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: Runtime Error)

#include<stdio.h>

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