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 n;
    scanf ("%d",&n);
    if (n<105)
    {
        printf ("0\n");
    }
    else if (n==105)
    {
        printf ("1\n");
    }
    else if (n>105)
    {
        int c=0,g=1;
        int i,j,k;
        for (i=106;i<=n;i++)
        {
            if (i%2!=0)
            {
                for (j=1;j<=n;j+=2)
                {
                    if (i%j==0)
                    {
                        c++;
                    }
                    if (c==8)
                    {
                        g++;
                    }
                }
            }
        }
        printf ("%d\n",g);
    }
    return 0;
}