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: Wrong Answer)

#include <stdio.h>

int main(){
    int numero, qtdNUM = 1, divisores; // qtdNUM = 1 pois já está incluindo o 105
    scanf("%d", &numero);
    if (numero < 105){
        printf("0\n");
    }
    else if (numero == 105){
        printf("1\n");
    }
    else{
        for (int i = 106; i < numero; i++){
            if (i % 2 != 0){ // se o número for par nem entra no loop
                divisores = 2; // pois todo número é divisivel por 1 e ele mesmo
                for (int j = 3; j <= i/2; j += 2){ // pois um número ímpar nunca será divisivel por um par
                    if (i % j == 0){
                        divisores += 1;
                    }
                }
                if (divisores == 8){
                    qtdNUM += 1;
                }
            }
        }
        printf("%d\n", qtdNUM);
    }
    return 0;
}