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<stdlib.h>

int div_num(int n){
    int i,counter=0;
    for(i=1;i<=n;i++){
        if(n%i==0){
            counter++;
        }
    }
    return counter;
}

int main(void){
    int counter=0;
    int i,N;
    scanf("%d",&N);
    for(i=1;i<=N;i++){
        if(div_num(i)==8 && i%2 == 1){
            counter++;
        }
    }
    printf("%d\n",counter);
    return 0;
}