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By flamestorm, 3 months ago, In English

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1722A - Spell Check

Idea: mesanu, MikeMirzayanov

Tutorial
Solution

1722B - Colourblindness

Idea: flamestorm

Tutorial
Solution

1722C - Word Game

Idea: flamestorm

Tutorial
Solution

1722D - Line

Idea: flamestorm

Tutorial
Solution

1722E - Counting Rectangles

Idea: mesanu

Tutorial
Solution

1722F - L-shapes

Idea: MikeMirzayanov

Tutorial
Solution

1722G - Even-Odd XOR

Idea: mesanu

Tutorial
Alternate Tutorial Sketch
Solution
 
 
 
 
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3 months ago, # |
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I didn't even think that Timur is lexicographically in order. What a brilliant solution!

Edit: It isn't, but it's still an interesting solution to think of.

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3 months ago, # |
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Problem D can also be solved in O(n) using two pointers

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    3 months ago, # ^ |
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    exactly

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    3 months ago, # ^ |
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    Can you give your solution here it would help

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      3 months ago, # ^ |
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      here if you need explanation feel free to ask

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        3 months ago, # ^ |
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        Hello, I saw your solution but I didn't understand this line . What is it for > for (k;k<=n;k++) ans[k]=s;

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          3 months ago, # ^ |
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          let's suppose that it was already optimal like RRLL so in the line you mentioned it just sets the remaining indexes (which could not be covered in while loop cuz of being optimal) they will just get the previous value as we will not be changing them...

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          3 months ago, # ^ |
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          Let's say the number of characters I needes to change was 8 and n is 10. I need go give an answer for all k such that 1<=k<=n so I just output for the remaining k

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          3 months ago, # ^ |
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          Check out my submission for problem D it might help you understand you the two pointer approach bettetb

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        3 months ago, # ^ |
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        Hi, I couldn't understand why you changed both the front and back pointer simultaneously in the while loop.

        Edit -> Nevermind, got it. Thanks.

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    3 months ago, # ^ |
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    I solved it in O(n) aswell, but using prefix array

    170271299

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      3 months ago, # ^ |
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      Can u pls explain,how did u do it using Prefix arrays

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        3 months ago, # ^ |
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        Take a look at the comments in the submission, if that doesn't help let me know once again.

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          3 months ago, # ^ |
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          Still i didn't get .M new to prefix array

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            3 months ago, # ^ |
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            You can have a look at others two-pointer solution. It's almost similar.

            Explanation
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    3 months ago, # ^ |
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    Yes, I solved it in the same way.

    here

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    3 months ago, # ^ |
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    What is two pointer approach... Can you share some resources from where I should learn... It would be a great help

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      3 months ago, # ^ |
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      Check Codeforces edu section. You can also read about it on USACO Guide

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      3 months ago, # ^ |
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      Two pointer is basically using two indices to traverse through an array on the basis of some condition that you can apply on both the pointers(pointing to a particular index), like in the D part of the question we take one pointer at very start and one pointer at very end of the array and then we can check if we got a left viewer with first pointer then changing it to R will give a better score till n/2-1 and checking the same thing for the right pointer for the right side viewer till n-n/2, we can do this till left pointer is less than right pointer.

      To understand two pointers you should try solving problems for it instead reading about it, this link will work for basic problems Two pointers and problems on it

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    3 months ago, # ^ |
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    i do it with queue :)

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      3 months ago, # ^ |
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      It can be done too, because there indices that must be changed before others right?

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      3 months ago, # ^ |
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      Super Cool Bro ,How did that idea come about?

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    2 months ago, # ^ |
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Why is problem A $$$O(n)$$$? You check if $$$n = 5$$$ in $$$O(1)$$$; If it is, then sort the string which in this case will always have length $$$n = 5$$$ so sorting and checking is done in $$$O(1)$$$ (since the length is upper bounded). So overall $$$O(1)$$$, or am I missing something?

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    3 months ago, # ^ |
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    I think that's true. Change it, flamestorm.

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      3 months ago, # ^ |
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      You check if "T" is in the string, that can be done in $$$O(1)$$$ (since it only has 5 elements). Then you check if "i", also $$$O(1)$$$ And so forth. In total $$$5 \cdot O(1) = O(1)$$$

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problem G:

topic 1

Regardless of your correct answer, when you choose some elements and take their XOR, then the value will be equal to the XOR of the remaining elements. Why?
Sample: $$$(4,2,1,5,0,6,7,3)$$$ is one of answer for $$$n=8$$$, and $$$4 \oplus 0 \oplus 3 = 2 \oplus 1 \oplus 5 \oplus 6 \oplus 7 = 7$$$

Answer
topic 2

There exists a randomized solution.

Explanation
topic 3

For $$$3 \le n \le 6$$$, the answers are provided in the sample, so let's use them as a prefix.

  • $$$n \equiv 0 \mod 4$$$ then prefix = $$$(2,1,3,0)$$$
  • $$$n \equiv 1 \mod 4$$$ then prefix = $$$(2,0,4,5,3)$$$
  • $$$n \equiv 2 \mod 4$$$ then prefix = $$$(4,1,2,12,3,8)$$$
  • $$$n \equiv 3 \mod 4$$$ then prefix = $$$(2,1,3)$$$

After that, add $$$(100,101,102,103,\dots)$$$ until the length of the sequence becomes $$$n$$$. (note that the length of added sequence is a multiple of $$$4$$$ ) Then, you can get a correct answer.
Submission: 170321109

Why?
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    3 months ago, # ^ |
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    master piece. i tried to construct a solution similar to topic 3, but unfortunately fell into tons of case work.

    however, instead we can take advantage of samples, XD

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    3 months ago, # ^ |
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    thanks a lot!!

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    3 months ago, # ^ |
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    Thanks alot. Can you share the proof of topic 3 ?

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      3 months ago, # ^ |
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      proof with an example
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    3 months ago, # ^ |
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    A1⊕A3⊕⋯=A2⊕A4⊕… ↔A1⊕A2⊕A3⊕A4⊕⋯=0

    can you swap randomly and still have the equality?

    if yes, can you do the same in

    X ^ Y < A ^ B -> x ^ B < A ^ Y ?

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The latex for $$$2^{30}$$$ is bugged in the tutorial of problem G.

Aside from that, problem G is a really cool problem!

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For anyone who is finding the solution of problem F, hard to understand/code, can take a look at my solution using DFS 170297341

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D can be solved using two pointers.

The main idea is that the first half of the line should look to the right and the second half to the left. We keep two pointers l and r on both ends and recalculate the sum whenever we encounter a position that needs to be changed.

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Can we solve problem E using binary search?

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Good contest, got sidetracked by D by trying to use two pointers. Your solution is much more elegant.

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Weak pretests for A , so many solutions got hacked.

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How to write the code for Problem G with the alternate tutorial, I am unable to understand the approach

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Simple implementation for problem G

void solve()
{
    int n;
    cin >> n;

    int xorr = 0;
    vector<int> ans;
    for (int i = 1; i <= n - 2; ++i)
    {
        ans.push_back(i);
        xorr ^= i;
    }
    if (xorr != 0)
    {
        ans.push_back(1 << 30);
        ans.push_back((xorr) ^ (1 << 30));
    }
    else
    {
        /*

        The maximum xor of all the elements from 1 to n-2 or n-3 will be less than 2**21.
        (So, you can choose any power of 2 from [21, 30) and replace n-2 with it)
        If we have XOR([1...n-2]) as zero, we can remove (n-2) and can push (1<<21)
        And now the XOR([1....n-3, (1<<21)]) will be some number having the 21st as set

        Now, we just need two more nos, that can be (1<<30) and XOR([1...n-3, (1<<21)])) ^ (1<<30)

        */
        xorr ^= (n - 2);
        ans.pop_back();
        ans.push_back(1 << 21);
        xorr ^= (1 << 21);
        ans.push_back(1 << 30);
        ans.push_back(xorr ^ (1 << 30));
    }

    for (auto &i : ans)
    {
        cout << i << " ";
    }

    cout << nline;
}
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Problem F can be done using

Surprise !

:D

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3 months ago, # |
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I had school so I wasn't able to do the contest, however I've solved them all and I present my solutions. They may or may not be worse quality than the official editorial.

A
B
C
D
E
F
G
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    3 months ago, # ^ |
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    Good!

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    3 months ago, # ^ |
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    Can you please tell me where my submission 170342106 for problem E is failing?

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      3 months ago, # ^ |
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      I think your code is quite overcomplicated

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    3 months ago, # ^ |
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    tgp07 Excellent editorial, better than official one! Could you please tell me if there is a way to solve E under the same time constraints for tighter dimensions of the rectangle say 1e5 or 1e9 or even higher?

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      3 months ago, # ^ |
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      That would be a lot tougher. I'm not immediately sure of any way to solve that.

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    3 months ago, # ^ |
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    your G one is really good Thanks for sharing.

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    I think an improvement for F would be to extend adjacency to include diagonals, i.e., each cell has up to 8 neighbors for the purposes of DFS. Then all we need to do is make sure each connected component has size 3 and forms an L shape.

    This way, there is no need to check if different blocks touch by edge or corner, since the DFS would've merged such cases into larger components.

    (also, checking whether three cells form an L-shape can be done by simply checking if, for each of the three pairs of cells, the difference between the x- and y-coordinates is at most 1)

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      3 months ago, # ^ |
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      This seems smart, have you submitted a solution with this approach?

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        3 months ago, # ^ |
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        I had a partially written code, but the contest ended and I didn't bother finishing it up. But your comment encouraged me to complete and submit it, so here it is: 170513958!

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Why I am getting runtime error in problem C .Here is my solution

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My first contest solving more than 1 problem :) I was excited to realize I could use a max heap to solve problem D.

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Can anyone tell me a testcase where my submission 170342106 is failing for problem E?

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What's hack ??! In the verdict of my submission it's written as hacked but earlier it was accepted any one care to explain what this hack is ??!

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    Some of the corner test cases are not added while you submit the solution. And if you have not handled those cases and someone else break your code for any of those test cases means your solution is hacked .

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Damn. I was not getting what was going wrong in the first question. Used a if condition, which updates a count and individually checked the letters with T i m u r. But, it gave me wrong results. If i had not lost time on this, 5 questions were quite easy to do

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When will my rating appear, yesterday was my 1st contest.

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    3 months ago, # ^ |
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    For div2, preliminary ratings are updated in 2-3 hours. For education rounds/Div 3/Div 4, we have an open hacking phase which lasts for 12 hours so yeah, expect waiting for atleast 18 hours.

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      3 months ago, # ^ |
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      Ok

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      3 months ago, # ^ |
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      bro I am new here.Can you tell me what is this open hacking phase?

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        3 months ago, # ^ |
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        In div3/4 and Educational round, we have an open hacking phase for 12 hours. This means you can view others' code and find a test case where their code fails. If you hack their solution, the system will use this test case provided by you after these 12 hours during the system test (after which we get our final result, I hope you know the difference between system test and pretests). Hacking others won't give you extra points in the Edu round or Div 3/4. The only purpose of Open hacking phase is to promote hacking others' solutions and to understand others' writing styles. Hacking others will give you extra points in Div 2/1. To know more in detail, please use the codeforces search box. Searching is a good habit when on the path to becoming a competitive programmer. All the best!

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      3 months ago, # ^ |
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      usually div. 3 gets preliminary rating updates by now right?

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        3 months ago, # ^ |
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        Max to Max 5PM IST (from what I have experienced) This is worst case.

        Edit: Looks like we have a new worst

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Hey!, can someone help me up solving question F,

My logic for the solution was that for each * there can be only and only 2 '*' neighbors in all 8 directions(if a cell exists there) no more, no less than that.

But it fails on test no. 4, 69th ticket. can someone help me find a test case where this fails. WA link — https://codeforces.com/contest/1722/submission/170290140

Any help would be very appreciated.

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    Check out my solution maybe it’ll help you

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    1
    3 3
    .*.
    *.*
    .*.

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      Thanks a lot, how do you come up with such cases, like how do you think of cases where our code might fail, how to develop that? any tips on that would be very helpful.

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        3 months ago, # ^ |
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        If it's after the contest, you can simply view the test cases in your submission. Often the input would be truncated, but you can often still find a way to reveal it. Modify your code by adding conditions to specifically check for the problematic test case, and then use it to print the result.

        In this case, for example, you can see that in Test #4, the first case has n = m = 3, unlike the earlier Test Suites. So you can write your program such that if the first test case begins with n = m = 3 (you can use scanf for this, if desynced with cin), then you call a different function that reads the first 68 cases while printing absolutely nothing (not even the answers), and then reads the 69th case to print the complete test case details. When submitting, this should pass Tests #1-3, and then print the problematic test case in #4 for you to view.

        This can be cumbersome sometimes, but at least it's a generally definite approach to finding the test case that breaks your program. I haven't yet encountered a situation where I was unable to expose the problematic test case this way.

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          3 months ago, # ^ |
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          That's an awesome way to find a test case, Thanks a lot

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1722A - Spell Check Hacking not done properly, go through it once

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My idea for G : XOR of 4 consecutive numbers starting from 0 is 0. 2^20 is greater than 2e5 which gave me sufficient power of 2's to play with.
Requires just some more case work for the remaining element. Approach

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for problem G

for n = 5,

the solution: 0 1 33554433 2 33554434

is valid or invalid???

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My code for C using sets concepts in maths.

t = int(input())
for _ in range(t):
    tot_words = int(input()) * 3
    s1 = set(x for x in input().split())
    s2 = set(x for x in input().split())
    s3 = set(x for x in input().split())

    only_in_s1 = len(s1 - s2 - s3)
    only_in_s2 = len(s2 - s1 - s3)
    only_in_s3 = len(s3 - s1 - s2)

    in_s1_s2_s3 = s1 & s2 & s3
    s1_s2 = len((s1 & s2) - in_s1_s2_s3)
    s2_s3 = len((s2 & s3) - in_s1_s2_s3)
    s3_s1 = len((s3 & s1) - in_s1_s2_s3)

    score_of_s1 = (only_in_s1 * 3) + (s1_s2 + s3_s1)
    score_of_s2 = (only_in_s2 * 3) + (s2_s3 + s1_s2)
    score_of_s3 = (only_in_s3 * 3) + (s3_s1 + s2_s3)
    print(score_of_s1, score_of_s2, score_of_s3)
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    Funnily I completed sets chapter only yesterday.

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fst

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Is it only Me or anybody else also noticed that F was comparatively easy E for those who till now havent encountered problem based on 2-D PREFIX Sum . I am feeling very frustrated as I spent more than one hour on E and also during the contest F have least successful submission so I just dont focused on F

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problem D can also be solved in O(n) time using two pointers using a greedy approach. in the first half of the string, we can get better answer by converting the leftmost 'L' to 'R' and in the second half of the string, we can get better answer by converting the rightmost 'R' to 'L'.

link to my code — 170257607

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A very simple and easy solution of G .

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Alternate solution for Problem G:
Observe that $$$ a \oplus b$$$ $$$=$$$ $$$\sim a \oplus \sim b. $$$ Now you can construct a sequence $$$a0$$$, $$$\sim a_0,$$$ $$$a_1,$$$ $$$\sim a_1,$$$ $$$a_2,$$$ $$$\sim a_2, \dots$$$ (upto n terms) where $$$n \equiv 0$$$ ($$$mod$$$ $$$4$$$).
For $$$n = 4k+1$$$ or $$$4k+2$$$ or $$$4k+3$$$, just take some prefixes of length $$$1$$$ (0), $$$6$$$ (4,1,2,12,3,8), and $$$3$$$ (2, 1,3) respectively from the given testcases itself and the remaining sequence will be of length $$$4k'$$$.

Note that $$$a_0$$$, $$$a_1$$$, $$$\dots$$$ can be taken to be $$$13,$$$ $$$14,$$$ $$$15,$$$ ... as none of the prefixes contain elements $$$>13$$$.
Also, $$$\sim a$$$ represents 1's compliment of $$$a$$$ inverting all 32 bits of $$$+a$$$, although (even the first 20 bits would work considering the constraints on $$$a_i$$$)

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On Problem 1722B Colourblindness I had the same Approach But still got WA on test case 3

#include<bits/stdc++.h> 
using namespace std; 
#define ll  long long;
#define vi vector<int> 
#define pb push_back
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(int i=a;i<b;i++)
 
string solver()
{
	int len;cin >> len; 
	char upper[len],lower[len]; 
	cin>>upper; 
	cin>>lower;
	rep(i,0,len)
	{
		if(upper[i] == 'R' && lower[i] != 'R')return "NO";
		if(lower[i] ==  'R' && upper[i] != 'R')return "NO";
	}
	return"YES";
}
int main(){
	ios::sync_with_stdio(1);cin.tie(0);cout.tie(0);
	#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin); 
    //freopen("output.txt","w",stdout); 
 	#endif
  int tc=0;cin >> tc;
  while(tc--)cout << solver() << endl;
  return 0; 
}
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what is the use of greater() in the sort function of the D problem?

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In the problem E, trivial $$$O(n q)$$$ solution passes due $$$\mathrm{TL} = 6 \mathrm{s}$$$: 170297542

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    I wonder how your fastIO works...My BF solution 170407110 could pass only when adding your fastIO code. (And on test 4, it is 15 times faster than simply using untie cin/cout). It seems impossible because IO shouldn't be the bottleneck of this problem. I'm really confused about this.

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      We need optimize any constant in time complexity of solution. And it is fastest I/O (on Windows) that I have been seen.

      In it, I completely abandoned small buffers (usually ~2KiB) that lead to a lot of file operations, and also decided to completely abandon non-system I/O functions due to overhead. Large buffers allow your program to use two file operations during work.

      And when -O3 is specified, gcc inlines my function ReadInt32, but no printf and std::cin.

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        3 months ago, # ^ |
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        I discuss this with some friends and get a weird conclusion. You can see the difference between 170435459 and 170435486. Seems that when we replace the last cin with a function (although we still use cin to read), the compiler will use SIMD to optimize the if statement. Replace function with inline function even macro gets the same result. But just using cin or scanf won't trigger optimization. I have no idea why this happens :(

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3 months ago, # |
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I solved 3 problems but my rating doesn't change. my rating point is still 0point

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3 months ago, # |
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I solved problem G quite differently from the editorial method utilising the property that XOR of consecutive numbers x and y is always 0 when x is an even number (if you didn't know this, it's easy to see why, by taking a few consecutive numbers and looking at their binary representation).

Let set a and b represent respectively the set of numbers at even positions and the set of numbers at odd positions. Then we break solution into 4 cases depending on the remainder of n on dividing by 4.

Case 1.) n%4 = 0
Case 2.) n%4 = 1
Case 3.) n%4 = 3
Case 4.) n%4 = 2

I think some of the cases can probably be combined for a simpler solution. Please let me know in the comments.

Not so elegant and repetitive implementation : 170369765

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3 months ago, # |
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In E, I'm getting MLE if I put prefix array in the heap memory, but the solution passing when I put the prefix array as a global array. Why is that so?

170374285

int main() {
    ll** pt = new ll*[1001];
    ll** dp = new ll*[1001];
    for(int i=0;i<=N;i++){
	pt[i] = new ll[1001];
	dp[i] = new ll[1001];
    }
   //......
}

vs

170372939

ll pt[1001][1001];
ll dp[1001][1001];
int main(){
   //.....
}
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3 months ago, # |
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A much simpler (in my opinion) and more readable solution for Problem F with necessary comments : 170377676

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3 months ago, # |
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Another way to solve F: L-shapes:

ref: https://codeforces.com/contest/1722/submission/170380611
Note that, a 2x2 grid will contain L shape if and only if there are exactly 3 '*' in it.

1)Firstly for all 2x2 subgrid that contain L-shape, check whether its surrounding has any '*' in it, if so, then the ans is NO. [except for the corner along void position(exactly one such position) of 2x2 grid ,
eg: '#' in below picture [and ^ is void position] ]

...#
.*^.
.**.
....

2)Now the only problem is, there can also be any other shapes than L-shapes. To find that,
In logic1, just whenever current 2x2 subgrid contain L,(ie:exactly 3 '*' in it) , mark those positions as #.
So that all L shapes will be marked, and if there exist any other shape than L shape then it will remain unmarked(ie:remains *).

So, atlast traverse the whole grid to find any such unmarked position,
if so, the ans is NO,
Otherwise YES.

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3 months ago, # |
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Can somebody explain solution E, I have never seen use of prefix sum like this. I dont quite get it what thay have exactly done here

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    3 months ago, # ^ |
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    suppose you have a data structure which stores 2D-prefix sums, such that sum of rectangles with height <= H and width <= W = pref[H][W]

    then if you wanna know the sum of areas of rectangles with height < Hb and width < Wb (condition — i) then that would equate to pref[Hb — 1][Wb — 1]

    now say you want the sum of areas of rectangles which have height <= Hs and width <= Ws(condition-ii)

    the rectangles which satisfy both condition i and ii is the required answer

    for learning about 2D-prefix sums I would suggest going through the USACO guide, they have detailed and easy to understand content, https://usaco.guide/silver/more-prefix-sums?lang=cpp#2d-prefix-sums

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3 months ago, # |
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Why didn't my ratings update ? :'(

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3 months ago, # |
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My Solution for F

did the same thing as mentioned in the editorial but is easier to understand.

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3 months ago, # |
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The first problem can be solved with a frequency array, by calculating the frequency of each character while taking inputs. Then check whether the frequency of each character of "Timur" is equal to one or not.

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3 months ago, # |
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I got wa on F test 3 170393551 can somebody say the problem

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3 months ago, # |
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There is O(n) (I believe it is, at least amortized) solution for problem D https://codeforces.com/contest/1722/submission/170241642

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3 months ago, # |
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Is there anywhere I can view solutions in java?

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3 months ago, # |
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It was a nice contest though F was quite exhaustive

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3 months ago, # |
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Here is a bit shorter code for F: 170427907

Spoiler
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3 months ago, # |
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In Spell Check Why do we need to sort name too? Because if I'm not sorting name then I'm getting an error Please can anyone explain what I'm missing here

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3 months ago, # |
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Is there any other way to solve #C without using a map?

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3 months ago, # |
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anyone , where can i get some greedy questions like "Line question", where need to handle edge cases and stuff like that?

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3 months ago, # |
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Here you can see the promble F be sloved by using BFS.

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3 months ago, # |
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Problem E can also be solved with offline queries and binary search on segment tree, which gives O(t*n*logn) that satisfied arbitrary large height and width of the rectangles.

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3 months ago, # |
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Simple Solution for 1722F -L-shapes problem without DFS or BFS.

Approach

C++ code: 170466848

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    2 weeks ago, # ^ |
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    Hello.

    A question : why isn't enough to check only if the cntstars ==3 ?

    what is the corner case ? 181822571

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3 months ago, # |
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My solution for G:

xor of 2x (even no.) and 2x+1 is always 1.

Case 1: n%4 == 0: A simple solution would be 20,20+2n,21,21+2n,22,22+2n,....

Case 2: n%4 == 1: Just add a 0 at the end of Case 1

Case 3: n%4 == 2: Just add 4 1 2 12 3 8 this sequence at the end of Case 1 (Provided in sample input for n=6)

Case 4: n%4 == 3: Just add 2 1 3 this sequence at the end of Case 1 (Provided in sample input for n=3)

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3 months ago, # |
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https://codeforces.com/contest/1722/submission/170472565 Can anybody tell me the mistake in my code for Problem F

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3 months ago, # |
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I got it

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3 months ago, # |
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D in linear time : 170362765

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3 months ago, # |
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Chekout the solution for Problem E 170565433

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3 months ago, # |
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Upvote if you love competitive programming by heart.

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3 months ago, # |
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Error Report:: For the 1722E ( Counting Rectangles ) Tutorial, the corners to consider should be top left and bottom right, instead of lower left and upper right!

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3 months ago, # |
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I am not able to solve 4 th question . can anybody help to solve it?????

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3 months ago, # |
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why does the another approach worked in G can anyone explain please?

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6 weeks ago, # |
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can u tell why m i getting runtime error  for qstn B ??
[submission:178089241]
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8 days ago, # |
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although it was 3 months later but I have to say that the tutorial of G has a little mistake

I think it should be $$$2 ^ {30}$$$ here but it's $$$2 ^ 30$$$ :)

anyway, good problems and good tutorial !