AtCoder Beginner Contest 139 Announcement

We will hold AtCoder Beginner Contest 139.

• Contest URL: https://atcoder.jp/contests/abc139
• Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20190901T2100&p1=248
• Duration: 100 minutes
• Number of Tasks: 6
• Writer: drafear, E869120, evima, gazelle, square1001, wo01, yuma000
• Rated range: ~ 1999

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

• 3 years ago, # ^ | No one gives a fuck about Chinese football matches.

• 3 years ago, # ^ | Seems that you're not very experienced in football, Chinese football matches aren't worth watching...

  • 3 years ago, # ^ | i don't think so, It's just that it's not as fascinating as it is in Europe

• 3 years ago, # ^ | The comment is hidden because of too negative feedback, click here to view it

• 3 years ago, # ^ | ← Rev. 3 →   0 LOL XD look at the down votes You can always watch the football match later after the game gets over

• 3 years ago, # ^ | It took me around 40 seconds to read and code problem A but only because the website was slow, I was able to submit at 3:47. So sad :(

• 3 years ago, # ^ | Indeed :))

• 3 years ago, # ^ | It is just we need to find the minimum number of power strips is needed, A is the number of sockets in each of them. By logic, the next power circuit is connected with the previous one, to get power.

• 3 years ago, # ^ | I think to publish it after the contest would be better.

• 3 years ago, # ^ | ← Rev. 2 →   +36 As a writer of this contest, please refrain to write some hints about the problem during the contest. The fact like "this problem can be solved with DP" or "this contest is already published 3 years ago" should be written after the contest. It might cause BAN or some bad thing.

• • 3 years ago, # ^ | Can you explain your method with more details?

• 3 years ago, # ^ | Maybe a connection or cache issue, I had no problem with atcoder and mozilla rather than at the begin of the contest.

• 3 years ago, # ^ | ← Rev. 2 →   +43 Not really, nowadays it's almost impossible to know all the problems from all platforms. In Poland, we're spending some time trying to find similar problems using Google. If we cannot find anything, then the task is fine.

• 3 years ago, # ^ | here is my code for D #include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; int a[n]; int x; if(n%2!=0){ x=n*(n-1)/2;} else x=(n/2)*(n-1); cout<<x; } can anyone identify the mistake? will be great help.

  • 3 years ago, # ^ | As x is of int data type, it would hold a max value of INT_MAX ( about 2*10^9) . Hence for bigger values of n, overflow would lead to WA.

• 3 years ago, # ^ | Thank you for your nice editorials and solution. Please continue this or someone should do this.

• 3 years ago, # ^ | For C, another approach is to let dp[i] be the number of moves you can achieve if you start from i.

• 3 years ago, # ^ | I'm not so lucky like you. QWQ I also used randomize + greedy, and I got WA.

• 3 years ago, # ^ | Just maybe your sorting function is so good. I actually generated testcase which naive hill climbing algorithm may fail. I think it's the case of chenxiaoyan. However, I expected that simulated annealing algorithm may pass. Within this constraints, I don't think that there exists a killer case for SA algorithm. SA is so powerful that even solves a problem in AGC.

• 3 years ago, # ^ | when b=1 ans should be 0

  • 3 years ago, # ^ | it gives 0 whenever B is 1

    • 3 years ago, # ^ | then check for a=2 and let b=5

    • 3 years ago, # ^ | check for a=2

• 3 years ago, # ^ | Nevermind, It worked, messed up the cycle check back then :( Check out my solution if anyone wants : https://atcoder.jp/contests/abc139/submissions/7293559

• 3 years ago, # ^ | Can you please explain the logic behind your approach, thank in you advance

  • 3 years ago, # ^ | First after we make the directed graph, and check the absence of cycles in it, to find minimum days, you must understand that the graph will be overlaps of some directed chains, and if you take the length of largest such chain, there's no other chain larger than it, and hence in those many days the rest of the chains can be allocated a day easily because at any point of intersection the topologically higher part will be contributing iff it is the longest subchain compared to the other chains and hence day assignment won't be a problem. Although I believe such is the solution, I cannot completely confirm it for any edgy test cases.

    • 3 years ago, # ^ | rohit_goyal, Thank you for your explanation ... I got some questions. How are you storing the graph (as adj lists or matrix) ? Can you send some links (tutorials, editorials and questions related to this problem) ?

      • 3 years ago, # ^ | I am storing the graph as adjacency list (instead of using pair as an element in list, I use compressed index for denoting pair (l, r) ). The only tutorial I can provide you is about how to check for cycle in graph : Link But if you want to check out another approach or more detailed solutions, check out this link below: Link

• 3 years ago, # ^ | From what I could comprehend, after you apply a socket wire ( or whatever it was called, having parameter A), the old socket is converted to a multi — socket of size A. So how many minimum we need to apply to at least get B sockets? You can refer the image for A = 3 and transitions: Link : https://ibb.co/bXcss0W See the numbererd nodes?They are our sockets after adding each wire at each step of size A.

  • 3 years ago, # ^ | now i understand clearly

• 3 years ago, # ^ | Initially, you have only 1 plug which is open. You want "B" open plugs. You have only "A" sized boards (which means each board has A open plugs.) Target is to find the minimum number of "A" sized boards required to get "B" open plugs. The key idea is Whenever you use a brand new board, you will have to use an empty plug from previous to connect this. So essentially whenever you add a board you add "A-1" open plugs and not "A". Just simulate this by adding boards until you reach B or more.

• 3 years ago, # ^ | Useful conclusion: a convex hull on integral points with and coordinates both in range contains at most points. In this problem, the maximum absolute value you can reach is , so at most points. Thus, even if you brute-force the convex hull after each vector, the total complexity will be (Let the size of the hull be ), which can fit the time limit. Of course this won't work for . In fact, we can scan the vectors with a half plane. Update the answer when their is a vector go into/out of the half plane. The total complexity is (imo). Maybe you can share your convex hull code here, because I didn't implement it :D

  • 3 years ago, # ^ | I think the convex hull on integral points with and coordinates both in range contains at most points, not points. That's because there are pair of which is and .

    • 3 years ago, # ^ | Can you link some proof regarding this claim, I am not able to find on google..

  • 3 years ago, # ^ | That is a very useful conclusion! While the result seems intuitive, is there a constructive proof? Here's my code from the contest: https://atcoder.jp/contests/abc139/submissions/7268385

• 3 years ago, # ^ | Wow, grats for actually thinking of this in contest fairly quickly. I saw it and immediately thought of https://codeforces.com/contest/995/problem/C, remembering that even though there was an elegant solution, randomization + greedy heuristic solved all cases and was hard to break. Turns out same was true for this one.

• 3 years ago, # ^ | Let's create a directed graph. Where each node is a match played between two players. Now let's add edges to this digraph. For an ith person's required order we add a directed each for each adjacent positions. That is for j in range (1..n-1) add directed edge from (j-1) to j. Also since match played between a&b and b&a is similar we merge these edges into one. Now we have a directed graph. We just check the topological ordering of this graph. If it's cyclic then we don't have an answer. Now to find the minimum days: For a node who has not yet finished (is still waiting on some nodes to finish), this node can only finish when the max days taking node it is still waiting for finishes and then it takes 1 more extra day to finish. Obviously, the nodes which don't wait for others to finish end on day 1. For other nodes, we update them using the previously done. And maintain maximum throughout.

  • 3 years ago, # ^ | What I have done is create a vector of set where each set in a vector represents the team that played on that particular day and at last output the size of vector (the number of days). How I did it? Iterate through all the existing sets of vectors, if neither of the two element(i and Aij) are present in that set, add both the elements to that set and if none of the set satifies this condition then add a new set to vector. Also, I have kept a value = temp1 which is equal to the kth vector(the vector in which last pasir of elements were added), this also helps to know if any two cases are contradicting.Can someone please help me out in finding error as I am getting WA in some test cases. Here is my solution:- https://atcoder.jp/contests/abc139/submissions/7296021

• 3 years ago, # ^ | In E : first observation was to see, among all the possible matches , which would be the first match . if we say the i th player played the first match with the j th player. then in the column of i , j would be the first element and in the column of j , i would be the first element . Now since on day 1, i th and j th player has played , so they can't play more matches on day 1. so we try to find some more pairs for day 1, if they satisfy the above criteria . we can do this while traversing the first column. after finding all the pairs which played on day 1 , we can start to find pairs who played on day 2. for finding the pairs what i did was to initialize n pointers to 1. if some pair satisfy the criteria , I incremented their respective pointer by 1.when we find no pair , we break out of the loop. at the end, if all pointers are at n , which means all have played , we simply print the number of days , else we print -1 . Hope it helps. for more clarity, you can see my code : Link

  • 3 years ago, # ^ | What is the time complexity of your solution? To me it looks O(n^3) in case where only one match is played in each day because your while loop will run n^2 and inner loop i O(n) or am I wrong?

    • 3 years ago, # ^ | Yup ,you are absolutely correct. It is my bad that i didn't calculate the worst time complexity of my solution. On a little thought , it can be optimised by using containers like map or set where finding such a pair is relatively easy.

• 3 years ago, # ^ | well, an well known problem appearing in any contest doesn't destroy the fun of problem solving as long as the problem is elegant/crafty, right? if someone knows the solution beforehand they can just solve it quickly and go rewatch some rick and mortys. On the other hand, it might turn out to be a wonderful contest with cool problems :)

• 3 years ago, # ^ | ← Rev. 2 →   0 Assuming you are asking 0-indexed numbering: First of all taking minimum and maximum is to avoid distinction between pairs like (1,2) and (2,1) or you can say we need unordered pairs in the problem. Secondly the numbering is one of possible numbers you can take. This numbering is just indicating, say in a grid, you are at (1,5) and grid size is 6*6, so the assigning numbers from first cell to the required cell, starting with 0,you get number 1*6 + 5(only for 0 based indexing ), so think it like this, at (i, j) cell, you need to move n*i number of cells along rows and then the jth cell is your requirement, hence the number n*i+j.

  • 3 years ago, # ^ | Thank u :D

    • 3 years ago, # ^ | Anytime mate!

• 3 years ago, # ^ | you guys must have some kind of magic wand!

• 3 years ago, # ^ | ← Rev. 2 →   0 Edit: Found the bug , was printing the last cnt instead of max value Got AC. Can you help me finding what is wrong with my solution? #include<bits/stdc++.h> using namespace std; int main() { long long n,i; cin>>n; long long int ara[n+2]; for(i=0;i<n;i++)cin>>ara[i]; long long cnt=0,mval=0,l=n-1; for(i=0;i<l;i++){ if(ara[i+1]>ara[i])cnt=0; else cnt++; if(cnt>mval)mval=cnt; } cout<<cnt; }

  • 3 years ago, # ^ | Congratulation!