tourist vs Benq Round

Tutorial of D is terribly explained. Worst.

Thanks

Can you please explain why is it a 0-1 knapsack.
Where are weights and capacities etc. which are parts of knapsack.
I can't understand.
Also, If possible suggest some resources(problems) to learn these types of implicit knapsack.

Worst Tutorial. Instead of giving explanations, the author has written 'This can be done', 'This is well known',etc.etc.

Actually I thought that the code was of this contest, but that was of some past contest. I thought that strings are unnecessary (due to my Assumption(wrong) that code was of this contest) even without looking at full code. Sorry Bro once again. Actually since there are a lot of cheaters, I get so angry at cheaters.

But sorry you are NOT a cheater. Good luck

Extremely sorry bro, I mistakenly thought you cheated. Sorry please forgive me. Sorry Again.

I can see properly. I can clearly see your color is RED.

Sorry, Actually I went to saw Sol of D in his profile (submissions) then I saw this soln. I did not realize that the soln was od another contest and presumed that he had done it to pass plagrasims. I am extremely sorry to have created this confusion.

Hey Cheater, why have you inserted unnecessarily string/char etc. in your code of D.

Hello cheater pranav2401, you started cheating from first contest!!!

There are lots of cheaters.

Congrats!!!

5

You're expert and solved only 3 problems of this contest. What a joke. Hahahahahahahaaa!!!

I don't understand your approach. Can you elaborate more?

The solution of C is poorly written. "Then, the sequence of numbers will have the form: (ai,bi,ai−bi,ai−2⋅bi,bi,ai−3⋅bi,ai−4⋅bi,bi,…) ." is wrong take ai = 5, bi = 3: sequence will be :5,3,2,1,1,0 clearly bi(3) is not appearing again and again as written int tutorial.

Proof of C is poorly written.

Here is a checker function.

bool check(vector <int> &arr,int x,int a,int b){
    int n = arr.size();
    int hh=0;
    for(int i=0;i<n;i++){
        int y = arr[i];
        int s=0;
        if(y>x){
            hh -= (y-x)/b;
        }
        else{
            hh += (x-y)/a;
            if((x-y)%a != 0){
                hh++;
            }
        }
    }           
    if(hh<=0) return true;
    else return false;
}