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: Wrong Answer)

#include<stdio.h>
#include<string.h>
int main(){
  int n,i,j,fc=0,c=1;
  scanf("%d",&n);
  if(n < 105){
    printf("0\n");
    return 0;
  }
  for(i=107;i<=n;i+=2){
    for(j=1;j<=i;j++){
      if(n%j==0)
        fc++;
    }
    if(fc==8){
      c++;
      puts("barse");
      fc=0;
    }
  }
  printf("%d\n",c);
  return 0;
}