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() {
  int i, j, number_n, flag, count = 0;
  scanf("%d", &number_n);
  for(i = 1; i <= number_n; i += 2) {
    flag = 0;
    for(j = 1; j <= i; j++) {
      if(i % j == 0) {
        flag++;
      }
    }
    if(flag == 8) count++;
  }
  printf("%d\n", count);
  return 0;
}