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;
    scanf("%d", &n);

    int counter = 0;
    for (int i = n; i >= 1; i--) {

        if (i % 2 == 1) {

            int cnt = 1;
            int half = i / 2;
            //printf("%d %d\n", i, half);
            for (int j = half; j >= 1; j--) {

                if (i % j == 0) {
                    cnt++;
                }

            }

            //printf("%d %d\n",i, cnt);

            if (cnt == 8) {
                counter++;
            }

        }

    }
    printf("%d", counter);

    return 0;
}