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>

int main(){
  int n, c = 0;

  scanf( "%d", &n );
  //scanf( "%d", &b );
  if ( n < 105 ){
    printf( "0\n" );
    return 0;
  }
   if ( n == 105 ){
    printf( "1\n" );
    return 0;
  }
  int i, j;
  c = 1;
  for ( i = n; i > 105; i-- ){
    int num = 0;
    if ( i % 2 == 0 ) continue;
    for ( j = 1; j <= i/* / 2 + 1*/; j += 2 ){
      if ( i % j == 0 ){
        num++;
      }
    }
    if ( num == 8 ){
      c++;
    }
  }
  printf( "%d\n", c );
  return 0;
}