C Exercises: Check if a given number is nearly prime or not

C Basic Declarations and Expressions: Exercise-93 with Solution

Write a C program to check if a given number is nearly prime or not.
Nearly prime number is a positive integer which is equal to product of two prime numbers.

Sample Solution:

C Code:

#include <stdio.h>

#define NUM_OF_PRIMES   3500

int is_prime(int num);

int main(void)
{
int primes[NUM_OF_PRIMES], num_of_primes = 0;    
primes[num_of_primes++] = 2;
for(int num = 3; num * num<= 1000000000; num++)
    {
int flag = 1;
for(int id = 0; id <num_of_primes; id++)
        {
if(num % primes[id] == 0)
            {
flag = 0;
break;
            }
        }
if(flag) primes[num_of_primes++] = num;
    }

int N, num;
scanf("%d", &num);
int flag = 0;
for(int j = 0; (j <num_of_primes) && (primes[j] * primes[j] <= num); j++)
        {
if(num % primes[j] == 0)
            {
num /= primes[j];
flag = 1;
break;
            }
        }
if(flag &&is_prime_num(num)) printf("It is a Nearly prime number.\n");
else printf("It is not a Nearly prime number.\n");

return 0;
}

int is_prime_num(int num)
{
if(num != 2 &&num % 2 == 0) return 0;
for(int factor = 3; factor * factor <= num ; factor += 2)
    {
if(num % factor == 0) return 0;
    }

return 1;
}

Sample Output:

It is not a Nearly prime number.

Flowchart:

C programming Code Editor:

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