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flamestorm's blog

Codeforces Round #742 Editorial
By flamestorm, 3 years ago, In English

Thank you for participating in our contest! We hope you enjoyed it. UPD: Implementations have been added.

1567A - Domino Disaster

Solution
Implementation (C++)
Video Solution

1567B - MEXor Mixup

Solution
Implementation (C++)
Video Solution

1567C - Carrying Conundrum

Solution (Observation)
Implementation (C++)
Video Solution
Solution (DP)
Implementation (C++)

1567D - Expression Evaluation Error

Solution
Implementation (C++)
Video Solution

1567E - Non-Decreasing Dilemma

Solution
Implementation (C++)
Video Solution

1567F - One-Four Overload

Solution
Implementation (C++)
Video Solution
  • Vote: I like it
  • +237
  • Vote: I do not like it

Comments Comments (118)
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3 years ago, # |
  Vote: I like it +57 Vote: I do not like it

Lightning Fast Editorial.

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

Thanks for the Video Solutions.

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3 years ago, # |
  Vote: I like it +44 Vote: I do not like it

Liked problem E and really disappointed about not solving it :(

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

This contest had a great difficulty curve. Kudos!

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

Weak test cases in problem B. Under the given condtions no precomputation should give TLE as no condition was given to limit the sum of all values of a.

Edit: Obviously I mean TLE only for those who have calculated XOR everytime in O(a).

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

    but one can clearly see that it will get tle if xor value is calculated every time.

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

      Hey, can you please tell me why do we get TLE if we precalculate the xor using for loop? Won't it be O(a)?

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

        We won't get TLE if we precalculate the XOR values. I saw your submission of B, You are getting TLE because you are calculating XOR values for every test case. The time complexity of calculating XOR will be $$$O(n)$$$, and you are doing this for every test case, so your time complexity is actually $$$O(t*n)$$$ and that's why the TLE.

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

          Oh okay I see now. But please tell me how can we precalculate the xor without getting the TLE?

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            3 years ago, # ^ |
              Vote: I like it +5 Vote: I do not like it

            You can make an array, where $i$th element will store the value, 0^1^2^....^i. You should do this before you start processing test cases.

            int arr[300005];
arr[0] = 0;
for(int i = 1; i <= 300004; i++)
{
     arr[i] = arr[i-1]^i;
}

// now you have precalculated the xor values.
// If you want to know what 0^1^2^...^i is, use arr[i].

while(t--) {

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

    I didn't note that need to precalculate too. But because of pragmas I get AC.

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

      Even with pragmas you wouldn't get AC in the worst case which will be around 1e10

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it +8 Vote: I do not like it

    There is a method that you can use, which can essentially answer each test case in O(1) without having to run any precomputation or use any pragma optimizations.

    It is based on the simple observation in the trend of the XOR value from 0 to a. It goes a little something like this.

    • If 'a' is a multiple of 4, the XOR of the sequence will be 'a' itself
    • If 'a' is a multiple of 2 but not a multiple of 4, the XOR of the sequence will be 'a' + 1
    • If 'a' is an odd number and the 'a' + 1 is both a multiple of 4 and 2, the XOR of the sequence will be 0
    • If 'a' is an odd number and the above constraints fail, the XOR of the sequence will be 1

    Which Roughly Translates to This:

    if(a % 2 == 0)
{
    if(a % 4 == 0) limit = a;
    else limit = a + 1;
}
else
{
    if((a + 1) % 2 == 0 and (a + 1) % 4 == 0) limit = 0;
    else limit = 1;
}

    Then you can simply perform the usual checks to get to the final answer.

    Here is my submission if you need any further clarifications. 128021227

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

Didnt got an approach to solve C. Were you able to solve C problem

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

seen/good

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

I knew how to solve E the moment I saw it, too bad I suck at implementation. Back to practice.

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

I think C is more difficult than D. Tutorial of problem D is much more difficult than mine. My solution works O(n)

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

    C is simple if u dont approach the dp way.Just divide the input string into even and odd parts.If u analyse the problem its actually simple addition on adjacent values.This observation is enough to solve the problem .

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

      Yeah i tried dp (couldn't solve), its very easy to make error, or think of wrong transitions equation.

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

Problem E is great but didn't understand the 11th base for problem D :(.

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

    basically 11th base was for hinting to the fact that you need to make n numbers such that they can achieve highest decimal place. example — to make 1000 with 2 numbers consider two cases, 900 100 990 10

    now you can see that in first case you are multiplying higher decimal places with even higher power of 11, 9*((11)^3) + 1*((11)^3) and in second case it is 9*((11)^3) + 9*((11)^2) + 1*((11)^2)

    so basically 1*((11)^3) > 9*((11)^2) + 1*((11)^2)

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3 years ago, # |
  Vote: I like it +27 Vote: I do not like it

For me at least, C is the type of problem that seems quite hard, then, when you see the editorial, you realise it was really easy and you were the one who complicated it.

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

    Yeah!thinking in that direction is hard! I tried one hour for it:)

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

      I actually came up relatively quickly with a solution, but it was a bit un-elegant, in my opinion: using backtracking, I fixed the digits that would have a carry, then I calculated how many pairs of numbers satisfy the fixed digits' carries.

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Wow! The solution for C is really cool! I got a kinda brute-force solution in O(2^log10(n)). Fixing which digits are going to add in such way that, it would create a carry. Then just checking how many pairs satisfy that kind of fix carried addition. Here is my submission link Link !

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

I thought of an algorithm to F, 127979513, that came up to mind when simulating filling the grid by hand. However, this algorithm failed Pretest #6, and I don't know where it went wrong, nor am I able to come up with a counter-testcase for it. The algorithm is as follows.

  • Create an array $$$ans$$$, which is our answer. Initially, all elements of $$$ans$$$ are $$$0$$$.

  • First, check if a marked cell has an odd number of unmarked cells. If true, then there must be no grid that satisfies the condition. Else, for each marked cell $$$(x,y)$$$ with $$$n$$$ neighbors, $$$ans[x][y] = \frac{5n}{2}$$$.

  • Create an array $$$req$$$. $$$req[i][j]$$$ is initially $$$0$$$ for unmarked cells and $$$ans[i][j]$$$ for marked cells. Think of $$$req$$$ as the sum which we still need to 'distribute' to adjacent cells. We will subtract from $$$req$$$ when we update the neighbors of marked cells such that at the end, $$$req[i][j] = 0$$$ for all $$$i$$$ and $$$j$$$.

  • Iterate through each cell in $$$ans$$$. For each unmarked cell with $$$ans=0$$$, update its value with $$$1$$$.

Updating a cell goes as follows:

  • To update an unmarked cell by $$$a$$$, set its value to $$$a$$$. Then, update all adjacent marked cells with $$$a$$$.

  • To update a marked cell at $$$(x, y)$$$ by $$$a$$$, subtract $$$a$$$ from $$$req[x][y]$$$. Then, if $$$req[x][y] = 1$$$ or $$$2$$$, we know that all remaining unmarked neighbors of $$$(x, y)$$$ should contain $$$1$$$. Thus, update them by $$$1$$$. The same thing happens when $$$req[x][y] = 4$$$ or $$$8$$$, where all its remaining neighbors should be $$$4$$$.

Does anybody have any idea on where or why it breaks down? Any help would be appreciated.

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

I suppose there is a mistake in explanation for problem C as $$$9 + 13 = 22$$$

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +5 Vote: I do not like it

    I don't know how to do addition, apparently. It will be fixed soon.

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

can anybody explain why we are doing (a + 1) * (b + 1) — 2 after spliting it to 'a' and 'b' ? in problem C

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

    There are a + 1 ways of getting the odd digits, there are b+1 ways of getting the even digits and they are independent. After that we have two solutions that dont work, (0,s) and (s,0), so we remove them with this "-2"

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

    if you're talking about problem C we can divide number n to 2 numbers a and b. by the position of their digits depending on odd or even. Then you have to make that number sum of 2 different numbers. we can 0+a,1+(a-1)+...a+1 in total a+1. it's the same for b. but the problem said positive integers which don't include 0. so we exclude the first number equals 0 and the second number equals zero which are 2. So the answer is (a+1) * (b+1) — 2. It might be different from some people's solution.

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3 years ago, # |
  Vote: I like it +19 Vote: I do not like it

I did D by simply printing the largest possible power of 10 (say x) such that (s-x) ≥ (n-1) in a loop while n is greater than 1. Finally printed whatever remained of s. This works in O(n).

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

    have you got any proofs why is this works? I will be very grateful if u explain it:D

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

      It's actually pretty simple. If the largest power of 10 we can put is x, then we will add 11^log10(x) to the answer. If we choose power of 10, we can get at most y*(11^log10(x/y)) in the sum, y being power of 10 lower or equal to x. Now we just need to prove that 11^log10(x) >= y*(11^log10(x/y)).

      11^log10(x) >= y*(11^log10(x/y))

      11^log10(x)/11^log10(x/y) >= y

      11^(log10(x)-log10(x/y)) >= y

      11^(log10(x)-log10(x) + log10(y)) >= y

      11^log10(y) >= 10^log10(y)

      11 >= 10

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3 years ago, # |
  Vote: I like it +90 Vote: I do not like it

E is a very standard problem. I don't see it appropriate for a Div2 Round. I guess it would be better in a D of an Educational Round.

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

Lightning-fast editorial with video solution. Nice Job

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

I know how to solve F, but I didn't read it during this round. What a pity. MySolution

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

Editorial of problem C is really good.

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

Wow, solution to C is incredible!!

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3 years ago, # |
  Vote: I like it +11 Vote: I do not like it

Did anyone solve C using brute force in o(2^n) like either consider the carry or don't consider the carry?

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

When I try to access C Submission link CF says: "You are not allowed to view the requested page"

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3 years ago, # |
  Vote: I like it +16 Vote: I do not like it

why Alice and Bob do weird things :(

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3 years ago, # |
  Vote: I like it -46 Vote: I do not like it

Can someone help me to debug this code Q(B) — MEX or Mixup

include <bits/stdc++.h>

using namespace std;

int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); long int t = 1; cin >> t; while(t--) { long long a,b; cin >> a >> b;

long long all = 0;

    a--;

    if (a % 4 == 0)
     all = a;

    if (a % 4 == 1)
      all = 1;

    if (a % 4 == 2)
     all =  a+1;

    if (a % 4 == 3)
     all =  0;


    a++;

    // cout << all; 
    if(all == b)
    {
        cout << a << '\n';
    }
    else
    {
        if(b^all != a)
        { 
            cout << a+1 << '\n';
        }
        else{
            cout << a+2 << '\n';
        }

    }
}

}

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

    in first else block declare a tmp variable of int type store the value of "b^all" and then compare with a i had the same problem lmao

    like else{ ll tmpvar = b^all; if(tmpvar != a){ cout << a+1 << endl; return; } cout << a+2 << endl; }

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

      why is it like this? , means what is the problem , can you explain little bit.

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

        It something related to buffer Sometimes without storing and directly using some function can cause error like 

        int a = 7;
double b = a/2;
int c = ceil(b);

// Above code gives wrong output for 'c'

// But not below one
int a = 5;
double b = a;
b = b/2;
int c = ceil(b);
 
For above code output will be as expected for 'c' which is 3
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          3 years ago, # ^ |
            Vote: I like it +3 Vote: I do not like it

          uhh, i dont think it is a buffer or whatever error,

          in the first code, when you do double b = a/2; a is considered int and a/2 is done in int division where we just cut everything after decimal.

          in second code b = b/2; is double division, so that gives correct output

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

    just enclose ^(xor) in a bracket, ^ has a higher precedence that = operator, that may be the error

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

      yes i think that was the issue (b^all == a) is treated as (b^(all == a)). so in my code it should be ((b^all) == a)

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

I can't see the implementations. Even after solving the problem it doesnt let me check them.

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

problem C was irritating to be honest

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

C question was really good and tricky and thank you for the contest and very fast tutorial

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

Question D was just amazing and liked the logic behind C and D. It was very good contest

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3 years ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

I can't view the codes in the tutorial. It says-'You are not allowed to view the respected page'.

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

In Problem D there is an ambiguity in the statement- the decimal number may not be uniquely interpreted in the 11-based system, for example, 100_{10} -> 100_{11} = 0 + 10*11 = 110_{10} or 100_{10} ->100_{11} = 0 + 0*11 + 1*121 = 121_{10}

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

    $$$100_{11}$$$ is never interpreted as $$$10 \cdot 11 + 0 \cdot 1$$$; the digit $$$10$$$ is typically represented by $$$A$$$.

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

      Why it is always optimal to divide the sum into a power of 10.

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

      Could you please explain why for the test case "999999 3" the answer "1 1 999997" is better then "800000 100000 99999"? It seems to me that they have the same sum in 11-based system

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

        They have the same sum, but your submission gets WA on test case 4: $$$10_{11}+90_{11}=A0_{11}$$$, but $$$1_{11} + 99_{11}= 9A_{11}$$$.

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

          Oh, I see. Thank you!

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

          For Problem D, regarding the idea of splitting the least power of 10: "we should split the smallest power 10". I can't seem to make a correct algorithm for input, say: 100 3.

          I would end up with the 3 following terms: 90, 9, 1.

          Starting with [100], splitting into [90, 10] and then, take the smallest power of 10, 10, ending up with [90, 9, 1].

          Have I misinterpreted your words?

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

            You should only split powers of $$$10$$$ when we have no other ways to split any of the numbers: $$$90_{11}$$$ can be split as $$$10_{11}$$$ and $$$80_{11}$$$ still, and so a better answer is $$$[80, 10, 10]$$$.

            When we add them:

            • $$$80_{11} + 10_{11} + 10_{11} = A0_{11}$$$.

            • $$$90_{11} + 9_{11} + 1_{11} = 9A_{11}$$$.

            As you can see, we should continue to split all numbers into powers of ten, and as soon as all numbers are powers of ten, then we can take the smallest one.

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

        they are same

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3 years ago, # |
  Vote: I like it +78 Vote: I do not like it

After seeing solution to E, I started to wonder what Segment trees can't do.

Maybe it can even end world hunger in O(nlogn)??

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

Someone please link memoisation based Dp solution for problem C. Mine is giving wrong answer at test 5.

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

    127946227

    ^ here is my submission. Hope it helps!

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

    a great exercise would be to make a dp based soln. for finding number of ways when we normally add 2 numbers. I know it is n + 1, but if you can make that dp then you can do that alternatively in this problem.

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

You can calculate XOR of first $$$n$$$ natural numbers in $$$O(1)$$$. You can learn more about that here.

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

    We were aware of this, but we didn't want to force people to google the solution. The editorial mentions this as well.

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

      It will be better if you add it in editorial. I didn't know about it before seeing this comment.

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

flame, please lmk how to be as orzosity as you

thanq

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

The video is really great

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

[submission:https://mirror.codeforces.com/contest/1567/submission/127980698] O(n) solution for problem D

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

I think F can be transformed into a 2-SAT problem (maybe easier to think for some people): If one marked cell has an odd number of adjacent empty cells, obviously there's no answer.

Otherwise if it has 2 adjacent cells, one must be filled with true (representing 1) and the other must be filled with false (representing 4). Let the two cells be $$$a$$$ and $$$b$$$, we have the relation $$$(a \lor b) \land (\neg a \lor \neg b)$$$.

If it has 4 adjacent empty cells, it's always better to fill the opposite empty cells with the same number, and we can get a similar relation as the above case.

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

    If it has 4 adjacent empty cells, it's always better to fill the opposite empty cells with the same number, and we can get a similar relation as the above case.

    Is this your assumption or can you prove that?

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

      Can't say I proved that but this is my rough idea:

      Cells that share common empty cells form a "connected component" in which the way of filling two diagonally adjacent cells determines how to fill the whole component.

      1. If some marked cell has only two adjacent empty cells, we should make the two empty cells different => opposite empty cells should be the same.
      2. Otherwise all marked cells have four adjacent empty cells => opposite empty cells can be the same or different.
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        3 years ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it

        After writing that I thought 2-SAT are just redundant. You can simply fill two diagonally adjacent empty cells and do a bfs/dfs to spread the coloring.

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

      I also solved the problem with this method, and my proof went something like this:

      Build a graph whose edges are pairs of unmarked cells that we want to force to be different. For marked cells with 2 adjacent unmarked cells, draw an edge between them; for marked cells with 4 adjacent unmarked cells, draw edges between (top, left) and (bottom, right).

      Any bipartite colouring of this graph yields a valid solution. To prove such a colouring exists, it suffices to show that there are no odd simple cycles in the graph.

      Let's assume, for the sake of contradiction, that such a graph with an odd cycle exists.

      There are two types of edges in this graph: "grid-aligned" and "diagonal". Notice that "grid-aligned" edges do not change the parity of coordinates, but "diagonal" edges change the parity of both coordinates. Therefore there must be an even number of "diagonal" edges, and so, an odd number of "grid-aligned" edges.

      If the edges are straight lines, the cycle is a boundary of a section of the grid. Notice that the only marked cells on the boundary are on the "grid-aligned" edges, so there are an odd number of them.

      Count the number of ordered pairs of marked cells (P, Q) where P and Q are adjacent and both on or inside the boundary. By double-counting, this number must be even.

      However, for a marked cell inside the boundary, we know that the number of adjacent marked cells on or inside the boundary must be even, and for a marked cell on the boundary, this number must be odd. As there is an odd number of marked cells on the boundary, the number of ordered pairs is odd, which is a contradiction.

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        3 years ago, # ^ |
        ← Rev. 2   Vote: I like it -26 Vote: I do not like it

        there is a simpler proof to the 2-coloring if you instead connect (top,right) and (bottom,left) for the diagonal edges .

        the difference between the x and y coordinate remains invariant under the diagonal edges since it ±1 to both coordinates , and grid aligned edges changes the difference by ±2 since you need a net change of 0 to the difference thus you need an even number of grid aligned edges (parity argument still holds for even diagonal edges)

        UPD : this is wrong I missed that for connecting two forced unmarked cells you can get diagonally left edges

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3 years ago, # |
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Was so close to solving B but missed the case when xor of a-1 and b is equal to a :(

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3 years ago, # |
← Rev. 17   Vote: I like it 0 Vote: I do not like it

I used a recursive algorithm for solving problem 1567C - Carrying Conundrum

128012196

The algorithm is based on the idea that as $$$2 \leq n \leq 10^9$$$, the 2-digit carry operation can be expressed in terms of the decimal digits and the carry bit using at most 10 linear equations.

Linear Equations

If the decimal representation of $$$n$$$ has $$$k$$$ digits, where $$$1 \leq k \leq 10$$$, then the equations relating the least signification two digits $$$n_0$$$ and $$$n_1$$$ do not have carry-in bit. Similarly, the equations relating the most significant two digits $$$n_{k-2}$$$ and $$$n_{k-1}$$$ do not have carry-out bit. Therefore, there are at most $$$2^{k-2}$$$ possible distinct states for the binary vector of $$$k-2$$$ carry-out bits.

It is noted that the $$$k$$$ equations are separable into two independent sets of linear equations according to the odd/even parity of the digit index. This leads to the same solution given in the editorial for partitioning $$$n$$$ into two numbers $$$a$$$ and $$$b$$$, with the conventional 1-digit carry, where the answer can be computed as $$$(a+1)(b+1)-2$$$.

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3 years ago, # |
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I am an absolute noob :’(. I got the idea of problem B but could not be able to find anything on the Internet about the formula for xor of 1…n. So bad. Anyway I love the editorials and thank you all for an awesome contest.

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3 years ago, # |
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In problem B, Should the elements of the be distinct?

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3 years ago, # |
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D can be done in $$$O(n*log_{10}(s))$$$ as well with simple recursion. https://mirror.codeforces.com/contest/1567/submission/127976931

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3 years ago, # |
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In problem C(Carrying Conundrum), Could anyone please tell Why we are subtracting 2?

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    3 years ago, # ^ |
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    The two cases of one of the number in pair being 0

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3 years ago, # |
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I feel like the Editorial to Problem D doesn't carry all cases, so I did a visual explanation.

Take $$$s=113$$$ and $$$n=7$$$. First we split the decimals into powers of $$$10$$$:

We could still need more numbers to fully get $$$n=7$$$. So we look for the smallest number $$$>1$$$:

An we split it into $$$10$$$ equal parts. To achieve this you can just replace this number with a tenth and then push back this value 9 times:

We repeat this step, until we have at least $$$n$$$ values. The editorial proposes a $$$O(n \log n)$$$ solution for this step using a priority queue, but this can also be done in $$$O(n)$$$ if we keep 2 separate arrays, one for numbers $$$>1$$$ and one for numbers $$$=1$$$. The former one will be automatically sorted at all times following this procedure.

In the next step we could have more than $$$n$$$ numbers. We just collect the overflow and add it to the last value:

And we are done:

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3 years ago, # |
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Special Thanks to flamestorm for problem-C !!!

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3 years ago, # |
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Thank you for that round, the problems were really interesting. BUT!!! Maybe C and D should be replaced)

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3 years ago, # |
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E is the hardest to me.

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3 years ago, # |
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Why are we doing -2 in (a+1)(b+1) in the editorial of problem C.

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    3 years ago, # ^ |
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    Because the pairs (0, n) and (n, 0) have also been included in (a+1)*(b+1). We need to remove them as we want both numbers to be greater than 0. Hope you got it. I also did not get it but I tried with n = 22 and I understood. Hope it helps you too :)

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3 years ago, # |
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can anyone explain query part of problem E

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

1567F : As I can see, those who got AC in the contest had simpler solutions than the author. The editorial is quite general, I think. Just curious how this problem can be extended.

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3 years ago, # |
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color[u] = (color[v] ^ 3); What is the purpose of this code?

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3 years ago, # |
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I also don't understand the aim of the below part of the code in the solution for F. Can someone explain further. Thanks.

// flip each cell appropriately, column by column
    for (int j = 1; j <= m; j++) {
        int curr = (j % 2 ? 4 : 1);
        res[1][j] = curr;
        int prev = color[val[1][j]];
        for (int i = 2; i <= n; i++) {
            if (grid[i][j] == '.') {
                if (color[val[i][j]] != prev) {curr = flip(curr);}
                res[i][j] = curr;
                prev = color[val[i][j]];
            }
        }
    }
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3 years ago, # |
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For problem E, the editorial's implementation is very suck.

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3 years ago, # |
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I think there might be some mistakes according to tutorial of problem F. In the tutorial it said: "However, the tricky case is to deal with cells with 0 unmarked neighbors". But it is obvious that cells with 0 unmarked neighbors will get the value of 0 in the end. So what is really tricky to deal with is those cells with 4 unmarked neighbors. I think the writer might confuses the definitions of marked and unmarked cell.

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    3 years ago, # ^ |
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    Yes, you are right; I updated the editorial some time ago, but for some reason, it didn't seem to update here. I am not entirely sure what's going on ¯\_(ツ)_/¯

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22 months ago, # |
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E editorial seems a bit impl heavy. My solution is a lot shorter(50 lines).

Basically, each node stores the following: answer for the range, longest non-decreasing prefix, longest non-decreasing suffix, size of range, 1st and last elements in the range. I won't explain the combine part as it's easy to understand from the code. Also, notice you could just store l,r for range instead to take care of the last 3 parts but I decided not to store the elements in an array.

Solution

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12 months ago, # |
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can someone tell me where this submission might be failing ?

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7 months ago, # |
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I am in a trouble with the last note of Problem C, "Note that pairs (a,b)and(b,a) are considered different if a≠b." Could anyone give me a proper explanation, please?

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