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>
#include<string.h>
#include<stdlib.h>
#include<math.h>

int main(){
  int n,x;
  int i,j,k;
  int cnt = 0;
  int cntd;

  scanf("%d", &n);

  for(i = 1;i <= n;i += 2){
    cntd = 0;
    for(j = 1;j <= i;j++){
      if(i % j == 0) cntd++;
    }

    if(cntd == 8) cnt++;
  }

  printf("%d\n", cnt);

  return 0;
}