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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  auto ok = [&](int x) {
    int ret = 0;         
    int until = sqrt(x) + 1;
    for (int i = 1; i <= until; i++) {
      if (x % i == 0) {
        ret += (x / i == i ? 1 : 2);
      }
    }
    return ret == 8;
  };
  int res = 0;
  for (int i = 1; i <= n; i += 2) {
    res += ok(i);
  }
  cout << res << '\n';
  return 0;
}