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, cont = 0;
    //int a[100];
    scanf("%d", &n);
    for (int i = 1; i < n+1; i += 2){
        int num_div = 0;
        for (int k = 1; k < i+1; ++k){
            if ((i%k) == 0) ++num_div;
        }
        if (num_div == 8){
            //a[cont] = i;
            ++cont;
        }
    }
    /*for(int i = 0; i < 100; ++i)
        printf("%d ",a[i]);
    printf("\n");*/
    printf("%d\n", cont);

}