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 <stdlib.h>
 
int main(){
  int N = 0;
  if(scanf("%d",&N) == 1 ) return 0;
  if((N < 0 ) || ( N > 200) || (N % 2 == 0)) return 0;
  int count = 0;
  int hoge = 0;
  for(int i = 1; i <= N; i++){
    for(int j = 1; j <= i; j++){
      if(i % j == 0) hoge++;
    }
    if(hoge == 8) count++;
  }
  printf("%d\n",count);
  return 0;
}