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){
    // Your code here!
    int j,N,count4,count = 0;
    scanf("%d",&N);
    for(int i = 1; i <= N; i=i+2)
    {
        j = 1;
        count4 = 0;
        while(j <= sqrt(i))
        {
            if(i%j == 0) count4++;
        
            j = j + 2;
        }
        if(count4 == 4) count++;
    }
    printf("%d\n",count);
}