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>

int main()
{
	int n,r=0,i,j,k,c;
	scanf("%d",&n);
	if(n<105) r=0;
	else if(n==105) r=1;
	else 
	{
		r=1;
		for(i=107;i<=n;i+=2)
		{
			c=0;
			for(j=1;j*j<i;j++)
			{
				if(j*j==i) c+=1;
				else if(i%j==0) c+=2;
			}
			if(c==8) r++;
		}
	} 
	printf("%d\n",r);
	return 0;
}