N states of prime number solution

The first one is ： The basic method

//4. Judge i Prime or not , Primes are numbers that can only be 1 And the number divisible by itself 
// lookup 1-100 The prime between 
#include <math.h>
#include <stdio.h>

int main()
{
    
	int i = 0;
	int count = 0;
	for (i = 1; i <= 100; i++)
	{
    
		// Trial division 
		int j = 0;
		for (j = 2; j <= i; j++)
		{
    
			if (i%j == 0)
			{
    
				break;
			}
		}
		if (j == i)
		{
    
			count++;
			printf("%d ", i);
		}
	}
	printf("\ncount = %d\n", count);
	return 0;
}

The second kind ： Modify the search method , Use the square method to reduce the number of searches

int main()
{
    
	int i = 0;
	int count = 0;
	for (i = 1; i <= 100; i++)
	{
    
		// Trial division 
		int j = 0;
		for (j = 2; j <= sqrt(i); j++)// Square root 
		{
    
			if (i%j == 0)
			{
    
				break;
			}
		}
		if (j >sqrt (i))
		{
    
			count++;
			printf("%d ", i);
		}
	}
	printf("\ncount = %d\n", count);
	return 0;
}

The third realm ： Look for prime numbers in odd numbers

int main()
{
    
	int i = 0;
	int count = 0;
	for (i = 1; i <= 100; i+=2)// Look for prime numbers in odd numbers 
	{
    
		// Trial division 
		int j = 0;
		for (j = 2; j <= i; j++)
		{
    
			if (i%j == 0)
			{
    
				break;
			}
		}
		if (j == i)
		{
    
			count++;
			printf("%d ", i);
		}
	}
	printf("\ncount = %d\n", count);
	return 0;
}