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 <stdlib.h>
#include <math.h>

//numberが素数ならば1，そうでなければ0を返す関数
int primality(int number);

int main(void){
  int N;
  int i, j, k;
  int count = 0, count2 = 0;

  scanf("%d", &N);

  for(i = 1; i <= (N+1)/2; i++){
    count = 0;
    for(j = 1; j <= 2*i-1; j++){
      if((2*i-1)%j == 0){
        count++;
      }
    }
    if(count == 8){
      count2++;
    }
  }

  printf("%d\n", count2);
  return 0;
}

int primality(int number){
  int i;

  if(number <= 1){
    return 0;
  }
  else{
    for(i = 2; i < number; i++){
      if(number%i == 0){
        return 0;
      }
    }
    return 1;
  }
}