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 divisor(int n){
	int i,count=1;
	for (i=1;i<n/2;i++){
		if(n%i==0) count+=1;
	}
	if(count==8) return 1;
	else return 0;
}

int main()
{
	int n,num=0,a,i;
	scanf("%d",&n);
	for (i=1;i<=n;i=i+2){
		a=divisor(i);
		if(a==1) num+=1;
	}
	printf("%d\n",num);
	return 0;
}