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>
#include <string.h>
#include <math.h>

int main() {
    int n, divisores = 0, num=0;
    scanf("%d", &n);

    for (int i=1; i<=n; ++i) {
        for (int j=1; j<=i; ++j) {
            if(divisores>8) {
                divisores = 0;
                break;
            }

            if(i%j==0 && j%2!=0)
                ++divisores;

            if(j==i && divisores==8)
                ++num;
            //printf("%d %d %d %d\n", num, i, j, divisores);

            if (j==i)
                divisores = 0;

        }
    }

    printf("%d\n", num);
    return 0;
}