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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>
 
int main(void)
{
   int n, i, f, s, x=0, r;
 
   scanf("%d",&n);
 
   for(i=1;i<=n;i=i+2){
      s=0;
      r=sqrt(i);
      for(f=1;f<=r;f++)
         if(i%f==0)
            s++;
      if((s==4)&&(i!=r*r)){
         x++;
      }
   }
 
   printf("%d\n",x);
   return 0;
}