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 REP1(i,n) for(int i = 1; i <= (n); ++i)
#define RREP(i, n) for(int i = (n) - 1; i >= 0; --i)
#define ALL(a) begin(a), end(a)
#define CEIL(a,b) ((a) / (b) + ((a) % (b) == 0 ? 0 : 1))
#define X first
#define Y second

using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<string>;

constexpr int MOD = int(1e9 + 7);
constexpr int INF = int(1e9 + 1);
constexpr ll LLINF = ll(4 * 1e18 + 1);
// constexpr int INF = 2147483647; // 2 * 1e9
// constexpr ll LLINF = 9223372036854775807; // 9 * 1e18

const int dx[] = {1, 0, -1, 0, 1, -1, -1, 1, 0};
const int dy[] = {0, 1, 0, -1, 1, 1, -1, -1, 0};

int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout << fixed << setprecision(10);
	int n;
	cin >> n;
	int ans = 0;
	REP1(i, n){
		int yakusuu = 1;
		int tmp = i;
		for(int j = 2; j <= tmp; ++j){
			int cnt = 0;
			while(!(tmp % j))tmp /= j, cnt++;
			yakusuu *= cnt + 1;
		}
		if(yakusuu == 8 && (i % 2))ans++;
	}
	cout << ans << "\n";
	return 0;
}