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 <math.h>

int main(void){
    int n;
    scanf("%d", &n);
    int list[5] = {3 * 5 * 7, 3 * 5 * 11, 3 * 5 * 13,
                (int)pow(3, 3) * 5, (int)pow(3, 3) * 7};
    int i, ans = 0;
    for(i = 0; i < 5; i++){
        if(n >= list[i]) ans++;
    }
    printf("%d\n", ans);
    return 0;
}