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 <cmath>
#include <algorithm>

using namespace std;

int N;

int division_count(int n){
    int cnt = 0;

    for(int j = 1; j <= sqrt(n); j++){
        if(n % j == 0) cnt++;
    }
    if(n > 1 && n == pow(sqrt(n),2)) cnt--; //n = 1の時に注意(1は平方数ではない)

    return cnt;
}

int solve(){
    int ans = 0;
    for(int i = 1; i <= N; i++)
        if(i % 2 == 1 && division_count(i) == 8) ans++;
    return ans;
}

int main(){
    cin >> N;
    // printf("%d\n",division_count(N));
    cout << solve() << endl;
}