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 n,i,j,num=0,count,check;
  scanf("%d",&n);
  for(i=105;i<=n;i=i+2){
    count=1;
    int x=i;
    for(j=3;j<=13;j=j+2){
      check=1;
      while(x%j==0 && x!=0){
        check++;
        x/=j;
      }
      count*=check;
    }
    if(count==8)
      num++;
  }
  printf("%d\n",num);
  return 0;
}