Function that returns count of prime factors of given number

Question

1st Example : 18 prime factors are 2*3*3 and it returns count =2 as prime factors includes 2 and 3 only or 2nd example: 27 =3*3*3 and return count =1.

Answer 1

int main()
{
    int numb,count1,i,j,count2;
    printf("Please enter the number who's prime factors and their count is to be determined\n");
    scanf("%d",&numb);

    count1=0;


    printf("\nThe prime factors are\n");
    for(i=2;i<numb;i++)
    {

        count2=0;
        j=2;

        while(((numb%i)==0)&&(j<i))
        {
            if((i%j)==0)
            {
                count2++;

            }

            j++;
        }


        if((count2==0)&&((numb%i)==0))
        {
            count1++;
            printf("%d\n",i);
        }
    }
    if(count1==0)
    {
        printf("None\n");
    }

    printf("So there are %d Prime factors\n", count1);

}

Ideone.com link : https://ideone.com/knarkA

Comments

Thanks Tejas

Answer 2

void primeFactors(int n)
{
    // Print the number of 2s that divide n
    while (n%2 == 0)
    {
        printf("%d ", 2);
        n = n/2;
    }

    // n must be odd at this point.  So we can skip one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i+2)
    {
        // While i divides n, print i and divide n
        while (n%i == 0)
        {
            printf("%d ", i);
            n = n/i;
        }
    }

    // This condition is to handle the case whien n is a prime number
    // greater than 2
    if (n > 2)
        printf ("%d ", n);
}

Credit: http://www.geeksforgeeks.org/print-all-prime-factors-of-a-given-number/

Comments

Hi,thanks for your ans but you didn't get my ques.  acc to your solution eg.if num is  18 then it prints   2 ,3,3  but I am interested only in prime factors count that is 2.

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But that is easy once u have prime factor, choose the unique count within the list...