### flamestorm's blog

By flamestorm, 3 weeks ago,

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

Solution
Implementation (C++)
Video Solution

• +237

 » 3 weeks ago, # |   +57 Lightning Fast Editorial.
 » 3 weeks ago, # |   +26 Thanks for the Video Solutions.
 » 3 weeks ago, # |   +44 Liked problem E and really disappointed about not solving it :(
 » 3 weeks ago, # |   -6 This contest had a great difficulty curve. Kudos!
 » 3 weeks ago, # | ← Rev. 2 →   0 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).
•  » » 3 weeks ago, # ^ |   0 Firstly I was also calculating XOR(n) for every test case and got TLE :/
•  » » » 3 weeks ago, # ^ |   0 Yep. I did the same mistake :(
•  » » 3 weeks ago, # ^ |   0 but one can clearly see that it will get tle if xor value is calculated every time.
•  » » » 3 weeks ago, # ^ |   0 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)?
•  » » » » 3 weeks ago, # ^ |   0 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.
•  » » » » » 3 weeks ago, # ^ |   0 Oh okay I see now. But please tell me how can we precalculate the xor without getting the TLE?
•  » » » » » » 3 weeks ago, # ^ |   +5 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--) { } 
•  » » » » » » » 3 weeks ago, # ^ |   0 Oh, yeah I get it now. Thank you so much :)
•  » » » » » » » 3 weeks ago, # ^ |   0 ll xorr; ll rem = (a — 1) % 4; if (rem == 0) xorr = a — 1; else if (rem == 1) xorr = 1; else if (rem == 2) xorr = a ; else if (rem == 3) xorr = 0;We can calculate xor upto a by this method also that taking modulo of a-1 with 4 and if remainder is o then xor will be a-1 else other condition. Time complexity =O(1)
•  » » » » 3 weeks ago, # ^ |   0 Precalulate zor values before inputting test cases then time complexity will be O(t)
•  » » » » 2 weeks ago, # ^ | ← Rev. 3 →   0 To calculate the xor value of all the numbers from 1 to n in O(1) time :Find the remainder rem = n % 4 If rem = 0, then xor = n If rem = 1, then xor = 1If rem = 2, then xor = n+1If rem = 3 ,then xor = 0This works because we get 0 as the XOR value just before a multiple of 4. This keeps repeating before every multiple of 4.
•  » » 3 weeks ago, # ^ |   0 I didn't note that need to precalculate too. But because of pragmas I get AC.
•  » » » 3 weeks ago, # ^ |   +4 Even with pragmas you wouldn't get AC in the worst case which will be around 1e10
•  » » » » 3 weeks ago, # ^ |   0 believe me, many got AC
•  » » » 3 weeks ago, # ^ |   0 Mine too accepted with pragmas + bruetforce
•  » » 3 weeks ago, # ^ |   0 I got tle :'( , then I had to use the properties of XOR. It was good tho
•  » » 3 weeks ago, # ^ | ← Rev. 3 →   +8 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
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 There is another way of computing xor in O(1) time complexity because xor of numbers from 1 to n follows a pattern and pattern repeats itself in the following manner lets say rem = n%4If rem = 0 xor = nIf rem = 1, xor = 1If rem = 2, xor = n+1If rem = 3, xor = 0and with slight modification we can get the required results.
•  » » » 2 weeks ago, # ^ |   0 well even looping through n and using pragma optimization gives AC:):)
 » 3 weeks ago, # |   0 Didnt got an approach to solve C. Were you able to solve C problem
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +1 -
•  » » » 3 weeks ago, # ^ |   +1 Thanks it get me a better understandimg
•  » » 3 weeks ago, # ^ |   0 yes
 » 3 weeks ago, # | ← Rev. 2 →   0 seen/good
•  » » 3 weeks ago, # ^ |   +1 Good development.
 » 3 weeks ago, # |   +1 I knew how to solve E the moment I saw it, too bad I suck at implementation. Back to practice.
 » 3 weeks ago, # | ← Rev. 2 →   0 I think C is more difficult than D. Tutorial of problem D is much more difficult than mine. My solution works O(n)
•  » » 3 weeks ago, # ^ |   0 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 .
•  » » » 2 weeks ago, # ^ |   0 But can you tell me why are we subtracting 2?
•  » » » » 2 weeks ago, # ^ |   0 yes as in question both numbers should be positive....hence if s=given sum,then we have to ignore two possibilities (s,0) and (0,s).
•  » » » 2 weeks ago, # ^ |   0 Yeah i tried dp (couldn't solve), its very easy to make error, or think of wrong transitions equation.
•  » » » » 2 weeks ago, # ^ |   0 Yes .
 » 3 weeks ago, # |   +3 Problem E is great but didn't understand the 11th base for problem D :(.
•  » » 3 weeks ago, # ^ |   +12 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 10now 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)
 » 3 weeks ago, # |   +27 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.
•  » » 3 weeks ago, # ^ |   0 Yeah!thinking in that direction is hard! I tried one hour for it:)
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   +1 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.
•  » » » » 3 weeks ago, # ^ |   0 Yeah! we do try to impliment dp for some problems and will get surprised looking at easy solution using some logic
•  » » » » 3 weeks ago, # ^ |   0 Yeah i also did the same backtracking thing considering two cases for each digit whether to give carry to next to next number or not. you can check my code here- https://codeforces.com/contest/1567/submission/127981331
•  » » 3 weeks ago, # ^ |   0 yes if someone approaches the DP way it gets complicated ,but if one thinks greedily its easy. Somehow i got the greedy idea on time:)
 » 3 weeks ago, # |   +8 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 !
 » 3 weeks ago, # | ← Rev. 2 →   0 Video tutorials are nice.
 » 3 weeks ago, # |   0 Very fast and amazing editorials with video solutions too.
 » 3 weeks ago, # |   +10 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.
 » 3 weeks ago, # |   +7 I suppose there is a mistake in explanation for problem C as $9 + 13 = 22$
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +5 I don't know how to do addition, apparently. It will be fixed soon.
 » 3 weeks ago, # |   0 The editorial came earlier than expected thank you.
 » 3 weeks ago, # | ← Rev. 2 →   0 For solution C: "Note that in every other column, the addition Alice performs is correct."This doesnt seem to be correct in the example in the question?
 » 3 weeks ago, # |   0 c was nice
 » 3 weeks ago, # | ← Rev. 2 →   0 can anybody explain why we are doing (a + 1) * (b + 1) — 2 after spliting it to 'a' and 'b' ? in problem C
•  » » 3 weeks ago, # ^ |   0 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"
•  » » 3 weeks ago, # ^ |   0 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.
•  » » 2 weeks ago, # ^ |   0 because the answer ask the pair of positive integer,answer can appear tow pairs (0,a+b)and(a+b,0),so you need to minus 2.
 » 3 weeks ago, # |   +19 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).
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 Wouldn't the time complexity be O(nlog10(s)) because of computing the power of 10 for each iteration?
•  » » 3 weeks ago, # ^ |   0 have you got any proofs why is this works? I will be very grateful if u explain it:D
•  » » » 3 weeks ago, # ^ |   +3 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) >= y11^(log10(x)-log10(x/y)) >= y11^(log10(x)-log10(x) + log10(y)) >= y11^log10(y) >= 10^log10(y)11 >= 10
•  » » » » 2 weeks ago, # ^ |   0 Thanks a lot!!!!
 » 3 weeks ago, # |   +90 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.
 » 3 weeks ago, # |   0 Lightning-fast editorial with video solution. Nice Job
 » 3 weeks ago, # |   0 I know how to solve F, but I didn't read it during this round. What a pity. MySolution
 » 3 weeks ago, # |   +1 Editorial of problem C is really good.
•  » » 3 weeks ago, # ^ |   0 Yeah, this time editorial were pretty good and damn fast
 » 3 weeks ago, # |   0 Wow, solution to C is incredible!!
 » 3 weeks ago, # |   +11 Did anyone solve C using brute force in o(2^n) like either consider the carry or don't consider the carry?
•  » » 3 weeks ago, # ^ |   0 I used that approach, check my submission here
•  » » » 3 weeks ago, # ^ |   0 Can you explain your approach?
•  » » 3 weeks ago, # ^ |   0 Yes,I did using backtracking all the possible cases.Link to my submission 128014364
•  » » » 3 weeks ago, # ^ |   0 Can you explain your approach?
•  » » » » 2 weeks ago, # ^ |   0 firstly i precalculated no.of ways to represent n(n<20) as a sum of two single digits. Now,start from the leftmost digit (let's say x) . It can be formed by two ways a) simply as a sum of two single digits which is ways[x] b) can be formed by a carry(which is 1) ,in this case x becomes x-1 and the digit next to the next to the current digit say y becomes 10+y; In each case the no.of ways the remaining digits can be formed can be calculated recursively.
 » 3 weeks ago, # |   +3 When I try to access C Submission link CF says: "You are not allowed to view the requested page"
•  » » 3 weeks ago, # ^ |   0 Fixed!
 » 3 weeks ago, # |   +16 why Alice and Bob do weird things :(
 » 3 weeks ago, # |   0 C was damn awesome after looking at editorial! Felt happy after looking at the editorial although I was unable to think it during contest
»
3 weeks ago, # |
-46

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';
}

}
}

}

•  » » 3 weeks ago, # ^ |   0 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; }
•  » » » 3 weeks ago, # ^ |   0 why is it like this? , means what is the problem , can you explain little bit.
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 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 
•  » » » » » 3 weeks ago, # ^ |   +3 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
•  » » 3 weeks ago, # ^ |   +3 just enclose ^(xor) in a bracket, ^ has a higher precedence that = operator, that may be the error
•  » » » 3 weeks ago, # ^ |   0 thank you
•  » » » 3 weeks ago, # ^ |   +1 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)
 » 3 weeks ago, # |   0 I can't see the implementations. Even after solving the problem it doesnt let me check them.
•  » » 3 weeks ago, # ^ |   0 Fixed!
 » 3 weeks ago, # |   +1 problem C was irritating to be honest
 » 3 weeks ago, # |   +1 C question was really good and tricky and thank you for the contest and very fast tutorial
 » 3 weeks ago, # |   0 I miss read the problem statement of B and got stuck and tried to find xor of every a-1 elements ;_;
 » 3 weeks ago, # |   0 Really fun contest. Unfortunately I was too dumb to realize the miniscule errors I made in B and D :'(But my rating increased by 69, so I guess that's nice.
 » 3 weeks ago, # |   0 Question D was just amazing and liked the logic behind C and D. It was very good contest
 » 3 weeks ago, # | ← Rev. 2 →   +1 I can't view the codes in the tutorial. It says-'You are not allowed to view the respected page'.
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 Oops, we are fixing it right now. UPD: It is fixed.
 » 3 weeks ago, # |   0 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}
•  » » 3 weeks ago, # ^ |   +1 $100_{11}$ is never interpreted as $10 \cdot 11 + 0 \cdot 1$; the digit $10$ is typically represented by $A$.
•  » » » 3 weeks ago, # ^ |   0 Why it is always optimal to divide the sum into a power of 10.
•  » » » 3 weeks ago, # ^ |   0 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
•  » » » » 3 weeks ago, # ^ |   +3 They have the same sum, but your submission gets WA on test case 4: $10_{11}+90_{11}=A0_{11}$, but $1_{11} + 99_{11}= 9A_{11}$.
•  » » » » » 3 weeks ago, # ^ |   0 Oh, I see. Thank you!
•  » » » » » 3 weeks ago, # ^ |   0 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?
•  » » » » » » 3 weeks ago, # ^ |   0 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.
•  » » » » » » » 3 weeks ago, # ^ |   0 I got you! Thank you for the time you took to reply.
•  » » » » 3 weeks ago, # ^ |   0 they are same
 » 3 weeks ago, # |   +78 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)??
•  » » 3 weeks ago, # ^ |   0 XD
 » 3 weeks ago, # |   0 Someone please link memoisation based Dp solution for problem C. Mine is giving wrong answer at test 5.
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 127946227^ here is my submission. Hope it helps!
•  » » » 3 weeks ago, # ^ |   0 Great solution . Thanks
•  » » 3 weeks ago, # ^ |   0 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.
 » 3 weeks ago, # |   0 You can calculate XOR of first $n$ natural numbers in $O(1)$. You can learn more about that here.
•  » » 3 weeks ago, # ^ |   0 We were aware of this, but we didn't want to force people to google the solution. The editorial mentions this as well.
•  » » » 3 weeks ago, # ^ |   +3 It will be better if you add it in editorial. I didn't know about it before seeing this comment.
•  » » » » 3 weeks ago, # ^ |   +4 Added.
 » 3 weeks ago, # |   0 https://codeforces.com/contest/1567/submission/127954668How does this solution for D split the 10s into 9s and 1s if there isnt enough digits?
 » 3 weeks ago, # |   0 Why does this get WA on test 3 in problem B https://codeforces.com/contest/1567/submission/127978046
 » 3 weeks ago, # |   0 Thank you so much
 » 3 weeks ago, # |   +6 flame, please lmk how to be as orzosity as youthanq
 » 3 weeks ago, # |   +3 The video is really great
 » 3 weeks ago, # |   0 [submission:https://codeforces.com/contest/1567/submission/127980698] O(n) solution for problem D
•  » » 3 weeks ago, # ^ |   0 127980698 here
 » 3 weeks ago, # | ← Rev. 2 →   0 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.
•  » » 3 weeks ago, # ^ |   +6 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?
•  » » » 3 weeks ago, # ^ |   0 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. 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. Otherwise all marked cells have four adjacent empty cells => opposite empty cells can be the same or different.
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 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.
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   +33 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.
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   -26 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
 » 3 weeks ago, # |   +3 Was so close to solving B but missed the case when xor of a-1 and b is equal to a :(
 » 3 weeks ago, # | ← Rev. 17 →   0 I used a recursive algorithm for solving problem 1567C - Carrying ConundrumThe 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$a_0 + b_0 = n_0 + 10~c_0$$a_1 + b_1 = n_1 + 10~c_1$$a_2 + b_2 + c_0 = n_2 + 10~c_2$$a_3 + b_3 + c_1 = n_3 + 10~c_3$$a_4 + b_4 + c_2 = n_4 + 10~c_4$$a_5 + b_5 + c_3 = n_5 + 10~c_5$$a_6 + b_6 + c_4 = n_6 + 10~c_6$$a_7 + b_7 + c_5 = n_7 + 10~c_7$$a_8 + b_8 + c_6 = n_8$$a_9 + b_9 + c_7 = n_9$where $0 \leq a_i, b_i, n_i \leq 9$ and $0 \leq c_i \leq 1$.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 $(a+1)(b+1)-2$.
 » 3 weeks ago, # |   +3 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.
•  » » 3 weeks ago, # ^ |   0 check in GFG
 » 3 weeks ago, # |   0 In problem B, Should the elements of the be distinct?
•  » » 3 weeks ago, # ^ |   0 no
 » 3 weeks ago, # |   0 Recursive Solution for problem C. Code#include using namespace std; typedef long long ll; ll solve(ll i,ll n,vector &a,vector &ways) { if(i==n-1) return ways[a[n-1]]; if(i==n-2) return ways[a[n-1]]*ways[a[n-2]]; ll ans=ways[a[i]]*solve(i+1,n,a,ways); if(a[i]==0) return ans; a[i+2]+=10; ll res=ways[a[i]-1]*solve(i+1,n,a,ways); a[i+2]-=10; return res+ans; } void fun(ll tt) { ll n; cin>>n; vector a; while(n>0) { a.push_back(n%10); n=n/10; } reverse(a.begin(),a.end()); vector ways(20,0); for(ll i=0;i<20;i++) { for(ll j=0;j<=i;j++) { ll k=i-j; if(j<10 && k<10) { ways[i]++; } } } n=a.size(); cout<>t; for(ll tt=1;tt<=t;tt++) { fun(tt); } return 0; } 
 » 3 weeks ago, # |   0 Can someone explain the DP solution for problem C in a detailed manner? Thanks!
 » 3 weeks ago, # |   0 In B, I never knew such formula existed for xor and I got TLE for using the naive approach. But then in the end, I formulated my own formula for xor and achieved O(1) time. I am so noob in coding, but to know that I constructed a formula that I never knew existed gives me some motivation. Its my dream to at least solve 3 problems during one contest
 » 3 weeks ago, # |   +3 D can be done in $O(n*log_{10}(s))$ as well with simple recursion. https://codeforces.com/contest/1567/submission/127976931
 » 3 weeks ago, # |   0 In problem C(Carrying Conundrum), Could anyone please tell Why we are subtracting 2?
•  » » 3 weeks ago, # ^ |   0 The two cases of one of the number in pair being 0
•  » » » 2 weeks ago, # ^ |   0 Got it. Thank You.
 » 3 weeks ago, # |   +21 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:
 » 3 weeks ago, # |   0 How 3+4+100+4=110 in base 11 in the third test case explanation? when 70+27=97 in base 11 in the first test case explanation? Am i missing any knowledge? Please share it! Thanks!
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 I got it now. Think like how we do normal base 10 addition9 + 1 = 10 where 1 is carry and 0 is the sum at that position. Similarly, In base 11, 4+4 = 8 + 3 = 9 + 2 = 9 + 1 + 1 = A + 1 = 10 where 1 is carry and 0 is sum at that position. Note: 9 + 1= A because in base 11, digits are {0, 1, 2,...9, A}See this https://math.tools/table/addition/base/11 Similarly think for binary addition (Base 2 addition). You will definitely understand it.
 » 3 weeks ago, # |   +5 Special Thanks to flamestorm for problem-C !!!
 » 3 weeks ago, # |   0 I really loved problem E. I have written an implementation of it by using two extra pieces of information at each node l and r, where l and r are the range that this node covers. I did this because I felt that the implementation of the query function in the editorial was hard for me to understand and hence I wrote this code which only changes the merge function and the query function is not changed: 128038196
 » 3 weeks ago, # |   0 Thank you for that round, the problems were really interesting. BUT!!! Maybe C and D should be replaced)
 » 3 weeks ago, # |   0 In Problem B, the facts that we use are for precalculating XOR from 1 to n are mismatching with the facts that I have.https://www.geeksforgeeks.org/calculate-xor-1-n/Can somebody confirm if it's wrong in the above link?
•  » » 3 weeks ago, # ^ |   0 I had messed up with a bracket in my code, giving me an erroneous result. rather than the above could you check how the 2 statements below are calculated differently?(d^b)!=ad^b!=a
 » 3 weeks ago, # |   0 E is the hardest to me.
•  » » 3 weeks ago, # ^ |   +9 E was pretty standard. You should practice some good segment tree problems that require custom merge operations. A good one is here : 739C - Alyona and towers.
 » 3 weeks ago, # |   0 Why are we doing -2 in (a+1)(b+1) in the editorial of problem C.
•  » » 3 weeks ago, # ^ |   +1 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 :)
•  » » » 3 weeks ago, # ^ |   0 got it thanx!!
 » 3 weeks ago, # |   0 I guess sol(obs) for C is wrong somewhere... because 15 * 90 = 1350 instead of 1950.Please correct me if I'm taking the solution in the wrong way!
 » 3 weeks ago, # |   0 can anyone explain query part of problem E
 » 3 weeks ago, # | ← Rev. 2 →   0 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.
 » 3 weeks ago, # |   0 Problem C Submission with complete Newbie Implementation https://codeforces.com/contest/1567/submission/128091033
•  » » 2 weeks ago, # ^ |   0 Hey, Can u please help in one sample test case of problem c the last one (10000) how its 99 ,I am getting only 81 cases with 2 case only Refer this image if u need According to me The blue ones underlined will have no contribution, and C means it will generate carry and NC means it will not generate carry Any help will be appreciated
•  » » » 2 weeks ago, # ^ |   0 Check the cases they should be 10X9+9X1=99
•  » » » » 2 weeks ago, # ^ |   0 Thanks man, appreciated
 » 2 weeks ago, # |   0 Nice round bro . I think E is easier than C
 » 2 weeks ago, # |   0 who can explain the C by using dp,the translate as if some trouble
 » 2 weeks ago, # |   0 Can someone explain D in simple words? I have looked at both the editorial and the video but don't seem to grasp the idea.
 » 2 weeks ago, # |   0 why the video solution can not connect
 » 2 weeks ago, # |   0 color[u] = (color[v] ^ 3); What is the purpose of this code?
 » 13 days ago, # |   0 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]]; } } } 
 » 12 days ago, # |   0 Problem E:My SubmissionGiving wrong answer on test case 3 128573992what's wrong in my code?