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 ans = 0;
    int ct;
    for (int i = 105; i <= N; i++){
        ct = 2;
        if (i % 2 == 0) continue;
        else{
            for (int j = 2; j < i; j++){
                if (i % j == 0) ct++;
                if (ct > 8) break;
            }
        }
        if (ct == 8) ans++;
    }

    printf("%d", ans);

    return 0;
}