Hello Codeforces community,

I am making a data science project for recommending memes, please fill one form with your preferences which will help in creating the machine learning project. One person should fill only one form.

Form A

Form B

Form C

Form D

Form E

A preview of a meme

Data collection is very very hard and I kindly request you all to please give 3 minutes and checkout the memes in the forms. Thank you all for the kind support.

DM me if you want to know more about the project as I have made the whole project using WALS factorization from scratch

P.S : ONE PERSON SHOULD ONLY FILL ONE FORM

Hi,

I wanted to explain you the method I used to upsolve the problem of [problem:https://codeforces.me/contest/1478/problem/C].

First of all they define a new kind of thing called symmetric array A[] of where there are n distinct positive integers and n additive inverse (multiplied by -1) of same integers. Whenever a problem define a new property or a thing , I find it better to make some specific and general example of it i.e A = {1, 2, 3, 4, -1, -2, -3, -4} or {a, b, c, d, -a, -b, -c, -d}

Second thing they define are D[] where each i-th term D[i] is the sum of (absolute differences between A[i] to all other possible A[j], j in [1,n]). For ex: n = 3, D[0] = |A[0] - A[0]| + |A[0] - (-A[0])| + |A[0] - A[1]| + |A[0] - (-A[1])| + |A[0] - A[2]| + |A[0] - (-A[2])| Similarly for all other D[1], D[2], D[3], D[4], D[5]. It is easy to see that this D value would be same for A[i] and -A[i].

Question is to tell whether its a valid D[] or not ?

I don't like absolute function but it has this property |negative value| = -1*(negative_value) = positive (basically distances between values or locations) Lets call it distance sum for future reference. so I try to visualize all numbers in sorted order since order of A[] doesn't matter I can freely choose to look at them in sorted way (make a 2d graph) so I get an idea which distance is going to have what value as it always equal to (larger value — smaller value) and its easy to see what is bigger and smaller in a picture.

We are working with a generic example of n = 4 A[] = {a, b, c, d, -a, -b, -c, -d}. I assumed a < b < c < d

Now lets choose the smallest one and start calculating the total distances from A to every other number. Now from the picture, it might be easy to see why distance sum value would be same for A[i] and -A[i].

1. Here you will see that distance sum for a is 2*(a + b + c + d) and same for -a. I drew the negative terms right below the positive terms for better clarity

You will notice that distance sum terms are

 2 * (a + b + c + d) , 2 * (b + b + c + d), 2 * (c + c + c + d), 2 * (d + d + d + d),  2 * (a + b + c + d) , 2 * (b + b + c + d), 2 * (c + c + c + d), 2 * (d + d + d + d)

They all are even and distinct (if you ignore the distances sum generated by negatives)

looks great to me which means that the largest distance sum is (d * 4) * 2 and in general let's say we sort all distance sum then D[2n] = 2 * n * A[n] => A[n] = D[2n] / (2n) where (D[2n] % (2n))=0 otherwise A[n] will be in floating points which contradicts that A[] are positive distinct integers.

Let's get back to our convenient example of {a, b, c, d, -a, -b, -c, -d} and distance sums as above. 2nd largest distance sum is 2 * (c + c + c + d) and we know d now so we can easily calculate c = (D[6] - 2 * d) / 2 * (3) but generally A[n - 1] = (D[2n - 2] - 2 * A[n] ) / 2 * (n - 1) Note [check if its perfectly dividing and positive after subtracting.] . (We use D[2n-2] because D[2n-1] is same as D[2n] , but you can use std:set and make implementation easy at log(n) insertion cost)

You can get the whole A[] back like this. Lastly check that they is no zero created as it also contradicts the properties of A[]. If you want to check my code 105777562. Yeah this was a long tutorial but I hope its understandable. If anyone has a suggestion on how to improve the reasoning Please share it. Thank you.

I watched a video of Errichto about stress testing [Youtube link (https://www.youtube.com/watch?v=JXTVOyQpSGM) It is very helpful if you want to stress test your solution but it lacks making random graphs(can be connected or disconnected) so you can check your graph theory solutions before submitting. Does anyone knows how to do it ?