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>

int main (){
  
 int a,b,k,i,j,count,ans;
  scanf("%d",&a);
  
  for(i = 1;i < a + 1;i++){
    //奇数だけにしている
  	if(i % 2 != 0){
      for(j = 1;j < i + 1;j++){
        if(i % j == 0){
        count++;
      	}
      }
      if(count == 8){
        ans++;
      }else{
        count = 0;
      }
   	 
 	}
  }
  
  printf("%d",ans);
    
  
      return 0;
  
}