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>
#define rep(i, n) for(int (i) = 0; (i) < (n); (i)++)
#define printd(x) printf("%.12f\n", ans);
#define print(x) cout << (x) << endl;
#define Yes() cout << "Yes" << endl;
#define No() cout << "No" << endl;
#define YES() cout << "YES" << endl;
#define NO() cout << "NO" << endl;
#define all(x) (x).begin(),(x).end()
#define printa(x, n) for(ll i = 0; i < n; i++) {cout << (x[i]) << " ";} cout << endl;
template<class T> bool chmin(T& a, const T& b) {if (a > b) {a = b; return 1;} return 0;}
template<class T> bool chmax(T& a, const T& b) {if (a < b) {a = b; return 1;} return 0;}
using namespace std;
using ll = long long;
using pll = pair<ll, ll>;

vector< ll > divisor(ll n) {
  vector< ll > ret;
  for(ll i = 1; i * i <= n; i++) {
    if(n % i == 0) {
      ret.push_back(i);
      if(i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}

int main(void) {
    int n;
    cin >> n;

    int ans = 0;
    for (int i = 1; i <= n; i+= 2) {
        vector<ll> tmp = divisor(i);
        if (tmp.size() == 8) ans++;
    }
    
    cout << ans << endl;
    return 0;
}