The Moroccan association for computer machinery MACM is getting ready to host the 5th edition of one of its most special events: Girls Code.

An event that aims to celebrate all the girls that are passionate about the world and the universe of computer programming and coding.

This event was normally supposed to take place in the previous month of March. However, due to the Coronavirus outbreak and the state of emergency in the country, it was decided that the event will be postponed to the 2nd of May.

Due to the same reasons stated earlier, the event will be taking place on an online platform, The MOI ARENA.

During the first week only, more than 250 candidates registered to participate in the competition.

Although all the obstacles that face the organization of such events, the Moroccan Association for computer machinery showed huge potential in adapting to new situations and overcoming obstacles. Girls and young ladies who are in love with computer programming will get a chance to show their talent and potential, from their houses, while completely respecting the regulations of the state of emergency.

The Moroccan association for computer machinery knows no limits and believes that talent and passion should be always celebrated.

The concept of the competition is as follow: each of the participants will have to deal individually with a number of programming problems that she will have to solve using one of these programming languages: C, C++, Java or Python.

The competition runs for hours and more than 30 volunteers work hand in hand for the event to run smoothly, and to create a better environment for participants.

For more information please visit the FaceBook page of the Association

Welcome to MOI Arena

print "Hello, to junior MOIers !"

La première épreuve de code arrive ! Rendez-vous ce Samedi 25 avril à 15:00 !

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Welcome to MOI Arena

print "Hello, to NCC!"

The first edition of the North Coding Contest (NCC 2020) is going to be held on Saturday, April, 11 at 20:00 Morocco time.

You will be given between 8 to 10 problems and 4 hours to solve them.

The contest will follow ICPC Rules : The penalty for each incorrect submission until the submission with a full solution is 10 minutes.

This contest will be RATED

More info at ECDH website

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Welcome to MOI Arena

print "Hello, Arena!"

We will be organizing the Arena round #7 on Friday, March 27th 2020 from 20:00 to 22:30 in Morocco time.

Problems are tenderly prepared and tested by :

• Hamza Mouhcine (BNL)

• Mohamed Elkhatri (medk)

We wish you a higher rating !

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Welcome to MOI Arena

print "Hello, Arena!"

We will be organizing the Arena round #5 on Saturday, January 18th 2020 from 20:00 to 22:00 in Morocco time.

Problems are tenderly prepared and tested by :

• Hamza Mouhcine (BNL)

• Aymane Lotfi

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Let \(Li\) the expected loss before choosing the next distinct number, given that \(i\) distinct numbers were chosen already. The answer would be \(An = \sum_{i=0}^{n-1}1-Li=n-\sum_{i=0}^{n-1}Li\)

For some \(i\) let's calculate \(Li\) :

In order to be able to perform a \(j^{th}\) pick while still having \(i\) distinct numbers chosen, all previous \(j-1\) picks should've resulted in a loss (one of the \(i\) already chosen numbers is picked again), meaning that the probability that the \(j^{th}\) pick results in a loss is \(P_j=(\frac{i}{n})^{j}\).

The expected amount to be lost in the \(j^{th}\) attempt is \((\frac{i}{n})^{j}\).

\(L_i\) is the sum of expected losses over all possible \(j^{th}\) attempts, thus \(L_i=\sum_{j=1}^{\infty} (\frac{i}{n})^{j}\).

Since that is just the sum of terms of a geometric series we can rewrite \(Li\) as : \(Li = \frac{i}{n}*\frac{1-(\frac{i}{n})^\infty}{1-\frac{i}{n}} = \frac{i}{n-i}\).

We get \(An=n-\sum_{i=0}^{n-1}\frac{i}{n-i}\).

Calculating the sum takes \(O(n)\) time, but we clearly have to answer each query in \(O(1)\) if we look at the constraints, so we should calculate all \(Ak\) and store them beforehand. To do that we find a recursive formula as follows:

\(A_{n+1} - A_{n} = 1 + \sum_{i=0}^{n-1}\frac{i}{n-i} - \sum_{i=1}^{n}\frac{i}{n-(i-1)} = 1 + \sum_{i=0}^{n-1}\frac{i}{n-i} - \sum_{i=0}^{n-1}\frac{i+1}{n-i} = 1 - \sum_{i=0}^{n-1}\frac{1}{n-i} = 1 - \sum_{i=1}^{n}\frac{1}{i}\).

We finally get : \(A_{n+1} = A_n+1 - \sum_{i=1}^{n}\frac{1}{i}\).

Since we know that \(A_1=1\) we can calculate all \(A_k\) in \(O(n)\) time and answer queries in \(O(1)\).

Welcome to MOI Arena

print "Hello, Arena!"

We will be organizing the first rated round of 2020 on Friday, January 3rd 2020 from 20:00 to 22:00 in Morocco time.

Problems are tenderly prepared and tested by :

• Omar Salim Moussa (osm97)

• Nabil Sahifa (nabil_shf)

• And myself (moncef)

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