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 <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <cmath>
#include <iomanip>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;

int main()
{
    int n, ans;
    cin >> n;

    if (n < 105)
    {
        cout << 0 << endl;
    }

    ans = 0;

    for (int i = 105; i <= n; i += 2)
    {
        int h = 0;
        for (int j = 1; j <= i; j++)
        {
            if (i % j == 0)
                h++;
        }

        if (h == 8)
        {
            ans++;
            //cout << i << endl;
        }
    }

    cout << ans << endl;

    return 0;
}