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 <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <limits.h>

int main(void){
    int n,ans=0;
    scanf("%d",&n);
    for (int i=1; i<=n; i++){
        if (i%2!=0){
            int count=0;
            for (int j=1; j<=i; j++){
                if (i%j==0){
                    count++;
                }
            }
            if (count==8){
                ans++;
            }
        }
    }
    printf("%d\n",ans);
	return 0 ;
}