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 main(void){
  int n;
  scanf("%d", &n);
  int ct = 0;
  int c;
  int i, j;
  if ( n < 105 ) {
    ct = 0;
  } else if ( n >= 105 ) {
    for ( i = 105; i <= n; i += 2 ) {
      c = 0;
      for ( j = 1; j <= i; j++ ) {
        if ( i % j == 0 ) {
          c += 1;
        }
      }
      if ( c == 8 ) {
        ct += 1;
      }
    }
  }
  printf("%d\n", ct);
  return 0;
}