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# Feedback Of February Challenge 2018

 4 I think codechef should start providing the test cases once the contest is finished. It will help us more in finding our mistakes. answered 12 Feb, 18:14 3★tony7 41●2 accept rate: 0% @admin contest was very good .. enjoyed solving it. but please provide test cases after the contest. (13 Feb, 00:32) 1 Hi, this was discussed many times particularly in this thread: https://discuss.codechef.com/questions/491/will-codechef-provide-test-cases . Codechef does not provide test cases since it spoils the real fun. :) (13 Feb, 00:59)
 3 can anyone tell how the bitset soln passd. Not only it has memory complexity N^2 but also time complexity of N^2. On the other hand i dont think testcases were very weak as my https://www.codechef.com/viewsolution/17342876 O(Nsqrt(N)log(sqrt(N))) algo failed. Or did i miss something? answered 13 Feb, 11:02 1.8k●6●14 accept rate: 21% Yeah, I thought of the solution using bitsets too, but disregarded it thinking the time complexity of the solution is $O(N^2/\text{WORD_SIZE})$ ~ $6.25*10^8$ operations which should give TLE. But I guess codechef judge is quite fast. (13 Feb, 18:30) I think overall complexity of using bitset was around 3.5 * 10^8 (including test cases) as the optimization factor is of 64 (N*N / 64) , correct me if i am wrong ? (14 Feb, 15:04) apptica5★
 2 Regarding the gradient of the contest , I feel the first 5 problems were good as per their difficulty level but 6th,8th,9th and 10th problems should have been a little harder than they were, especially the 9th problem which require only a greedy approach. answered 13 Feb, 11:44 5★abx_2109 275●1●11 accept rate: 0%
 2 I too agree with @abx_2109,probably that was because the alternative solution was much simpler than the desired solution(like bitmask for CHANOQ,greedy for 9th sum),if the memory limit were strict,then it could have been a more balanced contest. But overall,liked it. answered 13 Feb, 17:44 1.5k●2●9 accept rate: 23% I really didnt't think of an alternate solution for CHANOQ and that too by bitsets and a lot of people did it that way. Overall I too liked the contest. (14 Feb, 01:21) abx_21095★
 1 How come this solution to the question Chef and odd queries: https://www.codechef.com/viewsolution/17408003 that uses a bitset of total size 10^10 bits (i.e. 1250 MBs) passed, where normally memory limit is 256 MB or 512 MB in long challenge questions. Also, the memory limit is not mentioned at all in the problem! This solution solves the problem without any segment tree or sqrt decomposition but just using bitsets. I think this kind of solutions having O(N^2) memory are not supposed to pass for chef and odd queries. @admin @markysha answered 12 Feb, 16:07 53●4 accept rate: 0% codechef's memory limit is 1.5gb (12 Feb, 17:42) No! It is unlimited. This should be fixed soon. https://www.codechef.com/JAN18/status/MONSTER?sort_by=All&sorting_order=asc&language=All&status=All&handle=lboris&Submit=GO (12 Feb, 17:46) chemthan7★ Another one (Mine) https://www.codechef.com/viewsolution/17376783 Allocate M bitsets of size N, but still passes. (12 Feb, 17:51) 1 We have the issue in our mind. We are working on to fix it. (12 Feb, 18:01) admin ♦♦0★
 1 One of the best long challenges I've ever taken. Well balanced and nice problems. answered 12 Feb, 18:03 4★orlon 38●2 accept rate: 0%
 1 Hello Community! I wanted to contribute to the community but since I don't have enough karma points, I want you to upvote only if you like my unofficial editorial on PERMPAL : Contest link:https://www.codechef.com/FEB18/problems/PERMPAL My approach: 1.The answer will be -1 only when the count of any of 26 characters is odd and there exists more than one such character's example : abbc, here we can see both 'a' and 'c' have odd count in the given string. 2.In any other case, we will always find a answer, that is pallindrome. 3.Based on above point, we will deal with two cases : ---(A)When only one character comes with odd count and rest other characters with even count. ---(B)When all the character in the given string are of even count. Now the question comes How to solve this? So it is much clear for case 3(A) that the odd count characters should be in the middle of string in order to make it pallindrome, eg abbba n(b)=3 i.e odd Since the ordering of characters won't matter in case of even count, we will print index one by one from vector> v(26), I have used this for storing the index of each character(a-z) in the string.In the output I need to just print the even count characters index ,simultaneously from beginning and end. Keep in mind that the odd count character should be in the middle. If you give up, you can see my solution here:https://www.codechef.com/viewsolution/17309552 All suggestions are welcome! Thanks Community, Happy Coding! answered 13 Feb, 12:20 162●1●7 accept rate: 0%
 1 Nice contest! It would be really great if the search for Submissions for a particular problem allowed you to filter by "pts" as well as Language and Result (All/AC/WA/...). If you want to find the fastest 100pts solution, you may have to check through 100 pages of 10pts solutions first. Also, it would be nice if the Submissions search showed you the size of the program. So you could look for short programs, which could be helpful if you wanted to find the cleanest / most instructive answer. (I suppose this would be controversial, since it would give a little extra information during a contest, but it could be interesting/encouraging to know if there is a short solution.) If the Submissions search allowed you to download the results as a csv file or something, then people could make their own views of the solutions, e.g., a scattergram of (speed,size). answered 13 Feb, 19:57 192●4 accept rate: 25%
 1 The Competition was very well balanced!! One of the best long challenges till date!! answered 14 Feb, 00:48 11●1 accept rate: 0%
 0 Thanks for posting quick editorial :) answered 12 Feb, 15:52 20●3 accept rate: 0%
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