I was trying to solve this question. After trying enough i finally read the tutorial. Suppose the minimum number, is a[0]=600. Then we need to calculate how many elements of size 2*600-1=1999 can we form i.e. we simply need to divide each of the a[i] by 1999 and add each of them. But this code

include <bits/stdc++.h>

using namespace std;

define MULTI int _T; cin >> _T; while(_T--)

typedef long long ll; typedef pair<int, int> ii; typedef vector vi; typedef vector vii;

int main() { ios::sync_with_stdio(false); cin.tie(0);

MULTI
{
  int n;
  cin>>n;
  vector<int> a(n);
  long long sum = 0;
  for(int i=0;i<n;++i)
  {
    cin>>a[i];
  }
  for(int i = 0;i<n;++i)
  {
    sum+=(a[i])/(2*a[0]-1);
  }
  cout<<sum<<endl;
}

} is giving .