Printing prime numbers in c language - c

Question

I've been trying for last 3 days to overcome the problem but I'm failing continuously.
I'm trying to print all prime numbers from 1 to 300 in C.(Dev C++)
below is the code
#include <stdio.h>
#include <conio.h>
main() {
// 1 - 300
register int i, j;
for (i = 1; i <= 300; i++) {
for (j = 2; j < i; j++) {
if (i % j == 0) {
break;
} else {
printf("\n%d", i);
break;
}
}
}
}
please help me in this and also help me to clear the concept.
Thanks.

Answer 1

You are printing the numbers too early. Print the numbers at the end of the outer loop. ie, after the inner loop.
You could do
for(i=1; i<=300; i++)
{
for(j=2; j<i; j++)
{
if(i%j == 0)
{
break;
}
}
if(j>=i)
{
printf("\n%d",i);
}
}
If at the end of inner loop, the condition j<i still holds, the loop was terminated prematurely due to the break statement being executed and hence the number is not prime.
But if that condition is false, ie, if j>=i is true, the number is prime.
You could also do this with the help of a flag variable.
And you probably don't need to use the register storage class specifier. See this post.
And there's no need of the break statements. In the program you've posted,
if(i%j == 0)
{
break;
}
else
{
printf("\n%d",i);
break;
}
is same as
if(i%j == 0)
{
break;
}
printf("\n%d",i);
or just
if(i%j != 0)
{
printf("\n%d",i);
}

Answer 2

For primes up to 300, the Sieve of Eratosthenes algorithm is quite good.

Answer 3

This should do the work for now, but I am sure that there are more efficient ways for doing this job:
int main() {
int i, j;
char is_prime; // indicator for each number if it is a prime number
printf("2\n"); // first number in the series, and the only one witch is even
for (i = 3; i<=300; i += 2) // When I jump in 2 numbers each time, I save the check of all the even numbers, that we knows ahead that they are not prime numbers.
{
is_prime = 1; // Start with an assumption that the number is prime. If we will find out that it doesn't, we will turn this value to 0.
for (j = 3; j < i; j++)
{
if(i % j == 0)
{ // If this number fully divided by this `j` value, so it is not a prime number.
is_prime = 0; // Turn this variable to indicate that this is not a prime number.
break; // This loop is for testing the current number. Now we found out that it is not a prime number, so we can end this loop.
}
} // End of the inner loop that check if the current 'i' number is a prime number.
if (is_prime) { // If we found that the current 'i' number is a prime number (if is_prime != 0)
printf("%d\n", i); // Print the 'i' number.
}
}
return 0;
}

Answer 4

Generating prime numbers
This is the simplest way to achieve this
for(i=2 ; i <= n ; i++)
{
for(j=2 ; j<i ;j++)
{
if(i%j == 0)
break ;
}
if(i == j )
printf("%d\n",i);
}

Answer 5

#include<stdio.h>
void main()
{
int n,i,fact,j;
printf("Enter the Number");
scanf("%d",&n);
printf("Prime Numbers are: \n");
for(i=1; i<=n; i++)
{
fact=0;
for(j=1; j<=n; j++)
{
if(i%j==0)
fact++;
}
if(fact==2)
printf("%d " ,i);
}
getch();
}

Answer 6

Try to solve it your self, putting values of i and j respectively, you'll find the errors.
1.Declare a variable int and initialize it by 0 (int a=0).
2.Then in the inner for loop in the if statement increase the value of a for each division.
If(i÷j==0)
{a=a+1;//or a++
}
3.Get out of the loop now and look if the value of the a is still 0 then i is a prime else it's not!