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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(){
  
    int N;
    int i;
    int a;
    int b;

    a = 0;
    b = 0;
    

    scanf("%d", &N);

    for(N = 1; N <= 200; ++N){
      if(N%2==1){  
        for(i = 1; i <= 200; ++i){
          if(N%i==0){
            a = a + 1;
          }
        }
      }
    }
    for(N = 0; N <= 200; ++N){
      if(a == 8){
        b = b + 1;
      }
    }
    printf("%d\n", b);
      
return 0;
}