mesanu's blog

Codeforces Round #817 (Div. 4)
By mesanu, 2 years ago, In English

Hello Codeforces!

flamestorm, MikeMirzayanov and I want to invite you to Codeforces Round 817 (Div. 4).

It starts on Aug/30/2022 17:50 (Moscow time).

The format of the event will be identical to Div. 3 rounds:

  • 5-8 tasks;
  • ICPC rules with a penalty of 10 minutes for an incorrect submission;
  • 12-hour phase of open hacks after the end of the round (hacks do not give additional points)
  • after the end of the open hacking phase, all solutions will be tested on the updated set of tests, and the ratings recalculated
  • by default, only "trusted" participants are shown in the results table (but the rating will be recalculated for all with initial ratings less than 1400 or you are an unrated participant/newcomer).

We urge participants whose rating is 1400+ not to register new accounts for the purpose of narcissism but to take part unofficially. Please do not spoil the contest for the official participants.

Only trusted participants of the fourth division will be included in the official standings table. This is a forced measure for combating unsporting behaviour. To qualify as a trusted participant of the fourth division, you must:

  • take part in at least five rated rounds (and solve at least one problem in each of them),
  • do not have a point of 1400 or higher in the rating.

Regardless of whether you are a trusted participant of the fourth division or not, if your rating is less than 1400 (or you are a newcomer/unrated), then the round will be rated for you.

Many thanks to the testers: sandry24, SlavicG, tibinyte, Gheal, qwexd, haochenkang, TimDee.

We suggest reading all of the problems and hope you will find them interesting!

Good Luck!

UPD 1: The contest is delayed by 15 minutes.

UPD 2: The opinions of testers about the order of problems are very contradictory. Please do not rely heavily on the order of problems in the round and read all the problems. Good luck!

UPD 3: Editorial is out!

  • Vote: I like it
  • +172
  • Vote: I do not like it

Comments Comments (145)
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2 years ago, # |
  Vote: I like it +15 Vote: I do not like it

As a tester, I highly recommend everyone to participate <3

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    2 years ago, # ^ |
      Vote: I like it -11 Vote: I do not like it

    As a contestant, I highly recommend you not to test. bad pretest :(

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      2 years ago, # ^ |
      ← Rev. 3   Vote: I like it -21 Vote: I do not like it

      As a participant, I have participated.

      Spoiler for upvotes:
      → Reply
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2 years ago, # |
← Rev. 3   Vote: I like it -106 Vote: I do not like it

Why did you put a Div.4 contest on Tuesday? It would be better if it was on let's say, Sunday because Div.4 contests always have a ton of participants. This way a lot of people won't be able to participate because of their obligations.

And virtual participation is just not the same as live participation because you can see how many people solved each problem till the end of the contest, solutions etc.

I am not writing this because I am mad because I won't be able to participate in the contest (in fact, I will be able to participate) but because a lot of people won't participate because the contest is on Tuesday.

MikeMirzayanov or whoever decides the day of the contest

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    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it +15 Vote: I do not like it

    This is the same as saying "Why is there a div 2 on Thursday?". However, I haven't seen any contest being downvoted just for the day on which it will take place and also a div 4 round may be really good for low rated people to boost their CP confidence through educational problems.

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2 years ago, # |
← Rev. 2   Vote: I like it +9 Vote: I do not like it

Strange. Why all of the commments are downvoted???

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2 years ago, # |
← Rev. 4   Vote: I like it -29 Vote: I do not like it

Expecting some intresting problems

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

As a tester, I don't like sausages.

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    2 years ago, # ^ |
      Vote: I like it -53 Vote: I do not like it

    As a tester, I disrespectfully disagree with your opinion. Sausages could have increased round quality so much.

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    2 years ago, # ^ |
      Vote: I like it -22 Vote: I do not like it

    As a participant (out-of-competition), why?

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2 years ago, # |
  Vote: I like it -30 Vote: I do not like it
Why Top G wasn't mentioned in the blog? I thought he tested the round
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    2 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    they hate a user with the name andrewtate for no reason just because of the name. god these people

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2 years ago, # |
  Vote: I like it -11 Vote: I do not like it

Div 4 arrives... Every Specialist be like: First unrated contest for me!

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2 years ago, # |
  Vote: I like it -22 Vote: I do not like it

My first unrated contest

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2 years ago, # |
  Vote: I like it -12 Vote: I do not like it

:(

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2 years ago, # |
  Vote: I like it -13 Vote: I do not like it

As

Spoiler

a tester, mesanu should add me in this blogpost.

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2 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

What's wrong with the people who are downvoting every single comment? Is everyone supposed to not like these rounds just because you don't? This community is weird

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2 years ago, # |
  Vote: I like it -6 Vote: I do not like it

Particularly liked the part about urging people not to register new accounts solely for narcissistic purposes =)

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2 years ago, # |
  Vote: I like it +7 Vote: I do not like it

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2 years ago, # |
  Vote: I like it +43 Vote: I do not like it

Meme

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2 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Damnn! 5000+ unrated accounts are participating in this contest, isn't a huge number? what do you say?

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2 years ago, # |
← Rev. 2   Vote: I like it +12 Vote: I do not like it

Is the starting time changed?

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2 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Any reason for delay of 15 minutes?

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    This delay is a complete waste of time coz can't do anything good in this delayed time

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      2 years ago, # ^ |
      ← Rev. 3   Vote: I like it -15 Vote: I do not like it

      You can downvote my comment meanwhile

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        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Now most people don't waste their time and started downvote your comment ^^

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2 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Why contest postponed?

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Hope at least today I won't see the grid.

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2 years ago, # |
  Vote: I like it +26 Vote: I do not like it

My first coding competition ever! Kinda excited

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2 years ago, # |
  Vote: I like it +71 Vote: I do not like it

As a tester (I really tested it tonight!) I updated the post with the following text:

The opinions of testers about the order of problems are very contradictory. Please do not rely heavily on the order of problems in the round and read all the problems. Good luck!

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2 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Thank you for this contest!

Favourite problem is E it is simple to understand, but it isn't easy.

Nice Contest!

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2 years ago, # |
  Vote: I like it -34 Vote: I do not like it

Wtf problem F sick implementation?!Stop proposing this kind of problems!

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    2 years ago, # ^ |
      Vote: I like it +33 Vote: I do not like it

    it's not at all a bad problem for Div4. If you find there are lots of if-else, maybe you need to look for better observations.

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

how to solve E?

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    2 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    I don't want to spoil the whole solution ... so I'll give hints since height and width of rectangles are under 1000,

    hint: use a matrix such that matrix[l][r] = Area of all rectangles with height <= l && width <= r.

    Basically a prefix sum matrix.

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2 years ago, # |
  Vote: I like it +12 Vote: I do not like it

Contest was fun and tasks were great. Thank you for this Div4!

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2 years ago, # |
  Vote: I like it +18 Vote: I do not like it

if anyone is interested, I recorded myself doing all problems

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

if only for E it was hs<=hi<=hsb and ws<=wi<=wb, it would have been so much easier, well anyway can someone give me a hint, i stored areas of all rectangles , sorted them , stored prefix sum array, then searched idx_left=upper_bound(hs*ws) and idx_right=lower_bound(hb*wb)-1 on sorted areas, then got the sum with the perfix sum array, can i tweak this to fit the hs<hi<hsb and ws<wi<wb constraints? also brute force gave me TLE X_X

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why this code got runtime error at test 8?

https://codeforces.me/contest/1722/submission/170265006

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    2 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    for each h, can be more than 1000 width

    Runtime Error: This line is not true (line 26)

    vector<vector> pre(1001, vector(1001));

    Solution: You can resize each i in for loop of prefix sum 

    for(int i=1;i<=1000;i++) {
    pre[i].resize(v[i].size());
    for(int j=0;j<v[i].size();j++) {
        if(j == 0) pre[i][j] = i*v[i][j];
        else pre[i][j] = pre[i][j-1] + (i*v[i][j]);
    }
}
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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I found F easier than E. The time limit of 6 seconds on E allowed my O(q*1000*logn) solution to pass.

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

What is the point of $$$\mathrm{TL} = 6 \,\mathrm{s}$$$ in the problem E? To make stupid brute-force solution pass?

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    2 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    not every language is as fast as C++

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2 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

I was so close to solving 1722G - Even-Odd XOR (or I at least think so)! Can anybody please explain to me how to solve it?

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    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Here's my solution: Observation 1: If the xorOdd == xorEven, then xorOdd ^ xorEven == 0, which mean xor of all elements is equal to 0.

    This leads to observation 2: If n%4==0 then the answer is 0 to n-1. Consequentially, answer for n%4==3 is 1 to n.

    if n%4==2 then we take the answer for n-3, then we add 3 numbers of our choice (mine was 2^30, 2^29 and 2^30 + 2^29).

    if n%4==1 then we take answer for n-6, then we add 6 numbers.

    AC code: 170284159

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    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it +2 Vote: I do not like it

    I did it by adding incrementing numbers (from 0) until the array was length n-3. Then add 1 << 29 and 1 << 30. The final value is then the xor of the entire array (proven to be unique as it has both 1<<30 and 1<<29)

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You can check my code if you want

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    2 years ago, # ^ |
      Vote: I like it -10 Vote: I do not like it

    if a == b, then a ^ b == 0
    so you just need to make array xor = 0
    you can make ai = i
    and an = a0^a1...^an
    now the only problem is that an can be equal to other value
    if n is odd you can set 30th bit to 1 in all numbers except an
    if n is even you can set 30th bit to 1 in an and some other ai (if an==a0 choose a1, else a0 for example)

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Consider adding pairs of numbers which in binary are like $$$(x)0$$$ and $$$(x)1$$$, where $$$(x)$$$ are any other bits. Note that the xor will always be equal ignoring the least significant bit. When $$$n \equiv 0 \mod 4$$$, every $$$4$$$ numbers we will have the xor of the least significant bit be $$$0$$$ $$$(0 \oplus 0$$$ and $$$1 \oplus 1)$$$, so we see that the xors are indeed equal.

    We now only need to reduce other cases to this one. You don't even have to think about it because the sample test cases already give you answers for $$$n = 3, 5, 6$$$, so you put those numbers first and now you add the remaining numbers, starting from, say, $$$20, 21, 22, 23...$$$.

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    2 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    If $$$n \% 4 == 0$$$, I gave the even indices values from $$$1$$$ to $$$n/2$$$ and the odd indices the same value except by adding a fixed high power of 2. The XORs on this high bit position cancel to 0 since there are an even number of odd indices (since $$$n$$$ is a multiple of 4).

    For $$$n \% 4 == 1$$$, I did the same for the first $$$n - 1$$$ elements, then added an extra 0.

    For $$$n \% 4 == 3$$$, I did the same for the first $$$n - 3$$$ elements, then added three values in which two of them XOR to the third by utilizing two high powers of 2 ($$$10 \oplus 01 = 11$$$, except these are at some high bit positions)

    For $$$n \% 4 == 2$$$, I did the same for the first $$$n - 6$$$ elements, then added six values on high bit positions that get canceled out. This took considerably time, but I eventually came up with $$$1000 \oplus 0100 \oplus 0010 = 1001 \oplus 0110 \oplus 0001$$$ (these are four bits at some really high bit positions). I'm quite curious to see if there is a more elegant solution for this case.

    170258132

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    We can easily make the XOR equal in a group of 4 numbers (i, i + 1, 2^k — 1 — i, 2^k — 1 — i — 1).

    For example, k = 4: 1 (0001) ^ 14 (1110) = 2 (0010) ^ 13 (1101) = 15 (1111). We match the numbers so that the XOR result has k bits set.

    Let's pick k so that 2^k > 200000.

    What if the number of elements are not divisible by 4? Then we just need to add some numbers at the beginning of the sequence

    • n mod 4 == 1: begin with 0 because x ^ 0 = x
    • n mod 4 == 3: begin with 1 2 3 because 1 ^ 3 = 2

    How about n mod 4 == 2? Since XOR of the rest elements are equal, then these two numbers should be equal too, which means they are not distinct. So starting with 2 numbers is not possible. But we can start with a sequence of 6 numbers. In the contest I found a sequence which is 1 2 4 6 8 9 using brute force.

    Finally, subtract n by the numbers we have added to the sequence, we should have n divisible by 4 to generate the group of 4s.

    The initial value i should be at least 10 to be distinct with the starting numbers.

    Code: 170265768

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Thank you guys.

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You can just set the first n-1 values arbitrary, then the n_th value is the xor sum of the first n-1 values. It's almost impossible to hack.

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

when codeforces post its contest editorial ?

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone explain the error in my submission?170267520

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    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    You are using char array but you are not resetting them after every test case. So if test cases are like this — 

    2
10
abcdetIMUR
5
Timur

    Your code will output No for 2nd test case because strchr(str,"I") will return non null value

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone hack my solution for problem G. I think my answer might generate duplicate numbers. Link

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It won't generate duplicate numbers. Observe that all the numbers except the number you inserted, at last, are <= 2^30. The last number you inserted xor ^ temp, will actually be > 2^30, so it won't be duplicated

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2 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

For the problem D i thought of using prefix sum. I was not able to implement it. What you guys did for problem D?

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    you can use an array of maps 170197935

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    2 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    What made you think to use prefix sum? Anyway, I did it by just storing who wrote which word and then checked how many people wrote it. This can be simply done by using map (I used set aswell, it just made things easier for me)

    Edit: I just realised each person wrote only distinct word xD (but my solution also works for duplicates) , so there is no need for set actually, just store each occurrence

    170203831

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I think D can be done with dp. Here is my solution:

Every next answer is sum of previous answer and max points we can get by changing side of looking of some person. So ans[0] is score that we have in start. Max points we can add is actually biggest difference between points we can get from one person by its current looking and other side looking. So we can just make difference and sort it.

Here is code for my submission: https://codeforces.me/contest/1722/submission/170308742

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2 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

can someone please explain how to do E?hints please?

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Can you make use of h<=1000 and w<=1000 ?

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      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      the only thing that comes in my mind is brute force,i read about about people applying prefix sum but i dont have idea about prefix sum in 2d.

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        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        You can make such that for every h calculate prefix for all possible w

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2 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Why did my code gave WA test 2 on E?

https://codeforces.me/contest/1722/submission/170276269

Edit i saw it =(

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2 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Can someone hack my solution of G?

I randomly fill an array with N-2 numbers and then just add two numbers to make XOR's equal. If I fail (i.e. needed number already in the array), I randomly fill the array again.

https://codeforces.me/contest/1722/submission/170260049

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2 years ago, # |
← Rev. 2   Vote: I like it -8 Vote: I do not like it

.....

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    2 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Hi, no offence, but how did you solve problem G in just 3 minutes?

    I guess it was not that much easy. Even the highest-rated coders took 5-6 minutes as well.

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2 years ago, # |
← Rev. 2   Vote: I like it -11 Vote: I do not like it

Can anyone Explain Problem D? I used two pointer and greedy approach but don't know why is it failing for some test cases?

Here is my code

Code: 

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define loop(i,a,b) for(ll i=a;i<b;i++)
#define range(a)    a.begin(),a.end()
#define p(a)        push_back(a)
#define mp          make_pair 
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define w(x)            ll x; cin>>x; while(x--)
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define mii    map<int, int>    
#define umii   unordered_map<int, int>    
#define mll    map<ll, ll>    
#define umll   unordered_map<ll, ll>
#define pii    pair<int, int>    
#define pl     pair<ll, ll>
#define vi     vector<int>    
#define vl      vector<ll>    
#define si      set<int>    
#define sl      set<ll>    
#define vpii     vector<pii>    
#define vpl      vector<pl>    
#define vvi       vector<vi>    
#define  vvl       vector<vl>
#define fastinout ios_base::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);    
#define MOD 1000000007
ll binExpIter(ll a,ll b,ll mod){long long ans=1;while(b>0){if(b&1){ans=(ans*a)%mod;}a=(a*a)%mod;b>>=1;}return ans;}
ll gcd(ll a, ll b) {if (b > a) {return gcd(b, a);} if (b == 0) {return a;} return gcd(b, a % b);}
ll phin(ll n) {ll number = n; if (n % 2 == 0) {number /= 2; while (n % 2 == 0) n /= 2;} for (ll i = 3; i <= sqrt(n); i += 2) {if (n % i == 0) {while (n % i == 0)n /= i; number = (number / i * (i - 1));}} if (n > 1)number = (number / n * (n - 1)) ; return number;} //O(sqrt(N))
vector<ll> sieve(int n) {int*arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}


//************************************//
int main(){
   w(t){
      ll n;
      cin>>n;
      string s;
      cin>>s;
      int x=0;
      int y=n-1;
      for(int i=1;i<=n;i++){
         int c=0;
         int ans=0;
         while(x<=y){
            if(s[x]=='L'){
               s[x]='R';
               c++;
            }
            if(c==i){
               break;
            }
            if(s[y]=='R'){
               s[y]='L';
               c++;
            }
            if(c==i){
               break;
            }
            x++;
            y--;
         }
         for(int k=0;k<n;k++){
            if(s[k]=='L'){
               ans+=k;
            }
            else{
               ans+=(n-1-k);
            }
         }
         cout<<ans<<" ";
         
      }
      cout<<endl;
   } 
}
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2 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Today is Hacking day

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2 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Is there problems with hacking process, no one unable to hack this user for the same problem and it still in queue state?

Screenshot-2022-08-30-235249 Screenshot-2022-08-30-235323 Screenshot-2022-08-30-235358 Screenshot-2022-08-30-235433

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Too weak pretest for A. :(

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can D done with prefix sum?

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Different people said they solved it like that

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2 years ago, # |
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Hello everyone my solution for 1 problem is accepted during contest but now it's shows its hack and solution is wrong?? What is mean I didn't get. Can you please tell how I got my solution correct.

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It means a few things: 1. The round has open hacking, which continues (as far as I know) for 12 hours after the end of the contest; 2. Every participant(as far as I know) is able to hack anyone's solution. If you find a testcase that might make others' solutions fail, you can submit the testcase and hack their solution; 3. Someone saw your solution, found a testcase for it and hacked it successfully.

    After the hacking phase, all of the submissions will be re-judged with updated testcases, which will include testcases of successful hacks(as far as I know).

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Mize orz!!

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2 years ago, # |
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Can we have rating change quickly please

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2 years ago, # |
← Rev. 2   Vote: I like it -8 Vote: I do not like it

Attention! Your solution 170296922 for the problem 1722G significantly coincides with solutions Gaurav__Arora/170273818, om_arsa_2312/170279944, nihal_gupta/170293955, AhmedEl-Gohary/170296922. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.me/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

So my evidence that I didn't cheat or communicate with any one in the round is that i used a pre-written blog on geeks for geeks about finding an array of size n whose xor value is some k and I adjusted that function to solve yesterday's DIV4 G, you can take a look at my submission and compare for yourself.

https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/

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2 years ago, # |
  Vote: I like it -8 Vote: I do not like it

I have got a system message that my solution coincides with some other participant's solution for the question G, and as asked I am presenting here the common source published before the competition.

https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/

As I had seen this post earlier on geeks for geeks, I simply used it.

Please consider my participation for this contest.

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2 years ago, # |
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I have picked a portion of my code from an article of gfg: (https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/) You can verify the same. Please look into this and consider my participation. MikeMirzayanov

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2 years ago, # |
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Why ratings are still not updated?

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Was this round declared unrated?

I checked quite a few profiles from the comments and this round is marked as unrated for every single of those I checked. Any updates?

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2 years ago, # |
← Rev. 2   Vote: I like it -19 Vote: I do not like it

I have got a system message that my solution coincides with some other participant's solution for the question G, and as asked I am presenting here the common source published before the competition.

https://www.geeksforgeeks.org/program-binary-decimal-conversion/ Also used previosuly posted gfg function of binary to decimal conversion. As I had seen this post earlier on geeks for geeks, I simply used it.

Please consider my participation for this contest. Thank You

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2 years ago, # |
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Hello I've got a system message that my submission to problem G coincides with some participants, but I've used the code present on GFG link to the GFG article Kindly consider my participation in this round. Thank You.

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I have got a system message that my solution coincides with some other participant's solution for the question G, and as asked I am presenting here the common source published before the competition.

https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/

I have even put the link above in my submission. So please revert the decision.

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2 years ago, # |
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why is the rating update taking so long?

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2 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Your solution 170295384 for the problem 1722G significantly coincides with solutions Jeopardy_007/170181586, kshitijsabale/170235981, pravin_as/170236185, sksusha8853/170244580, daijoub_dattebayo/170253614, Ag.akhand29Dec/170260220, vipin/170268963, coomlhamdle/170268986, no-one/170270439, pritish_001/170280715, vamshikrishna7697/170283391, saranshgoel_20/170287297, O_QufaD/170287888, noozy/170289556, noob5367/170290799, anurag78_20/170293266, CodeR_SaaD/170294427, singham_20/170294546, anupam_singh20/170295384, blablablacksheep/170295836, abhirai24/170297702, xorhero_02/170299463. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.me/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

Hello! This question had a similar logic as https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/, which means the code was published before the round, so I thought to use it. I think that the coincidence has occurred due to the use of a common source published before the competition. I don't even know any of the person's mentioned above. Is it a violation of rules, if it isn't then I request you to please look into this and resolve this. Thank You

@MikeMirzayanov Why only I got the penalty, and not everyone, while no one else got it Only my rating became negative

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2 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Me looking at my code for F after getting it accepted. (170379863)

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    2 years ago, # ^ |
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    what is the idea behind your solution?

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      21 month(s) ago, # ^ |
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      Actually, I am checking out that, is there any connected component of size which is not equal to 3 (here two nodes are connected if they share a side or a corner). if so, then just print NO. Otherwise, I am viewing all the components which are connected and checking that whether they are L-shape or not.

      Slr, Actually I replied the answer to your question to the person below and just saw it today =(

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2 years ago, # |
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