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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define all(x) x.begin(), x.end()
#define rep(i, n) for (int i = 0; i < n; ++i)
#define reps(i, m, n) for (int i = m; i <= n; ++i)

typedef long long i64;
typedef pair<i64, i64> pii;

constexpr int INF = 0x3f3f3f3f;
constexpr i64 LINF = 0x3f3f3f3f3f3f3f3fLL;

map<i64, int> primeFactorize(i64 n) {
  map<i64, int> ret;
  for (i64 i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      ++ret[i];
      n /= i;
    }
  }
  if (n != 1) ret[n] = 1;
  return ret;
}

signed main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;
  cin >> N;
  int ans = 0;
  for (int i = 1; i <= N; i += 2) {
    int cnt = 0;
    for (auto e : primeFactorize(i)) {
      if (e.second == 1) ++cnt;
    }
    if (cnt == 3) ++ans;
  }
  cout << ans << "\n";
  return 0;
}