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 FOR(i, a, b) for (int i = (a); i < (b); i++)
#define whole(f, x, ...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x) // decltypeで型取得、引数があればva_argsのところに入れる
using namespace std;
typedef long long ll; // long longをllでかけるようにした
const int INF = 1e9;

int main(void){
    int n; cin >> n;
    int cnt = 0, ans = 0;
    for(int i = 1; i <= n; i += 2){
        cnt = 0;
        for(int j = 1; j <= i; j += 2){
            if(i % j == 0) cnt++;
        }
        if(cnt == 8) ans++;
    }
    cout << ans << endl;
}