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




int main()
{
  int n;
  scanf("%d",&n);
  if(n<105)
      printf("0");
  
  else if(n==105)
      printf("1");
  
  else
  {
    int i,cnt=0;
    for(i=106;i<=n;++i)
    {
       if(i%2!=0)
       {
         if(div(i)==1)
             cnt++;
       }
    }
    printf("%d",cnt+1);
  }
}