MUL_OR_DIV EDITORIAL

editorial
,
1

PROBLEM LINK

** PREREQUISITES**
OBSERVATION, MATH

EXPLANATION:
Our goal is to maximize the divide by 3 operations, so in each operation, we choose only one number which is a multiple of 3, and divide that number by 3, if no number which is a multiple of 3 is found we discard and stop there and print total operations, so it simply means the sum of log3(value) of each value which is a multiple of 3.

CODE:

#include<bits/stdc++.h>
using namespace std;
int fun(int n)
{
  int cnt = 0;
  while (n % 3 == 0)
  {
    cnt++;
    n /= 3;
  }
  return cnt;
}
void solve()
{
  int n; cin >> n;
  int ans=0;
  for(int i=0;i<n;i++)
  {
      int x;
      cin>>x;
      ans+=fun(x);
  }
  cout << ans<<"\n";
}

int main()
{
  int t = 1;
  cin >> t;

  for (int tt = 1; tt <= t; tt++)
  {
    solve();
  }

}