Problem:
Score: 200 points

Problem Statement

The number 105 is quite special - it is odd but still it has eight divisors.
Now, your task is this: how many odd numbers with exactly eight positive divisors are there between 1 and N (inclusive)?

Constraints

N is an integer between 1 and 200 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print the count.

Sample Input 1

105

Sample Output 1

1

Among the numbers between 1 and 105, the only number that is odd and has exactly eight divisors is 105.

Sample Input 2

7

Sample Output 2

0

1 has one divisor. 3, 5 and 7 are all prime and have two divisors. Thus, there is no number that satisfies the condition.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>

int main()
{
	int n,i,j,c=0;
	int a[]={105,135,165,189,195};
	scanf("%d",&n);
	if(n<105)
	{
		printf("%d",c);
	}
	else
	{
		for(i=105;i<=n;i++)
		{
			for(j=0;j<5;j++)
			{
				if(i==a[j])
				{
					c++;
					break;
				}
				
			}
			
		}
		printf("%d",c);
	}
	
	
    return 0;  
}