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 9.2.1) (judge verdict: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<vector>
#include<functional>
#include<algorithm>
#include<stdlib.h>
#include<string>
#include<string.h>
#define _USE_MATH_DEFINES
#include<math.h>
#include<deque>
#include<set>
#include<map>
#include<queue>
#include<list>
#include<iostream>
#include <bitset>
using namespace std;
typedef long long ll;
#define rep(i,a,b) for(auto i=a;i<b;i++)
#define rep2(i, a)for(auto i : a)
#define all(_x) _x.begin(), _x.end()
#define r_sort(_x) sort(_x, std::greater<int>())
#define vec_cnt(_a, _n) (upper_bound(all(_a), _n) - lower_bound(all(_a), _n))
#define vec_unique(_a) _a.erase(std::unique(all(_a)), _a.end());
#define vvec vector<vector<ll>>
ll gcd(ll a, ll b) { return a % b == 0 ? b : gcd(b, a % b); }
ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; }
#define INF 1 << 30
const int mod = 1000000007;
ll power(ll x, ll p) { ll a = 1; while (p > 0) { if (p % 2 == 0) { x *= x; p /= 2; } else { a *= x; p--; } }return a; }
ll mpower(ll x, ll p) { ll a = 1; while (p > 0) { if (p % 2 == 0) { x = x * x % mod; p /= 2; } else { a = a * x % mod; p--; } }return a; }
ll co(ll n, ll k) { ll a = 1; rep(i, 1ll, k) { a *= n - i + 1; a /= i; }return a; }
ll mc(ll n, ll m) { ll k = 1, l = 1; rep(i, n - m + 1, n + 1) k = k * i % mod; rep(i, 1, m + 1) l = l * i % mod; l = mpower(l, mod - 2); return k * l % mod; }
int main(void) {
	int c = 0, ans = 0;
	int k, _n;
	cin >> _n;
	for (int n = 3; n <= _n; n += 2) {
		c = 0;
		k = (int)sqrt(n);
		if (k * k == n)c++;
		rep(i, 1, k)if (n % i == 0)c += 2;
		if (c == 8)ans++;
	}
	printf("%d\n", ans);
	return 0;
}