 == 0)
            test = 1;
    else (
        for (i = 3; i <limit i + = 2, count + +)
          if (nb% i == 0)
            test = 1;
    )
    if (! test)
            printf ( "% d prime number, many iterations =% d \ n", nb, count);
  else
            printf ( "% d is not the first number, many iterations =% d \ n", nb, count);
    return 0;
  )

---