can understand a method in a c program about judging whether a number is prime

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hi guys

Here is a c program and I need your help!

For finding all prime Numbers between 100 to 200：

Procedure is as follows:

C++

#include<stdio.h>
#include<math.h>
void main()
{
  int m,k,i,n=0;
  for(m=101;m<=200;m=m+2)
   {
     k=sqrt(m);
     for(i=2;i<=k;i++)
        if(m%i==0)  break;
     if(i>=k+1)
        {
          printf("%d",m);
          n=n+1；
        }
     if(n%10==0)
        printf("\n");
    }
  printf("\n");
}

my question is:Program line 6, why m = m + 2, rather than m++? Why not judgment the value of m one by one?

I hope you coule help me !

Thanks!

Posted 22-Oct-13 19:55pm

STQ0809

Updated 22-Oct-13 19:58pm

nv3

v2

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nv3 23-Oct-13 2:00am

It wouldn't make sense to test even numbers for being prime, does it! That is why the loop progresses in increments of two.

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**Captain Price** · Answer 1 · 2013-10-22T20:29:00

This is a mathematical question rather a C programming question.

m is incremented by 2 because, every prime number is an odd number except number 2. For example take a loot at the prime numbers sequence below (Note that 1 is not considered prime):

2, 3, 5, 7, 11, ....

You can clearly see that all the prime numbers are odd expect 2, which is even.

So what advantage is there if you increment m by 1 ? Answer is None.

Thanks7872 · Answer 2 · 2013-10-22T20:09:00

See this:http://en.wikipedia.org/wiki/Prime_number[^]

It states that:

Quote:
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number. For example, 5 is prime because only 1 and 5 evenly divide it, whereas 6 is composite because it has the divisors 2 and 3 in addition to 1 and 6

Note: The only even prime number is 2.

Regards..

can understand a method in a c program about judging whether a number is prime

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