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 divisores(int i);
	
	int n;
	scanf("%d",&n);
	
	int y,d;
	
	
	for(int i = 1; i<=n; ++i){
		if(i%2 != 0){
			y = divisores(i);
			if( y == 8){
				d+=1;
			}
		}
	}	
	
	printf("%d\n",d);
	
	return 0;
}

int divisores(int i){
	
	int num = 0;
	
	for (int x = 1; x<=i; ++x){
		if(i%x == 0){
			num+=1;
		}
	}
	
	return num;
}