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,cnt,cnt8 = 0;
  scanf("%d",&N);
  
  int i,j;
  for(i = 1;i <= N; i += 2){
    cnt = 0;
    for(j = 1;j <= i; j += 2){
      if(!(i % j))	cnt++;
    }
    if(cnt == 8) ++cnt8;
  }
  printf("%d\n",cnt8);
  return 0;
}