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 <iostream>
#include <vector>
#define REP(i,n) for(int i=0;i<n;i++)
using namespace std;
int main(){
  int n;
  cin>>n;
  int d[]={2,3,5,7,11,13};
  vector<int> e(6,0);
  int cnt;
  int ans=0;
  REP(i,n){
    if((i+1)%2==0){
      continue;
    }
    int origin = i+1;
    REP(j,6){
      while(origin%d[j]==0){
        origin /= d[j];
        e[j]++;
      }
    }
    REP(j,6){
      cnt *= e[j]+1;
    }
    if(cnt==8){
      ans++;
    }
    REP(j,6){
      e[j]=0;
    }
    cnt=1;
  }
  
  cout<<ans<<endl;
  return 0;
}