1-gon's blog

By 1-gon, history, 13 months ago,

1345A - Puzzle Pieces

Tutorial

1345B - Card Constructions

Tutorial

1345C - Hilbert's Hotel

Tutorial

1345D - Monopole Magnets

Tutorial

1345E - Quantifier Question

Tutorial

1345F - Résumé Review

Tutorial

1344E - Train Tracks

Tutorial

1344F - Piet's Palette

Tutorial

• +586

 » 13 months ago, # |   +13
 » 13 months ago, # |   +39 Thank you for great round and fast editorial
 » 13 months ago, # | ← Rev. 4 →   +10 There is a typo in the editorial for D2C/D1A."This proves there is a collision if and only if all i + a_i are distinct" should be "This proves there is no collision if and only if all i + a_i are distinct".UPDATE: It has been fixed, but in a way different from what this comment indicates. Please read carefully before making further downvotes.
 » 13 months ago, # | ← Rev. 2 →   +185 I feel really sorry for the problemsetter.. He must have worked hard to create those problems..also this was his first round as a setter..The problems were great, thanks a lot :) :)
 » 13 months ago, # | ← Rev. 2 →   -11 For Div2B we can optimize the solution by iterating the loop from floor(sqrt(n)/2), as substituting this value in the inequality (3*h+1)*h/2 <= n always satisfies it. P.S: I don't think it affects the asymptotic complexity so it seems pretty useless now lol.
 » 13 months ago, # |   +379 I've read the problems and they're indeed very interesting. I congratulate you and thank you for your work, and really feel sorry about the issues that made this contest unrated.I want to leave an idea that has been growing on my mind. I think that given the fact that there are much fewer Div. 1 contests than Div. 2 or Div 3., in the case of "long queue" issues, one should give a priority to the judgement of Div 1. submissions. I know that if one interprets this as if it was "Div. 2 doesn't matter", it doesn't sound so nice. However, I think that there are much fewer participants in Div. 1 and it may be possible to at least try to "save that contest" when this issues occur.Only that. A suggestion. Cheers,Roberto Esquivel Cabrera.
•  » » 13 months ago, # ^ |   -75 It might be a good idea to limit the number of participants so that codeforces would work well (around 10000). Div 2 people will participate more rarely on average, say once a week instead of twice, but without long queues. The only effect on div 1 is reducing queues.
•  » » » 13 months ago, # ^ |   +71 Mike is eagerly waiting for a contest with 30k registrations.
•  » » » » 5 months ago, # ^ |   0 There we go :)
 » 13 months ago, # |   +2 tbh kind of disappointed with it going unated...tried so hard and finally got the 2nd question right after a hour.
 » 13 months ago, # |   0 what will be the quadratic formula for B?
•  » » 13 months ago, # ^ |   -6 For a given n:t = floor( ( -1 + sqrt( 1 + 24*n ) ) / 6 )then the tallest tower that can be built from n is : 2 + ( (t-1)*(3t+4) )/2keep subtracting this from n and recalculating 't' and tallest tower as long as n > 1
•  » » » 13 months ago, # ^ |   -9 Why did you not take t = floor( ( -1 — sqrt( 1 + 24*n ) ) / 6 )?
•  » » » » 13 months ago, # ^ | ← Rev. 2 →   -6 t cannot be negative, t represents the index of the sequence: 2 7 15 26 ... and floor to make sure that index doesn't point to a term > n
•  » » » » » 13 months ago, # ^ |   -10 Thanks for clarifying :)
•  » » » » » » 13 months ago, # ^ | ← Rev. 3 →   0 thought of a recursive solution and a less mathematical one for problem 1345B Just Precalculate till 10^9 and then binary search for the index of the biggest smaller number of cards than n in each iteration. Code
•  » » » 13 months ago, # ^ |   -10 How did you compe up with this formula?
•  » » » » 13 months ago, # ^ | ← Rev. 3 →   0 Series formula F(N) = N*(3*N + 1)/2..If you search you will able to find the series otherwise one can drive using the taking the test cases. I can able to find but made mistake in code I think so. Got wrong answer :'(.For reference of the seriesLink : https://oeis.org/A005449
•  » » » » » 13 months ago, # ^ |   0 Yeah thank you i understood
•  » » » » 13 months ago, # ^ |   0 More practical method is to use the recurrence from the problem and let https://www.wolframalpha.com/ solve it for you so you get n(h) (number of blocks needed to build pyramid of height h). Then you can also let it solve for the inverse and floor it to get the maximum height h' you can build using n blocks. This works because the function is increasing. Then you just do n -= n(h'). And you repeat this while keeping count.
•  » » » » 13 months ago, # ^ |   +2 If you look closely you can observe that in every structure there are complete triangles on every level except for surface. The no. of triangles are 1,1+2,1+2+3,1+2+3+4,....so on. so for nth structure there will be n(n+1)/2 no. of triangles, hence required card should be 3*n*(n+1)/2. But we know that we have to subtract base level no. of cards, hence final formula will be (3*n*(n+1)/2)-n. after rearranging this we get (3n^2+n)/2. Now you can reverse calculate highest tower n by making it equal to given number and solving the quadratic eqn.
•  » » » 13 months ago, # ^ |   -10 how those t and n equations obtained?
•  » » » 13 months ago, # ^ |   -10 Can you please explain how 2 + ( (t-1)*(3t+4) )/2
•  » » » » 13 months ago, # ^ |   -8 we can even subtract t*(3*t+1)/2 from given till given >1 right
•  » » » 13 months ago, # ^ |   0 t = floor( ( -1 + sqrt( 1 + 24*n ) ) / 6 ), i need proof of this formula,i didnt undrstand whole thing.
•  » » » 13 months ago, # ^ |   0 But this show TLE on 10^9 test case 4. I don't understand why this is happening.
•  » » » » 13 months ago, # ^ |   0 you need to change 24*n to 24LL*n.
•  » » » » » 13 months ago, # ^ |   0 Can you please elaborate upon how this works, what is the error in earlier choice It would be a great help
•  » » » » » » 13 months ago, # ^ |   0 Long in cpp has maximum value (2^31-1), so it's bad when n = 1e9 and 24*n is 1e10, it is leading to overflow. You should use long long for variables where you're not sure whether you might overflow using int/long.
•  » » » 13 months ago, # ^ |   0 why can't the tallest tower be t
•  » » » 10 months ago, # ^ |   0 floor isn't necessary as we are dealing with integers, in case of cpp implicit typecast will take place.
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 a much easier to understand formula is (trng(n)*3)-n where trng is the nth triangular number. Notice that there are 1, 3, 6, 10... triangles in the structure and each triangle requires 3 cards. Then we subtract n cards, as we do not need cards at the bottom.
•  » » 13 months ago, # ^ |   -11 I had a different solution for B, you can see that for pyramids of height 1,2,3,4.. you need 2,7,15,26..cards respectively and so on. Now if you see the difference between them it is 5,8,11-->this series is in A.P with Common difference of 3 so you can just precompute this array of needed cards and then start from back of this array, building the pyramid with 'ith' height if it is possible until you no longer can.
•  » » 13 months ago, # ^ |   0 I computed this: For the cards to be placed upright: n*(n+1) For the cards to be placed as base: n*(n-1)/2 So, to construct a pyramid of height n, cards required will be: n*(n+1) + n*(n-1)/2.
•  » » » 13 months ago, # ^ |   0 After this, I precomputed upto 1e5 and the performed binary search for each test, solving it in O(logn).
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 i know a better approach to find the formulafor a height h there will be 2*h no. of cards at the base there will be total of (h-1)*((h-1)+1)/2 no. of triangles now just run a loop till n<2 https://codeforces.com/contest/1345/submission/79162987 link to my submission
 » 13 months ago, # |   -13 Not expecting such a technical problem from codeforces server. Even still many submissions are not judged.
•  » » 13 months ago, # ^ |   -52 Go cry to your Mama. No one asked you to come here. If you are here, just be grateful that so many intelligent people are trying to make interesting contests free of cost.
•  » » » 13 months ago, # ^ |   -11 You are not on the list of those intelligent people so why don't you practice your own advice.
 » 13 months ago, # |   +14 loosed +160 delta
 » 13 months ago, # |   +5 The problem set was really nice, looking forward for more rounds from you.
 » 13 months ago, # |   -8 I am totally confuse dev2 A question.... Plz somebody plzz help me....
•  » » 13 months ago, # ^ |   +3 In this problem, when you take 2*3 grid you see that you can not solve that puzzle, therefore you can not solve similar grids which has higher dimensions than 2. That means, if either row or column is 1 it can be sokved and if both are 2 it can be solved. Otherwise not. I hope now you understand.Sorry for my bad english..
•  » » » 13 months ago, # ^ |   0 Chy_Chy.....thank brother..
 » 13 months ago, # |   +3 Really great round but I am sad for this nice problem became unrated.
 » 13 months ago, # |   +35 For F, I went with a much more natural approach to me, where for each position we have 4 variables $r_i, y_i, b_i, e_i$ (I hope their meaning is clear, e.g., $r_i=1$ if i-th position is red). YB operations etc are basically swapping variables. Since result of each mix operation is white if all colors occur with the same parity and a proper color if one of them has a different parity than the others two, each mix operations produces 2 equations modulo 2 corresponding to the difference between parities of some two colors. However we have a nasty condition that exactly one out of $r_i, y_i, b_i, e_i$ should be 1 that can't really be expressed modulo 2, which seems like a big problem. However, what happens if we put there an equation $r_i+y_i+b_i+e_i=1$? We can get a faulty solution where $3$ out of these bits are lit. BUT, this system of equations has a very peculiar property that for every $i$, every equations contains an even number of variables $r_i, y_i, b_i, e_i$, so if we flip all of them then this will still be a solution! Hence if 3 of these variables are true then we can change them so that 1 of them is true and if we do this for every $i$ where it is needed then we will get a solution of equation system corresponding to the valid coloring!
 » 13 months ago, # |   +67 The setting of 1344C - Quantifier Question is very nice. Thank you for the problemset!
 » 13 months ago, # |   +3 Can somebody explain div2C, I can't get my head around the proof ?
•  » » 13 months ago, # ^ |   +16 consider a subset of rooms, from room 0 to n-1. let n = 5. CASE 1 let's assume a0 = 0, a1 = 1, a2 = 2, a3 =3, a4 = 4. Now, the guest in room 0 will go to -> 0 + a0 = 0 similarly guest in room 1 will go to -> 1 + 1 = 2 room 2 -> 2 + 2 = 4; room 3 -> 3 + 3 = 6; room 4 -> 4 + 4 = 8;room 5 -> 5 + 0 = 5; room 6 -> 6 + 1 = 7; room 7 -> 7 + 2 = 9; room 8 -> 8 + 3 = 11; room 9 -> 9 + 4 = 13; .... and so on.No guests get the repeating room {explained later}CASE 2 let's consider a0 = 5, a1 = 7, a2 = 5, a3 = 8, a4 = 6; for room 0 -> 0 + 5 = 5; room 1 -> 1 + 7 = 8; room 2 -> 2 + 5 = 7; room 3 -> 3 + 8 = 11; room 4 -> 4 + 6 = 10;room 5 -> 5 + 5 = 10; {gets repeating room} ******in both cases, we can observe an arithmetic progression is forming . after every n__Now if we observe both the cases, we can see that if, for a particular value of 'i' b/w 0 to n-1 if there exists a j, such that aj + j lies in the arithmetic progression of (ai + i), it implies that there will be a collision. This is because, a time will come when becomes (aj + j), for example in case 2, for i = 0 we can see that ai = 5 and i = 0; so the first term of the arithmetic progression is ai + i = 5 + 0 = 5; now we'll check for j = 1 to n-1, whether there is such a j for which aj + j lies in the arithmetic progression of ai + i with difference = n; so for j = 4, aj = 6, and aj + j = 10; so aj + j clearly lies in the arithmetic progression of ai + i ( 5 + 1*5 = 5) {ai + i + (x-1)*n = xth;}this way we have to check for every i, that, whether there is a number which is in ap with ai + i and diff = n; this will take O(n*n). To optimize it we can just check whether there are more than one such numbers such that (ai+i)% n == (aj+j) % n because ai + i + (x-1)*n = aj + j; taking modulo both sides we'll end up with (ai+i)% n == (aj+j) % n and hence O(n).
•  » » » 13 months ago, # ^ |   0 Thanks a lot!! just one thing that arithmetic progression of n also means that one should also be a multiple of other. So we just need to check if the taking the mod of (i+a[i]) gives same value as any other index.
•  » » » » 13 months ago, # ^ | ← Rev. 3 →   +4 suppose we take n =3, a0 = 0, a1 = 1, a2 = 2; room 0 -> 0 + 0 = 0room 1 -> 1 + 1 = 2room 2 -> 2 + 2 = 4room 3 -> 3 + 0 = 3room 4 -> 4 + 1 = 5room 5 -> 5 + 2 = 7room 6 -> 6 + 3 = 9 ........ and so ontherefore room0, room3, room6, room9.....are in apsimilarly, room1, room4, room7, room10 ..... are in apand room2, room5, room8 .... are in apif any of the rooms in 2nd and 3rd row in this example are common with any room in first row {means are present in ap of first row,} Collison will occur. similarly if any rooms in 3rd row are common with any room in second row {means they are present in ap of 2nd row.} Collison will take place.
•  » » » » 13 months ago, # ^ |   0 yes
•  » » » 13 months ago, # ^ | ← Rev. 2 →   +9 This comment is by moha.jain credit to him for the explanation. and I re-wrote it to make it more readable. Consider a subset of rooms, from room $0$ to $n-1$. let $n = 5$. CASE 1: let's assume $a_{0} = 0$ , $a_{1} = 1$, $a_{2} = 2$, $a_{3} = 3$, $a_{4} = 4$. Now, the guest in room $0$ will go to $0 + a_{0} = 0$ similarly guest in room $1$ will go to $1 + 1 = 2$ room $2$ go to $2 + 2 = 4$; room $3$ go to $3 + 3 = 6$; room $4$ go to $4 + 4 = 8$;room $5$ go to $5 + 0 = 5$; room $6$ go to $6 + 1 = 7$; room $7$ go to $7 + 2 = 9$; room $8$ go to $8 + 3 = 11$; room $9$ go to $9 + 4 = 13$; .... and so on.No guests get the repeating room explained later CASE 2: let us consider $a_{0} = 5$, $a_{1} = 7$, $a_{2} = 5$, $a_{3} = 8$, $a_{4} = 6$; for room $0$ go to $0 + 5 = 5$; room $1$ go to $1 + 7 = 8$; room $2$ go to $2 + 5 = 7$; room $3$ go to $3 + 8 = 11$; room $4$ go to $4 + 6 = 10$;room $5$ go to $5 + 5 = 10$; gets repeating room In both cases, we can observe an arithmetic progression is forming . after every $n$Now if we observe both the cases, we can see that if, for a particular value of $i$ b/w $0$ to $n-1$ if there exists $a_{j}$, such that $a_{j} + j$ lies in the arithmetic progression of ($a_{i} + i$), it implies that there will be a collision. This is because, a time will come when becomes ($a_{j} + j$).For example in case 2:for $i = 0$ we can see that $a_{i} = 5$ and $i = 0$; so the first term of the arithmetic progression is $a_{i} + i = 5 + 0 = 5$; now we will check for $j = 1$ to $n-1$, whether there is such $a_{j}$ for which $a_{j} + j$ lies in the arithmetic progression of $a_{i} + i$ with difference equals to $n$; so for $j = 4$, $a_{j} = 6$, and $a_{j} + j = 10$; so $a_{j} + j$ clearly lies in the arithmetic progression of $a_{i} + i$ ($5 + 1 \times 5 = 10$) => $a_{i} + i + (x-1)\times n$ $=$ $xth$ this way we have to check for every $i$, that, whether there is a number which is in ap with $a_{i} + i$ and $diff = n$; this will take $O(n \times n)$. To optimize it we can just check whether there are more than one such numbers such that $(a_{i}+i)$ $mod$ $n$ $\equiv$ $(a_{j}+j)$ $mod$ $n$ because $a_{i} + i + (x-1) \times n = a_{j} + j$; taking modulo both sides we'll end up with $(a_{i}+i)$ $mod$ $n$ $\equiv$ $(a_{j}+j)$ $mod$ $n$ and hence $O(n)$.
•  » » » » 13 months ago, # ^ |   0 if n = 5, in your case 2, does not room 0 goes to room 0 + a[0] % 5 = room 0 ? Why does guest in room 0 goes to room 5?
•  » » » » » 13 months ago, # ^ |   0 I think you are missing that the mod operation is on the subscript and not on $a_{i}$if $n = 5$, then guest $0$ would go to $0 + a_{0 mod 5}$ $=$ $0 + a_{0}$ $=$ $0 + 5 = 5$
•  » » » » » » 13 months ago, # ^ |   0 Ahhh, I see.
•  » » » » 13 months ago, # ^ |   0 A little typo in this line $a_{i} + i (5 + 1 \times 5 = 5)$.It should be $a_{i} + i (5 + 1 \times 5 = 10)$.
•  » » » » » 13 months ago, # ^ |   0 thanks fixed
•  » » » 13 months ago, # ^ |   0 I solved by your approach but getting the wrong answer. Can you please tell me where am I wrong? https://codeforces.com/contest/1344/submission/79319693
•  » » » » 13 months ago, # ^ |   0 you have to take into account the mod of negative number with na[i] = a[i] + i; a[i] = (a[i])%n; if(a[i]<0){a[i] = a[i] + n;}
•  » » » » » 13 months ago, # ^ |   0 still wrong
•  » » » » » » 13 months ago, # ^ |   +1 int main() { int t;cin>>t; while(t--) { long long int n; cin>>n; long long int a[n],i; unordered_sets; for(i=0;i>a[i];a[i] = i + a[i];} for(i=0;i
•  » » » » » » » 13 months ago, # ^ |   0 thanks a ton.
•  » » » » » » » 12 months ago, # ^ |   0 if(a[i]<0){a[i]=a[i]+n;} can you explain this step plz
•  » » » » » » » » 12 months ago, # ^ |   0 modulus of a negative number can be written as a %m = (a + m)%m example: -1 % 20 == (-1 + 20)%20; =19;
•  » » 13 months ago, # ^ |   0 Consider the number line divided into segments of size n i.e [-1*(n) -2 -1] [0 1 2 ...n-1] [n n+1 n+2...2*n — 1] so now when you add to i + a(imodn) each i converts to some j which will be a index of one of these segments definitely.. What i am trying to say is that.. For example 1)n==3 Consider applying shuffling operation of segment [0..2] and we have a1,a2,a3 such that after shuffling 0->3,1->7,2->11 now 3 is actually at 0th position in its segment [3,4,5] 7 at 1st in [6,7,8] and 11 at 2nd position in [9,10,11] so in this example above we can see that each element results into an unique positions ,whatever be the segment it settles So for such kind of example always the answer is YES!!But consider this example 2) n==3 and we have a1,a2,a3 such that 0->3,1->6,2->11 so now 11 is still the 2nd position but 3,6 both are ones at 0th position Therefore such array a will result in generation of elements in which no 1st position element is ever created but the 1st position elements get converted to 0th position So Vacancy will be definitely created.. So for such kind of example the answer is NO!! * HOPE YOU UNDERSTAND..
•  » » 5 weeks ago, # ^ | ← Rev. 3 →   0 I know it's late but I think this is easier (it might help someone) We know that a collision occur if for some $0 \leq i,j \leq n-1$ where $i \neq j$, the following condition holds true $i + k_1n + a_i = j + k_2n + a_j$ because $x+yn \equiv x \mod n$ where $0 \leq x \leq n-1$ Now, $i+a_i-j-a_j = (k_2-k_1)n$$\implies i+a_i-j-a_j \equiv 0 \mod n$ $\implies i+a_i \equiv j+a_j \mod n$
 » 13 months ago, # |   0 Can anyone explain why all numbers being distinct will not affect values of k>n?
•  » » 13 months ago, # ^ |   0 k refers to any integer and not any index of the array
•  » » » 13 months ago, # ^ |   +3 As because the rooms were infinite, my question was whether rooms with k>n would collide with those with k
 » 13 months ago, # |   0 Can someone explain C again?The shuffle isn't k + a[k mod n]?
•  » » 13 months ago, # ^ |   -9 The shuffle is k + (a_k mod n). So, for example, if a_1 = 12 for n = 7, room 1 will go to room 1 + (12 mod 7) which is room 6. Room 8 will also go to 8 + (a_1 mod 7) since 8 = 1 mod 7, so room 8 will go to room 8 + (12 mod 7) which is room 13. In general, room 1 + kn will go to room 1 + kn + (a_1 mod 7) = 1 + kn + 5 = 6 + kn. Hope this was helpful.
•  » » » 13 months ago, # ^ | ← Rev. 2 →   0 Oops, I made a slight error. It is actually k + a_(k mod n). So in the example I gave where a_1 = 12 and n = 7, room 1 goes to room 1 + a_(1 mod 7) = 1 + a_1 = 13, and room 8 goes to room 8 + a_(8 mod 7) = 8 + a_1 = 20, and in general room 1 + kn goes to room 1 + kn + a_((1 + kn) mod n) = kn + 1 + a_1 = kn + 13.
•  » » » » 13 months ago, # ^ | ← Rev. 2 →   0 I think it is room 1 + kn + a_((1 + kn) mod n)
•  » » » » » 13 months ago, # ^ |   0 Yes, that's what I meant :).
•  » » 13 months ago, # ^ |   0 Yes, but notice that he is taking $0 \leq i < n$ so we have $i$ $mod$ $n = i$. The same goes for $j$.
•  » » » 13 months ago, # ^ |   0 I know that, but that proof from the tutorial: i+ai≡j+aj(modn), why everything mod n ?
 » 13 months ago, # |   -8 Solved 3 problems before round was announced to be unranked. Never started that good :(
•  » » 13 months ago, # ^ |   -8 Same with me. Solved A,B,C within 20 minutes. I had a significant increase in rating even if I hadn't solved any other problem.
•  » » » 13 months ago, # ^ |   -8 same to me haha,long time to escape from the level 'pupil' qvq
 » 13 months ago, # |   0 In Div2C, I have a doubt. What I did was, for each a[i], found (k+a[k mod n]), till all are positive, i.e., for every number a[i], found what (k+a[k mod n]) will be if positive and kept pushing these values to a vector. Now,if the vector has a GCD greater than 1, it means, there will be duplicates. Else, no. Can you tell what is the flaw with this?
 » 13 months ago, # |   +11 The decision of making this round unrated is highly appreciable. Many of us waited to check if our solution was correct before solving a new one. Feeling sorry for the problem setter/s. He/they must have worked so hard but a bug ruined all. Hope mike will find a way through this.
 » 13 months ago, # |   0 Could anyone please tell the approach using binary search for 1345B?(card construction)
•  » » 13 months ago, # ^ |   0 Idea : I will try to find the max height possible with given cards. If all of the cards are used my answer is 1, else I will find nearest height to make possible. It's not hard to derive the formula for number of cards vs height. I think my implementation is very understandable so check out my solution, I have used goto function.https://codeforces.com/contest/1345/submission/79190502
•  » » » 13 months ago, # ^ |   0 Thanks for the solution! Got the idea:)
 » 13 months ago, # |   0 Thanks a lot for a quick editorial :)
 » 13 months ago, # |   0 Can someone explain E in simpler terms? The concept of topological sorting is confusing me here. I understood the part of the graph being acyclic, but didn't get what they did next.
•  » » 13 months ago, # ^ |   +19 Topological sorting isn't needed in this problem. The idea is, look at the numbers from a_1 to a_N in order. When you get to an index k, how do we know if we can put an A there? It's only possible if a_k is "not comparable" to any other a_i where i
•  » » » 13 months ago, # ^ |   0 why we can not fill all cell with south pole question D/B Div2/Div1
•  » » » » 13 months ago, # ^ |   0 I think you replied to the wrong comment. But to answer your question:Your solution has to have the property that all white squares are unreachable. If you put south pole magnets everywhere, then every square will be reachable by a north pole magnet, which is bad.
•  » » » » » 13 months ago, # ^ |   0 thanks for your reply i got my mistake . yes i replied to wrong comment because before this i asked two times and no one replied so i find random people who was online ,and u was the first one
•  » » » 13 months ago, # ^ |   0 Thank you so much. That was a great explanation.
•  » » » 13 months ago, # ^ |   0 Thanks, I was thinking the same way but topological sorting got me really confused that what if I'm wrong. Now it's clear.
 » 13 months ago, # |   0 Div 2 CExample:33 2 1In the fourth test case, guests 0 and 1 are both assigned to room 3.20 1In the test case, guests 1 and 2 are both assigned to room 2.Why we have guest 2? I see only 2 rooms — 0 and 1. If k=0 and n=2, k+a[k mod n]=0+a[0]=0, so guest 0 still sit in room 0. And other guest assigned to room 2. Where is my mistake?
•  » » 13 months ago, # ^ |   0 Read the problem statement carefully. There are infinite rooms.
•  » » » 13 months ago, # ^ |   0 guests 1 and 2 are both assigned to room 2 It means, that we have 3 guests: a[0],a[1],a[2], but at the beginning we have only a[0] that will be assigned to room 0 and a[1] that will be assigned to room 2.Each of the two guests has their own room so we must return YES, but in example true answer in NO,because extra 3rd guest are in room 2 with guest 1
•  » » » » 13 months ago, # ^ |   0 in fact,we can assume that we have infinite room,each room contain a guest at the beginning.(we could think that there are infinite guests as well},so once we use the constraint to re-organize the position,all of the guest (infinite of course) will change there position by k+a[k mod n].so the example is ok whatever
 » 13 months ago, # |   0 Can someone help me in figuring out why Div2D test 50 fails? Seems like I'm counting connected components incorrectly, but I can't seem to figure it out.. .https://codeforces.com/contest/1345/submission/79197607
•  » » 13 months ago, # ^ |   +1 try f.second+1 < m
•  » » » 13 months ago, # ^ |   0 Ahh thank you for that suggestion :D
 » 13 months ago, # |   +9 In div2 E, let's say we add egde from j's->k's(in_degree[j's]++. Assuming resulting graph is acyclic then we can make those vertices as universal if their in_degree is equal to 0. Can someone tell me what's wrong with my logic? Below is the link to my code. https://codeforces.com/contest/1345/submission/79209006
•  » » 13 months ago, # ^ |   +15 Suppose I have a graph with two nodes and an edge $2\to 1$. Node $2$ has indegree $0$, yet the statement $\exists x_1,\forall x_2, x_2 < x_1$ is false.
•  » » » 13 months ago, # ^ |   0 Thanks,I didn't focused on order of xi's.
 » 13 months ago, # |   +3 For Div2.C, I tried a (supposed) N^2 solution and it got accepted: https://codeforces.com/contest/1345/submission/79212988 I suppose it can be proved that it will never actually do so many iterations. Otherwise, maybe test cases are not strong enough.
•  » » 13 months ago, # ^ | ← Rev. 3 →   +12 Hacked it with testcase: 1 200000 0 0 0 0 0 ... 
•  » » » 13 months ago, # ^ |   0 haha really
 » 13 months ago, # |   +3 Does anyone know why in div1C,the answer of the testcase below: 3 2 1 2 3 2 is 1 AEE ?shouldn't it be 2 AEA???
•  » » 13 months ago, # ^ |   +5 The problem is that, x3 < x2 Now you have fixed x2, (As There exists x2 comes first in order) , so not all x3 holds x3 < x2.A requirement for universality is that the variable is only comparable with larger-indexed variables. It is mentioned in the 2nd paragraph of editorial.
•  » » » 13 months ago, # ^ |   0 thanks and I missed that part of requirement XD
 » 13 months ago, # |   0 In the sol. for Div2E/1C, I don't understand thisFor each variable, we can find the minimum index of a node comparable to it by doing DP in forward and reverse topological order.Please explain this.
•  » » 13 months ago, # ^ | ← Rev. 2 →   +15 The requirement for a variable x(i) to be universal is that, In the directed acyclic graph, No node x(j) with j < i can reach x(i) and x(i) also cant reach such x(j). In other words, all nodes reachable from x(i) and reachable to x(i) should have higher labels.To check that, we do DP and find minimum below and minimum above in reversed graph.
•  » » » 13 months ago, # ^ | ← Rev. 2 →   0 Can you explain this testcase:5 41 21 35 14 1The correct answer is: AEEEE.I can make x2 and x3 anything and set x1 as min(x2,x3), I can set x4 and x5 according to x1 , So won't the answer be 2 EAAEE. What am I missing here?
•  » » » » 13 months ago, # ^ |   +6 Problem Statement says : Note that the order the variables appear in the statement is fixed. For example, if f(x1,x2):=(x1
•  » » » » » 13 months ago, # ^ |   0 Oh I see.Thanks a lot for this!!
•  » » » 13 months ago, # ^ |   0 Can I rephrase it as The element with the smallest index in every chain of the poset gets 'A' rest all 'E'?
 » 13 months ago, # | ← Rev. 2 →   0 Dang, I really like Div2F. I had the right idea for it but didn't implement because I thought it was wrong and it was unrated anyways.Great problems...too bad there were technical difficulties.
•  » » 13 months ago, # ^ |   0 Could you please explain the idea behind this question?
•  » » » 13 months ago, # ^ |   0 Which part of the tutorial don't you understand?
•  » » » » 13 months ago, # ^ |   0 Simply binary search on the value of A so that we increment exactly k times. To compute the cutoffs for the x values, we can either use the quadratic formula or do another binary search. There may be ties for the Δi(x) values, but this can be handled without too much trouble. -> this part..
 » 13 months ago, # | ← Rev. 2 →   0 Started well off however for some reason, for div2D my submission solves all the test cases correctly on my IDE however does not seem to even get the correct answer for the very first test case when I submit it to the online judge. Any help would be appreciated — cheershttps://codeforces.com/contest/1345/submission/79218763
•  » » 13 months ago, # ^ | ← Rev. 4 →   0 I get two warnings when compiling that, both which might lead to undefined behavior (I got the answer 6 when running it on my computer). One is that size is ambiguous (size is a std function you included) and the second is that k isn't initialized. I compile with -Wall -Werror -Wpedantic (and I have it so Vim + YouCompleteMe highlights such errors).
 » 13 months ago, # | ← Rev. 2 →   0 can someone tell why is this not optimal solution Question D div 2
•  » » 13 months ago, # ^ |   0 please some one reply
•  » » 13 months ago, # ^ |   0 What's your meaning, The 3*3 grid is all colored white?
•  » » » 13 months ago, # ^ |   0 that image is solution (according to me) of first sample test case
•  » » » » 13 months ago, # ^ |   0 This is wrong answer. Because it's not possible for a north magnet to occupy black cell after some sequence of operations from the initial placement.
•  » » » » » 13 months ago, # ^ |   0 yup i got my mistake thanks
•  » » » » » 13 months ago, # ^ | ← Rev. 3 →   0 Why in the first place do we need to place north poles in the grid? Or did I missed something? Can't the minimum no. of north poles be 0 in all cases except -1?
•  » » » » » » 13 months ago, # ^ |   0 The 0 is a acceptable answer. We need minimum the number of North poles.
 » 13 months ago, # |   +10 Poor Monogon.But the problems are really great.Thanks!
 » 13 months ago, # | ← Rev. 5 →   0 I think there is something wrong with the Div2 C for it's type. $(k+a_k){\bmod n}$ instead of $k+a_{k\bmod n}$And does anybody tell me why k is in[0,n — 1]?
•  » » 13 months ago, # ^ |   0 actually k is included in R.because we have infinite room.once we reorganize the position,we are able to let the k pos guest move to k+a[k mod n] position
•  » » » 13 months ago, # ^ |   0 o,It's my fault. But I cannot understand this code why need we mod n in (i + a[i] % n + n) % n; and why i in [0, n — 1]
•  » » » » 13 months ago, # ^ | ← Rev. 5 →   0 Let's divide the R into segments,each segment contain n elements;We find that for each ith element,it transforms to the (i + a[i] % n + n) % n th in one segment;if every i in [0,n-1] obey the rule -- each transformation make unique pos in those segment,then for each new position of element,it won't be the same place(not violate the rule the problem statement show)
•  » » » » » 13 months ago, # ^ | ← Rev. 2 →   0 as for i in [0,n-1],we can make sure that the other segment(such as [n,2n-1])is also follow the ruleactually,the range of i we choose could be the set which contain n successive integer
 » 13 months ago, # |   0 I had the right idea for Div2 C. Not sure where my error is, I'm pretty sure this checks if all the rooms are distinct, and it works for almost all the test cases, but fails on a couple for some reason...http://codeforces.com/contest/1345/submission/79207288Any help would be appreciated.
•  » » 13 months ago, # ^ | ← Rev. 3 →   0 Fixed your code the number you module is negative,which may cause problems
•  » » » 13 months ago, # ^ |   0 Thank you! That was in fact the problem. I fixed it by replacing the insert line withtoAdd += j; toAdd %= n; if(toAdd < 0) toAdd += n; a.insert(toAdd);I'm still a little confused though, because I thought my conditional (toAdd + j < 0) ? ((toAdd + j) % n) + n : (toAdd + j) % n took care of negatives. I'm new to C++ so maybe I'm misunderstanding what that expression does or how the % operator works. How was it possible for my previous code to return a negative mod?
•  » » » » 13 months ago, # ^ |   0 % doesn’t do the mathematical mod, it does remainder. you have to add in the modulus to get it to be the mathematical modulus.
•  » » » » » 13 months ago, # ^ |   0 What do you mean by add in the modulus? Isn't n the modulus there? Sorry I'm a little new to programming lol.
•  » » » » » » 13 months ago, # ^ |   0 The modulus is the number that you're "modding" by. It's like the divisor in division; when you do $15\ /\ 5$, you call $5$ the divisor.What I mean by adding it in, is consider $-9\ \%\ 5$. The result of this expression in C++ is $-4$, but in math terms, what we want is $1 \equiv -9 \pmod{5}$. So what we can do, is when we can add $5$ to the original expression (that resulted in $-4$) to get what we want, which is $1$.Basically, if you want to compute $a \pmod{b}$, where $a$ can be negative, you write ((a % b) + b) % b
•  » » » » » » » 13 months ago, # ^ |   0 But the expression a % b only returns the wrong value if a is negative, right? So if I have the conditional (toAdd + j < 0) ? ((toAdd + j) % n) + n : (toAdd + j) % n it checks if a is negative (in this case a is toAdd + j), and returns the bolded part where I do in fact add n (the modulus). If it's not negative I don't need to add it, I think. I'm still slightly confused. But thanks for your help!
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 Like this? 79238791.I guess the problem is toAdd + j can't be negative, which makes the statements behind that meaningless.
 » 13 months ago, # | ← Rev. 2 →   0 Div2E/Div1C:WA on pretest 5 -I tried my to make a directed graph and check for a cycle. If it is acyclic, if some vertex has non-zero indegree, I give it an E, else it has an A. What is wrong with this logic? https://codeforces.com/contest/1345/submission/79197644
•  » » 13 months ago, # ^ |   0 Check this out. Link
•  » » 13 months ago, # ^ |   0 I also WA on pretest5 in contest. When I look other's AC code, I am wonder about test 15. In my program I get 3 AEAEA Look at pretest 3, and 1 2 is can be ignored. I wonder why test15's answer is like 2 AEAEE 
•  » » » 13 months ago, # ^ |   0 Well, now I know why. wwwwwwwwwwwwwwwww
 » 13 months ago, # |   +10 So unfortunate that the contest had to be unrated. The problems felt good and interesting. Hope the problem gets fixed today
 » 13 months ago, # |   +3 In Div1B/ Div2D, why can't we fill all cells with South pole? 1-gon
•  » » 13 months ago, # ^ |   0 yes i have the same question
•  » » 13 months ago, # ^ | ← Rev. 2 →   +1 We can put a south pole magnet in any black cell. Out of the white cells, we can only put south pole magnets at intersections of completely white rows with completely white columns. If a row (or column) contains at least one black cell and we put a south pole magnet in one of its white cells, then the south pole magnet will be able to pull a north pole magnet from the black cell to the white area at some point, since each black cell must be reachable by a north pole magnet by the problem requirement. While all white cells must be unreachable by north pole magnets in any possible sequence of moves.
•  » » » 13 months ago, # ^ |   0 Nicely explained!!
•  » » » 13 months ago, # ^ |   0 Why in the first place do we need to place north poles in the grid? Or did I missed something? Can't the minimum no. of north poles be 0 in all cases except -1?
•  » » » » 13 months ago, # ^ |   0 Because every black cell must be reachable by a north pole magnet, it is rule #2 from the problem statement. So we need at least one north pole in each black connected component (area composed of adjacent black cells). The only case when 0 north pole magnets is the right solution is when the entire grid is white.
•  » » » » » 13 months ago, # ^ |   0 Thanx for responding. It was written that it is possible for the N pole to reach the black cell and not must, and that's why I had such doubt. Don't you think the problem statement was a bit confusing in the way that it used possible instead of must there?
•  » » » » » » 12 months ago, # ^ |   -16 Such a shit statement. I fucking urge contest writers to learn some basic English before writing problems. Time wasters like this really annoys me.
•  » » 13 months ago, # ^ |   +1 Because North pole can't occupy white cell, and North pole can reach the cell which be occupied by South pole if they are in same Column or Row.
•  » » 13 months ago, # ^ |   +1 Consider this situation. Spoiler....#....If you fill all cells with South pole and put a North pole in the centre cell, you cannot satisfy rule 3.Actually pretest 2 is a good example and you should check it out.
•  » » 13 months ago, # ^ |   0 because a sequence of operations must exist for every black cell such that it ends up being occupied by a north magnet
 » 13 months ago, # |   +1 I think it's better to add a link below the contest announcement. Just personal advice.
 » 13 months ago, # |   0 Resume Review has some problem in explanation"However, we can observe that this process increments the values as long as Δi(x)≥A for some constant A"Shouldn't it be "However, we can observe that this process increments the values as long as Δi(A)≥0 for some constant A." I.E. We find a value for 'x' such that the change function Δi(x), is positive till that value of x
•  » » 13 months ago, # ^ | ← Rev. 3 →   0 I TOO DIDN'T UNDERSTAND THE TUTORIAL CLEARLY
•  » » 13 months ago, # ^ |   0 See whatever i observe i think it is right for some constant A not 0 exactly cause there is this constraint sum of all b[i]'s should be exactly k , that's y As can be seen in the first test case example the value of A comes out to be -32 so it justifies
 » 13 months ago, # |   0 I don't understand this statement. Please help me."For each variable, we can find the minimum index of a node comparable to it by doing DP in forward and reverse topological order. Then for every variable not comparable to a smaller indexed variable, let it be universal. All other variables must be existential. Our requirement of universality proves this is optimal."
•  » » 13 months ago, # ^ |   +1 I'm assuming you are familiar with the application of dynamic programming on DAG. First, you need to find a topological sort of the graph (it exists only if the graph doesn't contain any cycle). Now do a forward traversal on the topological sort and for every value 'i' calculate the minimum value of the node that appeared on a path that ends at vertex 'i' (while moving forward). This can be calculated easily by DP. Now do a backward traversal on the topological sort (also reverse the edges) and separately calculate the minimum value of the node that appeared on a path that ends at vertex 'i' (while moving backward). Now for each node, you have the minimum value of the node that is somehow connected to this node. If this minimum value is less than the value of the node then you cannot assign a universal quantifier to the node (let's say a node is 2 and the min value is 1, then either x1 < x2 or x2 < x1. In either case you cannot assign A to 2), otherwise you can assign the universal quantifier to this node. See 79647809
•  » » » 13 months ago, # ^ |   0 Thanks
 » 13 months ago, # |   0 Though I got the tutorial and the logic for prob B but I'm not able to come up with the code to implement the logic. If anyone could provide their code if they have done it the way described in the tutorial.
 » 13 months ago, # | ← Rev. 2 →   -7 [contest:Codeforces Round 639 Div B] 1345B - Card Constructions I have a very easy approach for this problem !!formula for no. of cards with height (h) = (n*(3*n + 1))/2 This formula takes you to the status accepted : How ?Link to my solution : Do watch the solution , it's damn easy with efficient approach !!!
•  » » 13 months ago, # ^ |   0 That ide is trash. Can't even see the full solution.
•  » » » 13 months ago, # ^ |   0 79240516 check this out
 » 13 months ago, # |   0 https://codeforces.com/contest/1345/submission/79177038 I got wa in this and it gets accepted as soon as I put (n+(i+a[i]%n))%n here https://codeforces.com/contest/1345/submission/79242347 can anyone tell me why is that I mean n+(i+a[i])%n should always give the same value as (n+(i+a[i])%n)%n can anyone help me with this
•  » » 13 months ago, # ^ |   0 what if n = 3, i = 1, a[i] = 2?
•  » » » 13 months ago, # ^ |   0 it will only go in the condition when (i+a[i]) is less than zero
•  » » » » 13 months ago, # ^ | ← Rev. 2 →   0 Ok what about i=1, a[i]=-4, n=3? My point is that you are ignoring the case when I+a[I] is divisible by n.
•  » » » » » 13 months ago, # ^ | ← Rev. 3 →   0 (i+a[i])%n cannot be greater than equal to zero
•  » » 13 months ago, # ^ |   0 In C++ (-1%5) will be -1.That is why you have to compute ((-1%5) + 5)%5 which will give 4 as output.More on it here
•  » » » 13 months ago, # ^ |   0 but 5+(-1%5) will give the same output
•  » » » » 13 months ago, # ^ |   0 This is what you have done one the wrong submission.m[n + (i + a[i]) % n];This is your code from the correct submissionm[(n + (i + a[i]) % n)%n];Assume n=10,a[i]=10,i=0.You would end up getting 10 for the incorrect code, and 0 for the correct one :)
•  » » » » » 13 months ago, # ^ |   0 it is only going in that condition when i+a[i] is less than zero
•  » » » » » » 13 months ago, # ^ |   0 In that case assume i=0, n=10 and a[i]=-100 :)
•  » » » » » » » 13 months ago, # ^ |   0 oh yeah sry i got it
 » 13 months ago, # |   0 Thanks for your great problems. Your probs and solutions are wonderful. Keeps it up!
 » 13 months ago, # |   +2 Is it only me or did someone else also found div2C confusing? I loved the question but took me a long long time to decode the meaning.
 » 13 months ago, # |   +5 can any one help me out why i am getting wa on 17th test case??79245621
•  » » 13 months ago, # ^ |   +1 I think you have not considered the case where some rows and columns might be empty Test case3 5 ##... ..... ...## answer should be 2 fill the empty middle row by S.
•  » » » 13 months ago, # ^ |   0 **Thanks a lot it helps **
 » 13 months ago, # |   0 Then the shuffling works as follows. There is an array of n integers a0,a1,…,an−1. Then for each integer k, the guest in room k is moved to room number k+a [k]modn. this is written in problem C, does this mean (k+a[k]) mod n, k + a[k] mod n, or k + a[k mod n]?. I read the editorial, it's supposed to be (k+a[k]) mod n. but k mod n has the same font size which makes it a bit confusing.
 » 13 months ago, # |   +5 I am not able to find the error in my solution for 1345D. My Solution Link = https://codeforces.com/contest/1345/submission/79249181It is giving WA on test 17 Please Help Thanks
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 I think you have not considered the case where some rows and columns might be empty Test case3 6 ##.... ...... ....## answer should be 2
•  » » » 13 months ago, # ^ |   0 Okay Thankx a lot brother
 » 13 months ago, # |   +1 can someone explain me fifth test case of input given at div2 C problem. How the first and second guest are having the same room no. 2 after suffling?
•  » » 13 months ago, # ^ |   +1 guest 1: $1 + a_{1 \bmod 2} = 1 + a_1 = 1 + 1 = 2$guest 2: $2 + a_{2 \bmod 2} = 2 + a_0 = 2 + 0 = 2$
 » 13 months ago, # |   0 I am asking about the fifth test case of the first test case which is given in question?
 » 13 months ago, # |   0 I got stack overflow error in Div2D during the contest, then I tried using thread as I read it somewhere. That too isn't working for me.Maximum recursion depth in my code is 10^6. Can anybody tell me how to resolve this in java!
 » 13 months ago, # |   0 that description of fifth test case of 1st test case is given wrong in question i think!!
 » 13 months ago, # |   +4 Div2 C detail explanation with example and code here in case anyone need :)
•  » » 13 months ago, # ^ |   +1 Amazing buddy it helped so much to solve this problem :) Keep going on like this for every difficult problem . thanks for this
 » 13 months ago, # |   0 Why is this MLE? 1-gonIsin't the answer same as the editorial : 79185068
 » 13 months ago, # |   0 Div2F ai−3x2+3x−1!=ai-3x(x+1)-1 but why do work ai-3x(x+1)-1?? anyone help please
 » 13 months ago, # |   0 Can someone please explain div 2 problem F?
•  » » 13 months ago, # ^ |   0 I also need the detailed explanation on div2F/div1D. Please someone explain this problem..
 » 13 months ago, # |   0 https://codeforces.com/contest/1344/submission/79310940 (1345D Monopole Magnets)Does anyone know what exit code -1 means? I'm not sure what is causing the runtime error. Please help. I seem to be outputting the correct answer so I am very confused.
•  » » 13 months ago, # ^ |   0 Change return -1 to return 0. Exit code is whatever the main function returns. If it does not return 0, then it becomes a runtime error.
•  » » » 13 months ago, # ^ |   0 Omg, thank you! I don't think I would've figured that out, and the internet wasn't being much help.
 » 13 months ago, # |   0 For the problem — Monopole Magnets, it is written:- "every row and every column has exactly one segment of black cells or is all-white"From what I'm understanding, there should be continuous black cells within a row or a column. No white between them.So, if I have a 3x3 square with the middle cell (1,1) white and all other cells black. Then I can have a solution with four south magnets at the four corners ((0,0), (0,2), (2,0), and (2,2)).Can anyone tell me if it's correct or I'm making some mistake?
•  » » 13 months ago, # ^ |   +1 You are forgetting that there has to be at least one south magnet in every row and every column. In your solution, there is no south magnet in neither row 1 nor column 1. Also, both of these have two segments of black cells, so answer should be -1.
•  » » » 13 months ago, # ^ |   +3 Yeah, thanks. I don't know why I was forgetting this condition for this test case.
 » 13 months ago, # |   0 Thanks for the great contest And yes, feeling sad for Monogon as this contest became unrated. Sorry for the comment. BUT Am i the only one who got think the problem div2C was written wrong? As it was written in problem that guest in room k will be shifted to room k+a[k mod n] but after seeing the tutorial, i come to know that it should have been ((k+a[k]) mod n). Sorry for poor english. Plz reply Monogon
 » 13 months ago, # |   0 In Div2F/ Div1D Resume Review Problem, """Simply binary search on the value of A so that we increment exactly k times""". How exactly the binary search is giving us the optimal solution? Can anyone please explain it with some example testcases?
•  » » 13 months ago, # ^ |   +1 You can think of it as what is the minimum increase we are willing to use a move for. If it's too high, than we are not using all our moves, and if it's too low, then we are wasting moves.
•  » » » 13 months ago, # ^ |   0 For example, if we take a = 5 what value of (bi) I should take and why because it is always decreasing and there is no minimum ? What is the basis of minimum ? Is it value of K?
•  » » » » 13 months ago, # ^ | ← Rev. 2 →   0 The value of $b_i$ depends on $a_i$ for other types also. Try the sample test case. For each type of project, find the score for each value of $b_i$. Then find the answer for each value of $k$ from $1$ to $\sum a_i$.for each value of k find the minimum increase you got from a project you added. You'll then understand how binary search was used here.
 » 13 months ago, # | ← Rev. 2 →   0 why am i getting TLE ?? https://codeforces.com/contest/1345/submission/79351507
 » 13 months ago, # |   0 I think problem B can be solved more easily. I just formed a formula, 3(n+1)n/2-n. Suppose n=1, then the 1st highest pyramid will consist of 3(1+1)1-1 = 2 cards. Here n is the number of i'th pyramid. The 2nd highest pyramid will consist of 3(2+1)2-2 = 7. You can see my code and will get the general idea. https://codeforces.com/contest/1345/submission/79398607
•  » » 13 months ago, # ^ |   0 If you calculated this,then that is great but it can also be found on https://oeis.org/search?q=2+7+15+26&language=english&go=Search and this website is great for finding any type of pattern.
 » 13 months ago, # |   0 Can anyone explain why in div 2 problem c solution after getting i + a[i] , we find modulus of this term? Why is it required even one number is negative it will be on left side and not repeated please if anyone can explain this properly.
 » 13 months ago, # |   0 In Div 2 F, what if there isn't an $A$ which can make $b_i$ increase exactly $k$ times?
 » 13 months ago, # |   0 In question Hilbert and Hotel What's the difference in value of q between q=((i+p)%n+n)%n; and q=(i+p)%n;when I am writing q=((i+p)%n+n)%n it passes all cases but with q=(i+p)%n it is not passing all test cases.This is my Code unordered_map m; ll p; int flag=1; for (ll i=0;i>p; ll q=(i+p)%n; if (m[q]==1)flag=0; else m[q]=1; if (flag)cout<<"YES\n"; else cout<<"NO\n";Thanks in Advance!!
 » 13 months ago, # |   0 Hello! I'm new here. Don't know if this is the best place to post this but I'm having wrong answer on test case 43 for D. Haven't been able to debug. Thanks for any help! https://codeforces.com/contest/1345/submission/79673927
 » 13 months ago, # |   0 This was a very nice contest with very nice questions. Here are my solutions. I have two questions for 1-gon The editorial for $E$ was a little confusing. Am I right in saying that the number of universal identifiers is equal to the number of connected components as each connected component can have exactly 1 universal identifier ? Can you share the process behind how you came up with the Div $2$ $F$ ? How you formulated it and arrived at the quality function ? It was quite nice.
 » 13 months ago, # |   0 Great Problems Well done Monogon Please write more problems on dsa you are a great setter!
 » 13 months ago, # | ← Rev. 2 →   0 In Hilbert's Hotel problem.... I looked at the fifth case where n=2 and the guest array is [0, 1]here after reshuffling both do not fall on the same room no 2 i think see the calculation below: ____ for guest 0: new position is 0+ a[0 mod 2] = 0 + a[0] = 0 + 0 = 0___ for guest 1: new position is 1+ a[1 mod 2] = 1 + a[1] = 1 + 1 = 2___ie both are not falling in the same room.but the answer given is NO. why? they both are in 2 different rooms... the answer must be YES. please help me
 » 13 months ago, # | ← Rev. 2 →   0 Simplified logic for Div2C:BasicallyThere are infinite elements-INF to INFlet K goes from (-INF to INF){ let x = k + a[k % n]; which means room k goes to room x; }If (every k goes to different x) ans="YES" else ans="NO"which means k1==>x and k2==>x then ans="NO"vi b(n);Let us consider K running from 0 to n-1{ b[k] = k + a[k]; since k % n = k }case1: if b[i]==b[j]: x values are same in the range (0 to n-1) x1==x2 where x1=k1+a[k1%n] where k1 lies in (0 to n-1) x2=k2+a[k2%n] where k2 lies in (0 to n-1) ans=NO case2: if b[i]%n == b[j]%n : b[i]=i+a[i] b[j]=j+a[j] if (b[i]%n == b[j]%n): (i+a[i])%n = (j+a[j])%n (i+a[i]) = i+a[i]=j+a[j]+tmp*n (i+a[i]) = (j+tmp*n)+a[(j+tmp*n)%n] (j+tmp*n)+a[(j+tmp*n)%n] is x value for k=(j+tmp*n) (i+a[i]) is x value for k=i same x value for two k values ==> ans=NO case3:  other than case1 and case2: ans=YES x1==x2 is not possible `
•  » » 13 months ago, # ^ |   0 thanks
 » 12 months ago, # |   0 1345/problem/B-----> if n=15 the I can make of h=3 and h=2 but in the test case I the output is given as 1...
 » 12 months ago, # | ← Rev. 2 →   0 in C of div 2, it is mandatory to check that if i randomly take contiguous segment of n numbers then they should be shifted such that their rooms are still in contiguous allocation(mod n), (which shows the first condition (i+ai)%n) am i right??whats about second condition??
 » 11 months ago, # |   0 My solution is failing on test 50 for Monopole Magnets. LINK
 » 6 weeks ago, # |   0 Thanks for the beautiful tutorial.