### vovuh's blog

By vovuh, history, 4 years ago,

All problem ideas belong to MikeMirzayanov. And he prepared all problems himself. I just helped with testing, reviewing and tutorial writing.

1352A - Sum of Round Numbers

Tutorial
Solution

1352B - Same Parity Summands

Tutorial
Solution

1352C - K-th Not Divisible by n

Tutorial
Solution

1352D - Alice, Bob and Candies

Tutorial
Solution in O(n)
Solution in O(n^2)

1352E - Special Elements

Tutorial
Solution

1352F - Binary String Reconstruction

Tutorial
Solution

1352G - Special Permutation

Tutorial
Solution
• +133

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 » 4 years ago, # | ← Rev. 3 →   +159 so some of the participants are going to reach 1800+ ratings ? i believe this shouldn't be happening. CHECK THIS OUT![ ]()edit : this is what cf predictor is saying.
•  » » 4 years ago, # ^ |   +17 Obviously many experienced users made new accounts just to increase their ratings and Mike asked people not to do this but they won't listen. This is pretty bad and unintended.
•  » » » 4 years ago, # ^ |   0 The contest is rated for only those people who have given at least 2 rated rounds I guess. Please correct me If I am wrong
•  » » » » 4 years ago, # ^ |   +4 It is rated for unrated people too
•  » » » » » 4 years ago, # ^ |   +3 bro but i have not been rated and iu solved 1 problem
•  » » » » » » 4 years ago, # ^ |   +3 The ratings haven't been updated yet
•  » » 4 years ago, # ^ |   +52 I think there should be some limit on the max increase possible for this round. Otherwise, it is very unfair for participants with rating >= 1400.
•  » » » 4 years ago, # ^ | ← Rev. 4 →   -27 Removed, I realize I am wrong
•  » » » » 4 years ago, # ^ |   +15 "unrated" starts off with 1500, then he gets 375, Add them up, what do you think?
•  » » » » » 4 years ago, # ^ | ← Rev. 7 →   -33 Removed, I realized rating matters.
•  » » » » » » 4 years ago, # ^ |   +14 "Regardless of whether you are a trusted participant of the fourth division or not, if your rating is less than 1400 (or you are a newcomer/unrated), then the round will be rated for you." - mike
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   +3 i agree that the big boost is bad but i don't think anyone who didn't deserve at least 1600-1700 rating could solve all the questions in like 30 mins. but 1800 rating is a bit yikes (which is the +300 rating), so i agree that the max gain should be lowered to around 250 at MOST (for when you solve it as fast as the reds). the problem lies more in smurfing though
•  » » » 4 years ago, # ^ |   +4 Also an important thing to notice is that out of the top hundred official participants, there are more than 60 users who have created a new account and these users increase as we go for further ranks, so this makes it even more difficult for the green/gray coders to increase their ratings.
•  » » 4 years ago, # ^ |   0 extension not working
•  » » 4 years ago, # ^ |   +6 Just wanna know, how accurate this CF PREDICTOR have been in the past ??
•  » » » 4 years ago, # ^ |   +8 Completely!
•  » » » 4 years ago, # ^ |   +1 VERY. like only a difference of <=+5 or >=-5 is what I observed from the actual rating change in every contest I sit in.
•  » » 4 years ago, # ^ |   +8 But even if someone with a low rating became Expert after this round, their rating will fall again drastically after a div 2 round due to the number of problems they will solve. So it doesn't make that much of a difference I guess!
•  » » » 4 years ago, # ^ |   +8 I agree with you,rating is not all,what the most important is we can improve ourselves by the codeforces
 » 4 years ago, # |   0 Can anyone explain the binary search solution for problem C?
•  » » 4 years ago, # ^ |   +2 The number of integers not divisible by $n$ among numbers $1, 2, \dots, x$ is $x - \lfloor\frac{x}{n}\rfloor$, so you can do binary search to find suitable value of $x$ (such smallest $x$ that $x - \lfloor\frac{x}{n}\rfloor < k$).
•  » » 4 years ago, # ^ |   0 You may see my binary search implemented solution. 79500682
•  » » 4 years ago, # ^ |   +1 Binary search Code...  int n,k; cin>>n>>k; int lo=1,hi=1e10; int res; while(lo=k){ if(mid%n!=0) res=mid; hi=mid; } else{ lo = mid+1; } } cout<
•  » » 4 years ago, # ^ |   0 Hi, I have solved the ques in O(1) time without binary search. You can see here 79680203
•  » » » 4 years ago, # ^ |   0
 » 4 years ago, # |   +1 In problem C, how need=(k-1)/(n-1) ?
•  » » 4 years ago, # ^ |   +6 This can be explained more easily if you understand that before each multiplier of $n$ there are $n-1$ integers not divisible by $n$, so we need to "skip" some number of such blocks. $\lfloor\frac{k}{n-1}\rfloor$ doesn't match because when $k$ is divisible by $n-1$ then we don't need to count the last block.
•  » » » 4 years ago, # ^ |   0 won't number of such multipliers be k / n ? if n = 3 and k = 8 there are 8 / 3 multipliers of 3
•  » » » » 4 years ago, # ^ |   0 Yeah, but we want the numbers that are NOT divisible by n.
•  » » » » » 4 years ago, # ^ | ← Rev. 3 →   0 for n=3, k=81,2,4,5,7,8,10,11 k-1/n-1 +k gives Ans=8+3=11However, a simple k/n gives Ans=8+2=10I guess, there is a more complex logic of using k-1/n-1. Can anyone explain it in a simple way? Stuck at this for quite some time.
•  » » » » » » 4 years ago, # ^ |   +3 okay so idk how simple this is but like n = 3 k = 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15so k/(n-1) is kinda like accounting for the extra cycle, because you need 1 less N to traverse a cycle (a cycle is 3 numbers, you only need 2 indices in this case to traverse the cycle). then k/(n-1) accounts for that extra N every time it cycles through (if u write it out on paper it makes sense lol)and if the cycle perfectly lands (when k%(n-1) ==0) then you gotta subtract one because you overcounted a cycle. for example, in the n= 3 and k = 8 case, if you did this cycle counting normally you'll end up with 12, but because the cycle cleanly divides you gotta subtract 1. at least i think that's how you do it idk it made sense when i wrote it out and went through casesand of course catch edge cases k == n and k < n
•  » » » » » » 4 years ago, # ^ | ← Rev. 2 →   +1 From 1 to N, there are N-1 numbers not divisible by N. In every block of N numbers, N-1 are not divisible by N. So if we want K numbers, there must be at least $\frac{K}{N-1}$ blocks. Each block with N numbers. Let the number of blocks be B. This will take you to $B\times N$. So, from 1 to $BN$, you have $B\times(N-1)$ bad numbers. Count how many more do you need and add that to your current value. If you are already complete, then $BN-1$ is your $K^{th}$ Number.I hope I'm not making this complex. Code For Reference
•  » » » » » » » 4 years ago, # ^ |   0 i wonder how vovuhs solution is correct :)
•  » » » » » » » 4 years ago, # ^ |   0 nice dude, beautiful explaination...thank you!!!
•  » » » » » » 4 years ago, # ^ |   0 I got stuck at this for an hour and ruined my contest. D was definitely easier than C and it is unfair to put it after C.
•  » » » » » » » 4 years ago, # ^ |   0 I think the difficulty is subjective. I couldn't do B for a long time, switched to C, and I felt it was easier. I totally agree that D was easy to understand but the implementation can get a bit complicated. C was thinking based and if you get the trick, it becomes really easy to implement.
•  » » » » » 4 years ago, # ^ |   0 "Each multiplier of n shifts our answer by 1. The number of such multipliers is need = k — 1 / n — 1" so he finds numbers such that are divisible by n
 » 4 years ago, # |   0 I attempted to solve F using dp[n1][n2][n3][ends] : answer to the original problem when you have n1,n2,n3 and the resulting string ends in "ends" ("0" or "1"). My Python solution seems to TLE. Is it because this solution is basically O(10^6) and the number of test cases <= O(10^3), overall O(10^9)? or am I doing something silly Spoilerdef recurse(x, y, z, end, dp): if x == 1 and y == 0 and z == 0: if end == "0": return 0 else: return None if x == 0 and y == 1 and z == 0: if end == "1": return 1 else: return 10 if x == 0 and y == 0 and z == 1: if end == "0": return None else: return 11 if (x, y, z) in dp: return dp[(x, y, z)] res = None if end == "0": if x > 0: res = recurse(x - 1, y, z, "0", dp) if res != None: res = res * 10 if y > 0: res = recurse(x, y - 1, z, "1", dp) if res != None: res = res * 10 else: if y > 0: res = recurse(x, y - 1, z, "0", dp) if res != None: res = res * 10 + 1 if z > 0: res = recurse(x, y, z - 1, "1", dp) if res != None: res = res * 10 + 1 dp[(x, y, z)] = res return res for _ in range(int(input())): x, y, z = map(int, input().split()) res = recurse(x, y, z, "0", {}) if res == None: res = recurse(x, y, z, "1", {}) if len(str(res)) - 1 != x + y + z: diff = x + y + z + 1 - len(str(res)) print("0" * diff + str(res)) else: print(res) 
•  » » 4 years ago, # ^ |   0 It's even worse than that, because there can be $c_0+c_1+c_2$ digits in each dp state, so you get something like $O(T \cdot (c_0+c_1+c_2) \cdot c_0c_1c_2) \approx 3 \cdot 10^{11}$, which will very clearly not pass.
•  » » » 4 years ago, # ^ |   0 it can be done using DP, see this: 79628831
•  » » » » 4 years ago, # ^ |   0 nice solution, just one doubt, inside the main function, how do I know whether to call solve(n0,n1,n2,1) or solve(n0,n1,n2,0), I mean both are equally likely to happen(except when we only want 0s or only want 1s in our desired string). Sometimes, I stuck with such doubts and hence unable to decide which way to call recursive dp functions. Can you explain why in main(), you called solve(n0,n1,n2,0) first and then when s.length() was 0, then you decided to call solve(n0,n1,n2,1). Rest is fine to understand.
•  » » » » » 4 years ago, # ^ |   0 I assume you understand DP state, using 0 means string with ending at 0 which has n0,n1 and n2 count, but if there is no such string let's calculate one that ends with 1
•  » » » » » » 4 years ago, # ^ |   0 ok, so you're first guessing whether it can be done with string ending at 0, if not then you try it with string ending with 1. Right?
 » 4 years ago, # | ← Rev. 2 →   +1 this round was so good and i had so much fun as a pupil and learnt new things i hope u do more div4 as u can <3
 » 4 years ago, # |   0 Can we do Div4-E with segment trees?
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 You dont need to use it and it will most probably give MLE . Segment trees take huge amount of space and building and quering them would take additional logn time which would prove to be costly .
•  » » » 4 years ago, # ^ | ← Rev. 2 →   +4 Segment trees take O(n) space. MLE won't be an issue, I guess. It is just that you don't wanna use a sword for cutting lemons.
•  » » » » 4 years ago, # ^ |   0 But TLE might be an issue for sure . MLE not sure but still a bad choice for 64MB question XD.
•  » » » 4 years ago, # ^ |   +9 Segment trees take 4*n space. So it won't be a problem with MLE.
 » 4 years ago, # |   0 How to solve problem G using DFS (Depth First Search) ? I tried by joining numbers whose absolute difference satisfies the given condition and then traversing. How can I modify to make it work.?
•  » » 4 years ago, # ^ |   +8 Checking all possible paths would lead to an exponential solution I think. There can be cycles too [2->4->6->2] (>_<).Maybe as someone mentioned above, don't cut lemons with swords.
•  » » 4 years ago, # ^ | ← Rev. 2 →   -7 removed
 » 4 years ago, # |   0 Could someone try to hack 79593227? It does O(n * n * log(n)), which should be cut by the strict time limit but I couldn't hack it.
 » 4 years ago, # |   0 Can anyone explain the tutorial of the B problem in a much simpler way?
•  » » 4 years ago, # ^ |   0 essentially, odd numbers can only be made with odd numbers, and an odd amount of numbers at that.even numbers can be made with any amount of even numbers and an even amount of odd numbers.then you set all the numbers to 1 or 2 (depending on if you want the array to be even or odd based on those rules) and then the last number is the amount that ur left with after subtracting all of the 1s and 2s
 » 4 years ago, # |   +3 I think my solution to G is a bit interesting. For all quadruples of natural numbers till the floor(n/4)th tuple can be arranged in order : 2 4 1 3, eg. 2 4 1 3 6 8 5 7 10 12.... and for the last incomplete tuple, it is easy enough to insert them in between elements of the last complete tuple.
 » 4 years ago, # |   0 can anyone tell me, how to solve G using bfs/dfs?
•  » » 4 years ago, # ^ | ← Rev. 4 →   0 The vertex set can be represented as the set of all numbers from 1 to n. The adjacent numbers for a node 'x' would be {x-4,x-3,x-2,x+2,x+3,x+4} provided they are in between 1 and n inclusive. Now you can perform DFS. The answer would be the DFS which satisfies the given condition in the problem. (NOTE: Edited: It will not be the DFS which visits all vertices...since the graph formed is connected, each DFS will visit all vertices) Even if you do a DFS from each number, the complexity would be O(n^2) (this include the time for checking the validity of the current DFS), which I guess would pass the problem constraints.
•  » » » 4 years ago, # ^ |   0 i tried it earlier but its not working.
•  » » » » 4 years ago, # ^ | ← Rev. 3 →   0
•  » » » » » 4 years ago, # ^ |   0 Saw your code, inside the dfs() for a given node, you are try all possible relevant nodes to jump. I saw that you are only pushing nodes into your path array. Why ain't you popping the elements off, if you don't find the relevant path? Also, why in your main function are you checking val<2 && val>4, ain't you pushing the relevant nodes into you path array. You yourself first built the graph with reachable nodes, why is that condition checking even necessary?Kindly Help.
•  » » » » » » 4 years ago, # ^ |   +3 I am not trying all possible nodes to jump. Take for example, N = 4, if you draw the graph the edges will be 1-3, 1-4, 2-4. If I start a DFS from node 1, the path array will contain elements in the following order {1,3,4,2}, which is not a valid answer. The correct answer would be the DFS that starts from node 2, {2,4,1,3}. A DFS generates a Spanning Tree, not a path. The path array is just storing the sequence in which the depth first traversal occurs and it may or may not be a path. In order to check whether it is a valid path (both in terms of the definition of being a path and the given condition of the problem), by just checking the given condition of the problem, I would be able to arrive at the solution.
•  » » » » » » » 4 years ago, # ^ |   0 Wouldn't the if(val<2 || val>4) always hold false because you built the graph in way that only relevant nodes are traversed and after the dfs() call is over, you are again checking the same thing. Shouldn't you rather be checking if the lenght of path array is n or not. Because if it is n, then we always know that the path will be relevant(again because that's how you built the graph array and we will never be able to traverse irrelevant nodes(that do not hold the condition given in the question)). If you agree with me, then I still do not understand why without checking the path length that your solution is passing plus why that rechecking of (val<2 || val>4) condition when that's how your dfs function will traverse the nodes? Kindly explain.
•  » » » » » » » » 4 years ago, # ^ |   +3 Please refer to my previous reply. In the example I have given, both have a length of n, but only DFS from node 2 is a valid answer. Refer to this link for better understanding.
•  » » » » » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 You didn't get me, you stored nodes in the adjacency list such that they atleast differ by 2 and atmost by 4, then why are you doing checking it again after the dfs call. The dfs traversal will take care of jumping only to the relevant nodes. You said that the path array may contain {1,3,2,4} can you tell me how did you path store 2 after 3, when you don't have 2 in the adjacency list for 3?
•  » » 4 years ago, # ^ |   +2 Check it out! 79806425
 » 4 years ago, # |   -8 In the tutorial of problem D it is mentioned that we could solve the problem in O(n^2) but isn't the required complexity O(n) ? Correct me if i am wrong.
•  » » 4 years ago, # ^ |   0 O(n^2) is the time complexity and O(n) is space complexity. In this question space is unusually restricted so space complexity should not exceed O(n) but time complexity of O(n^2) works just fine
•  » » » 4 years ago, # ^ |   0 I am talking about problem D not E.
•  » » » » 4 years ago, # ^ |   0 I think you overestimated the running time of O(n^2) in this problem.Simply (n^2)*T or (SUMn)^2 might not give the correct complexity in this problem. Just try using lagrange or some differential method and we get that worst case is when T=200 and all Ni = 1000, which can be optimized to 1e8 executions:Link
 » 4 years ago, # |   0 The solution of G in tutorial is very good but how to come up with this solution? I mean the intuition for the tutorial solution.
•  » » 4 years ago, # ^ |   0 I played with small permutations, trying to find a working block small enough so I could replicate it, and built the big N-elements answer. For me, it turned out to be a block of the first 4 numbers:2 4 1 3For N = 4 this is good. So as I said, I tried to make it work like a "brick" to build the answer, making chains of this block. So, if you have N = 8, you can "link" two of these structures:2 4 1 3 — 6 8 5 7And so on. Then I worked out on the case when N = 5 (more generally, when N%4 = 1), and I noticed that simply adding n to the sequence is ok.For N = 6, or N%4 = 2, you have to pop the last element (3) of your sequence, and mix the remaining two elements:2 4 1 5 3 6And lastly for N = 7 or N%4 = 3, an extra step, also removing the last element and mixing the remaining three elements:2 4 1 5 7 3 6
 » 4 years ago, # |   0 Am I the only one who solved problem G by graph?
•  » » 4 years ago, # ^ |   0 Can you explain your approach
•  » » » 4 years ago, # ^ |   0 Make a graph of N nodes numbered 1 to N. Then for each node i add bidirectional edges between node i and i+2,i+3,i+4. After that Run dfs from each node separately, and check only for the straight path going from the node to any leaf node, if that path covers all N nodes then answer is the node of the path in the same order, otherwise if don't get any such path at the end, then answer doesn't exist.
•  » » » » 4 years ago, # ^ |   0 I saw in your code that in your dfs(), you have used break statement. If we don't get a relevant path by traversing from the given node then don't you think that then we should try for another node rather than giving yup our hopes that we won't find the right path if the very first node is unvisited. Moreover, in that case the path vector (in your code-v) will be not remain global but rather be passed in the function(in non-reference way)Kindly correct me, as to why your code used break?
•  » » 4 years ago, # ^ |   0 I just saw your solution for G, where you visit a node if it's not visited and then break out of the loop. What if you start from Node 1, then visit Node 3 and find out that permutation starting from 1->3 is not possible but permutation starting from 1->4 is possible ? You are not visiting node 4 from 1 as you have a break statement.
•  » » » 4 years ago, # ^ |   0 What you are saying is theoretically not possible, can you give any test case?
•  » » » » 4 years ago, # ^ | ← Rev. 3 →   0 Yes, it seems like it's not possible after checking few cases. But can you prove that your solution always finds a solution if it exists? Also why is it theoretically not possible ?
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 ok, nice, just one question, is that due to the order you store nodes in the adjacency list for a given node or what? Will that still hold if you change the order of placing elements in some random order inside the adjacency list?
•  » » » » » 4 years ago, # ^ |   0 Did you figure it out ?
•  » » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 Actually, changing the order of elements does not change the answer. But, I still didn't get the break; in the dfs in the for loop. If you got that, kindly reply.
 » 4 years ago, # |   -8 is this contest unrated?
•  » » 4 years ago, # ^ |   +5 Rated for participants below 1400 rating
 » 4 years ago, # |   -8 Can anyone explain in Problem E why am I getting TLE when i am using map to count the frequency and find them 79611958 and it accepts the solution when i m using unordered map 79613165 i think in both cases overall time complexity is O(n*n). vovuh?
•  » » 4 years ago, # ^ |   0 Map has a time complexity of O(nlogn).
•  » » » 4 years ago, # ^ |   0 still smaller than n*n
•  » » » » 4 years ago, # ^ |   0 NO it's O(n^2logn)...
•  » » 4 years ago, # ^ | ← Rev. 2 →   +1 The total complexity becomes $n^2*\log{n}$, which is higher than the expected complexity. The extra $\log{n}$ factor arises because of using map
•  » » » 4 years ago, # ^ |   0 I used unordered_set instead of map. Which should be have constant Time Complexity for inserting and finding elements. Still TLE on subtask 3.
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 I think unordered_set has a very high constant time factor because it uses hash table. Even though insertion could be O(1), the hash table updating is very slow. Therefore, the number of insertion operations that can be done in 1 second is lesser than that of typical constant time operations.
•  » » » » 4 years ago, # ^ |   0 Unordered (hash-based) sets/maps have average constant time complexity, but since the standard hash function is deterministic, a test case can be crafted specifically to cause a high number of items to be placed in the same bucket, degenerating it to a linked list.
 » 4 years ago, # |   -8 Can anyone explain the case when n1 = 0 for problem F. Weren't we supposed to output a SINGLE string??????? I assumed that n1 = 0 would implicitly mean that either n2 = 0 || n0 = 0. Wasn't that supposed to be the case?????????
 » 4 years ago, # |   -9 can anyone explain how case n0=1 , n1=0 , n2=1 will be handled in question F ??
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 there is no valid answer for cases like n1=0 and n0 and n2 being both being non zero .your submission will not be evaluated on these type of cases.
•  » » » 4 years ago, # ^ |   0 when will the ratings be updated
•  » » » » 4 years ago, # ^ |   0 After system testing is complete.
 » 4 years ago, # |   0 Is there a way of doing E using sliding windows that doesn't cause TLE? or particularly any other way of doing E, I don't like editorials' solution for E, and was thinking if there is some other way.
•  » » 4 years ago, # ^ |   0 I got accepted with two pointers (which you can relate to sliding window). My approach was going through each element of the array and finding a subsegment with a sum equal to the element, as long as the difference between the two indices $l$ and $r$ is higher than 1. Two pointers has a complexity of $O(2n)$...so you can simplfy it to $O(n)$, and doing it for each element gives a complexity of $O(n^2)$. Here's my submission: 79549243Just to be sure, I also used printf and scanf instead of cin, cout and FAST I/O, as it can be even faster with large inputs. Also note that I could have also saved the result for previous numbers to make it faster when there are repetitions, but in the worst case scenario where that wouldn't happen, it wouldn't be of much help.
•  » » » 3 months ago, # ^ |   0 had the same exact idea but had an off by one error. took a look at your code was able to fix mine. thanks for posting!
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 use this function for each elements of array ,increment ans by 1 if this function return true,take care of the function it return true for subarray of length one also but in our case it should be at least two. overall time complexity O(n*n),space complexity O(n). my sol
 » 4 years ago, # |   0 did not receive a rating even after solving a question in the contest div.4
•  » » 4 years ago, # ^ |   0 Rating will be updated soon!
•  » » » 4 years ago, # ^ |   0 I can't find my username in the final standings list.
•  » » » » 4 years ago, # ^ |   0 click on show unofficial
•  » » » » » 4 years ago, # ^ |   0 Not there either
•  » » » » » » 4 years ago, # ^ |   0 You are ranked on position 9100.
 » 4 years ago, # |   0  ll two,one,zero; cin>>zero>>one>>two; string ans = ""; bool started=false; char prev; while(two--){ if(!started){ started = true; ans += "11"; prev = '1'; }else{ ans += "1"; } } while(one--){ if(!started){ started = true; ans += "10"; prev = '0'; }else{ if(prev=='0'){ ans += "1"; prev = '1'; }else{ ans += "0"; prev = '0'; } } } while(zero--){ if(!started){ started = true; ans += "00"; prev = '0'; }else{ if(prev=='1'){ ans += "00"; prev = '0'; }else{ ans += '0'; } } } cout<
•  » » 4 years ago, # ^ |   0 1 1 8 1
•  » » » 4 years ago, # ^ |   0 What do you mean by this Eureka17?
•  » » » » 4 years ago, # ^ |   0 Your code will output 110101010100. However there is 9 pairs that have 1 "1". The reason is that you probably didn't consider the case where there is even number of pair that has 1 "1". When this happens, before you add only zeros, the last digit of sequence will be 1 (the reason is you started with 0 and you add even number of digits). Because the last digit is 1, when you add zeros, one more pair that has 1 "1" appear.
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 For a case 1 2 0, your code is producing a output 10100. In your output n2 = 3, but it should be 2.
 » 4 years ago, # |   0 Can someone help me out, in problem A my code is working fine with testcase1 in Hackerrank editor and my terminal but it is giving different output when I submit it in CForces. Submission: https://codeforces.com/contest/1352/submission/79486521
•  » » 4 years ago, # ^ |   0 submit GNU C++14 or GNU C++17
•  » » 4 years ago, # ^ |   0 I had the same problem during the round. I fixed it by replacing pow function with a variable starting in 1 outside the loop, and multiplying by 10 each cycle.
 » 4 years ago, # |   0 I have used a hashmap in problem E but still TLE.Can anyone explain why i cant get it Accepted with this solution(https://codeforces.com/contest/1352/submission/79563004).I use java,can this be a factor as it is slower than c++.
 » 4 years ago, # |   0 Consider case n1=0 separately and print the sting of n0+1 zeros or n2+1 ones correspondingly.won't this make a 01 appear in the string while there should be no such pair in the string?
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 (I kind of misunderstood what you said, so I'm rewriting my comment.)That sentence means how to handle cases like 3 0 0 or 0 0 4.For 3 0 0 you need 0000, and for 0 0 4 you should output 11111.Testcases such as 3 0 3 will not appear at all, since there must exist "01" or "10" somewhere so there is no solution for this case. The problem statement says that impossible testcases will not be given.
 » 4 years ago, # |   0 nice problems for rating<=1400....i learned a lot (mainly difference b/w sets and arrays)
 » 4 years ago, # |   0 Скажите пожалуйста почему в задаче C количество множителея не равно [n/k]?
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4 years ago, # |
0

Can anyone tell me the problem with my Codeforces Round 640 (Div. 4) Ques B. Here is my code:

# include<bits/stdc++.h>

using namespace std; int main(){ int t; cin>>t; while(t-->0){ int n,k; cin>>n>>k; if(k==1){ cout<<"YES"<<endl; cout<<n; } else if(n==k){ cout<<"YES"<<endl; for(int i=1;i<=k;i++) cout<<'1'<<' '; } else{ if(n%2==0){ if(k<=n/2){ cout<<"YES"<<endl; for(int i=1;i<k;i++) cout<<'2'<<' '; int number = n — (k-1)*2; cout<<number; } else cout<<"NO"; } else{ if(k%2==0) cout<<"NO"; else if(k>n) cout<<"NO"; else{ cout<<"YES"<<endl; for(int i=1;i<k;i++) cout<<'1'<<' '; int number = n — (k-1)*1; cout<<number; } } } cout<<endl; } return 0; }

•  » » 4 years ago, # ^ |   0 In your submission of that code, the judge says it failed in case 54 of the 2nd test, as you didn't give an answer to it, although it exists. Luckily that line of the test case is shown, and it seems that fails with 6 4. And as you can see, you can represent it with numbers 1 1 1 3.
 » 4 years ago, # |   +3 Anyone still struggling about writing a good comment?I do :(
 » 4 years ago, # |   0 My solution for 1352E had O(n^2) time and got TLE. Approach is quite similar to the one described in editorial. Perhaps the time limits are too stricthttps://codeforces.com/contest/1352/submission/79526875__
•  » » 4 years ago, # ^ |   0 The issue with your code is that you are using map, and increases the complexity of your solution to $O(n^2*n log n)$, giving TLE.
 » 4 years ago, # |   0 Can anyone tell me if there is any mathematical intuition behind the derivation of formula ⌊k−1/n−1⌋ in .1352C - K-th Not Divisible by n
•  » » 4 years ago, # ^ |   +1 for every n integers , there are n-1 integers before it which are not divisible by n; say n=4, then there are n-1=3 integers 1,2,3 which are not divisible by 4.so for after every n-1 positions,there is multiple of n . So our ans is shifted by 1; then for k postions, your ans is shifted by k/(n-1);but if k is divisble by n-1 , we dont need to count the final shift.So we can check for only k-1 pos , i.e shift= k-1 / (n-1)shift=(k-1)/(n-1); ans= k + shift ; print ans;
 » 4 years ago, # |   0 #include #include #include #include #include #include #include using namespace std; typedef long long ll; const int N = 8010; int a[N]; int cal[N]; bool is[N]; int sum[N]; int main() { int t; cin >> t; while(t --) { memset(is, 0 , sizeof is); memset(cal, 0, sizeof cal); memset(sum, 0, sizeof sum); int n; scanf("%d", &n); for(int i = 1; i <= n; i ++) scanf("%d", &a[i]); for(int i = 1; i <= n - 1; i ++) { sum[i] = a[i] + a[i + 1]; is[sum[i]] = 1; //here } for(int i = 3; i <= n; i ++) { for(int j = 1; j <= n - i + 1; j ++) { if(sum[j] != -1) sum[j] += a[i + j - 1]; if(sum[j] > n) sum[j] = -1; else is[sum[j]] = 1; } } int ans = 0; for(int i = 1; i <= n; i ++) { if(is[a[i]]) { ans ++; // cout << i << ' '; } } cout << ans << endl; } return 0; } why this code is accepted? where i mark "here",sum is out of bounds.
•  » » 4 years ago, # ^ |   0 problem E
•  » » 4 years ago, # ^ |   0 Arrays in C or c++ are allocated in contiguous memory. Let your array size is 6 and you store 8 elements in the array. Then the two extra elements may or may not be stored in the array. It depends on your memory location. If the two extra elements stored lie in another's memory location (i.e. stored in those locations where memory is already allocated) then the compiler will show some error, but if those additional elements get some unallocated memory or free memory then there will be no problem while running the program.In your case, the additional elements are getting free unallocated memory in the memory allocation (i.e. program's virtual memory), so your program is not showing any error. It varies compiler to compiler.
•  » » » 4 years ago, # ^ |   0 thank you very much
 » 4 years ago, # |   0 I feel div4B harder than 1329B - Dreamoon Likes Sequences, which is div1B. Div4B was too dirty kind of problem.
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 That depends on the perspective and the mathematical background. For me 1329B is more or less unsolvable. I remember having read the editorial some days ago but still could not instantly tell how it works.
 » 4 years ago, # |   0 I am getting WA on test case 2 of problem E Can someone help me know which test case can it go wrong? Link : https://codeforces.com/contest/1352/submission/79717403
 » 4 years ago, # | ← Rev. 2 →   -8 say please, why the solution of C isn't k+[k/n]?
•  » » 4 years ago, # ^ |   0 You Can Refer This Explaination
•  » » » 4 years ago, # ^ |   0 Thank you
 » 4 years ago, # |   0 In problem E. Special Elements, if I compute a prefix array and run two loops to take out sum and use hashing as used above to count the special elements, why does it gives TLE?When this also takes O(n^2) and the above mentioned solution also takes O(n^2) ?Does it shows TLE even if memory limit exceeds ? If not then what does the judgement says ?
•  » » 4 years ago, # ^ |   +1 You did not check the sum int k = pre[j]-pre[i]; to be in bounds of the array, but then write at that position. h[k]=0;
•  » » » 4 years ago, # ^ |   0 Thanks, got an AC.Could you please state the reason as why does not checking the bounds gives TLE and not Runtime error ?Accessing something out of bounds gives Runtime error , right? Please correct me.
•  » » » » 4 years ago, # ^ |   0 Thats the undefined part of undefined behaviour
 » 4 years ago, # |   0 For Problem F .. The constrain (0≤n0,n1,n2≤100; n0+n1+n2>0 ) . But n2 must be n2>=1 . if n0 = 1 n1 = 0 n2 = 2 output is = 00 . Why? and How !!!!!!!if n2==0 there will no such solution that can satisfy the condition.
•  » » 4 years ago, # ^ |   0 It is guaranteed that the answer for given n0,n1,n2 exists.
 » 4 years ago, # |   0 Can Anybody tell me the logic behind why l<=r and not l
•  » » 4 years ago, # ^ |   0 Because either of them will eat the candy at the index l=r. If you use l
•  » » » 4 years ago, # ^ |   0 BhavyaBhatt Suppose if Alice has eaten that candy , and According to the loop's Condition(l<=r) bob can eat the candy but he shouldn't be eating that candy because Alice has already eaten it.Am I right?
•  » » » » 4 years ago, # ^ |   0 l is the index of leftmost uneaten candy and r is the index of rightmost uneaten candy. so when l=r happens, that means all candies from index 0 to index l-1 are eaten by alice and all candies from index r+1 to n-1 are eaten by bob. so candy at index l=r will be uneaten so in the loop we should proceed. Now if alice's turn is to eat candy then l++ will be done and all candies are eaten (l>r happens) and similarly if bob's turn is to eat candy then r-- will be done and again all candies are eaten (l>r happens).
•  » » » » » 4 years ago, # ^ |   0 Thank you for the reply BhavyaBhatt.I know the concept you are talking about , but Suppose Alice ate the candy at l=r then we should have not let bob eat that already eaten candy , because if bob also eats the candy then output will be wrong because it will be counted!
•  » » » » » » 4 years ago, # ^ |   0 Yes, Bob should not be able to eat that candy and the code given in editorial won't allow Bob to eat that candy as there is (l<=r) condition in the while loops inside the if-else condition.
 » 4 years ago, # | ← Rev. 4 →   0 For F(Binary string construction), if n0=1, n1=0, n2=1, the editorial solution, gives output 00. Some other codes gives output 0011, which is incorrect. What should be the correct answer since no valid string is possible for this case.Update- Missed the statement " It is guaranteed that a solution exists."
 » 4 years ago, # |   0 Can someone explain why am I getting TLE in question D, I know its an easy problem but I checked my while loops and none of them are recurring, idk why my code suddenly stops any bad practices you see,Thanks My Solution
•  » » 4 years ago, # ^ |   0 This loop? for(k = 0;k<=i;j++) { u+=x[k]; } 
 » 4 years ago, # |   0 Problem 1352F Can someone please explain why am I getting TLE in this solutionThis is my submissionSince maximum number of elements inside the string will be always less than 500 and number of independent test case is 1000 so we can solve the question in O(t*n) time right?? But why I am getting TLE here?
 » 2 years ago, # | ← Rev. 2 →   0 my Approach for G : start from x=2-(n%2); x x+2 x+4 x+6 x+8 .. .. .. .. n (n-3) (n-1) n-5 n-7 n-9 n-13 n-15 notice that we swapped the two values after n mysubmission
 » 2 years ago, # | ← Rev. 2 →   0 In problem C, why is $need = ⌊(k-1)÷(n-1)⌋$, and not $need = ⌊k÷n⌋$?
 » 6 months ago, # |   0 Shittiest constraints ever for problem E.
 » 5 months ago, # |   0 how can I solve "C" with binary search?
 » 3 months ago, # |   0 i did not understand this part of this 1352C -** K-th Not Divisible by n** The number of such multipliers is need=⌊k−1/n−1⌋ , where ⌊x/y⌋ is x divided by y rounded downcan someone explain?