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 divisor(int n){
    int a = 0;
    for(int i = 1; i <= n; i= i+2){
        if(n%i == 0){
            a++; 
        }
    }
    return a;
}

int main(void){
    int N;
    int n = 0;
    scanf("%d",&N);
    if(N>=105){
        for(int i  =105; i <= N; i=i+2){
            if(divisor(i) == 8){
                n++;
                printf("i = %d, div = %d\n",i, divisor(i));
            }
        }
    }
    printf("%d\n",n);
    return 0;
}