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

int counter(int num){
  int x = 0;
  for(int i=1;i<=num;i++){
    if(i%2==1 && divisor(i)==8){
      x++;
    }
  }
  return x;
}

int main(void){
  int n = 0;
  scanf("%d", &n);
  printf("%d", counter(n));
}