Check whether an integer is a prime number in C

[Projectkmo]

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

/************************** \ 
* * 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;
)

[Projectkmo]

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

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

   * * 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; 
  )

[Projectkmo]

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

Dividing up the odd square root of N will be tested

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

   * * 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; 
  )