### Golovanov399's blog

By Golovanov399, 4 months ago,

Sorry for the issues with a couple of problems

Problem A of tc/div2 (Finding Sasuke)
Problem B of tc/div2 (A New Technique)
Problem C of tc/C div2/A div1 (Perform Easily)
Problem D of tc/D div2/B div1 (Shurikens)
Problem E of tc/E div2/C div1 (Solo mid Oracle)
Problem F of tc/D div1 (Roads and Ramen)
Problem E of div1 (A Convex Game)

• +146

 » 4 months ago, # |   0 almost got the idea of DIV2 C :(
•  » » 4 months ago, # ^ | ← Rev. 3 →   0 todays contest was really good....watched ecnerwala stream....
•  » » 4 months ago, # ^ | ← Rev. 2 →   +11 It's really interesting who was the author. I will say that it's really interesting to know which problem was made by whom. Golovanov399, could you please show us!
•  » » » 4 months ago, # ^ |   +1 YES! Please Golovanov399! please!
•  » » » » 4 months ago, # ^ |   +57 The author of Div1A and Div1C is Golovanov399, I am the author of all other problems.
•  » » » » » 4 months ago, # ^ |   +11 Thank you!
 » 4 months ago, # |   +62 div2 D was way easier than div2 C !! anyways great contest indeed !!
•  » » 3 months ago, # ^ |   0 Can't agree with this. I solved Div2 C in less than 15 minutes, and couldn't solve Div2 D at all. IMHO, the second part of solution in editorial is more difficult than it could be, because after sorting the pair array, we can simply use two pointer method
 » 4 months ago, # | ← Rev. 2 →   -28 Explanation for some questions are not upto the mark.
 » 4 months ago, # |   +83 I think proof of Div1 D looks incomplete. You should also consider a case when diameter and optimal path intersect, I think they are a bit different.
•  » » 4 months ago, # ^ |   -155 Please mark your comment with spoilers. Some participants might want to upsolve without spoilers.
•  » » » 4 months ago, # ^ |   +411 Well, reading comments under editorial is a bit reckless if you want to solve problems without spoilers
•  » » » 4 months ago, # ^ |   -67 ahahaah he destroyed you nub
•  » » 4 months ago, # ^ |   +20 Yeah, little bit.
•  » » 4 months ago, # ^ |   0 Correct me if I'm wrong, but I think it still holds when $C = D$ (optimal path intersects diameter) and when $D = F$ (optimal path begins somewhere on diameter), so that should cover all the cases. The proof does not make assumptions on edges existing between nodes in the diagram, so setting paths have zero length should be fine.
•  » » » 4 months ago, # ^ |   0 If C=D then everything is fine, but if diameter and considered path have some edges in common then it is a bit different
•  » » » » 4 months ago, # ^ |   0 ah, i see. i agree that it's incomplete then, there's no way to cover that without another case.
 » 4 months ago, # | ← Rev. 2 →   -50 Golovanov399 why my sol for div2 D getting tle https://codeforces.com/contest/1435/submission/96683823 please help...
•  » » 4 months ago, # ^ |   +18 Try taken.lower_bound() instead of lower_bound(all(taken))
•  » » » 4 months ago, # ^ |   0 Why does this work?
•  » » » » 4 months ago, # ^ |   +13 In simple words lower_bound(all(v)) uses random access iterators and works well only for vectors, arrays etc. whereas s.lower_bound is a function built to handle binary search for sets, multiset etc where elements cannot be randomly accessed. there is a blog about this, try searching if interested further.
•  » » » » » 4 months ago, # ^ |   0 Do you have that link?
•  » » » » » » 4 months ago, # ^ |   0
•  » » » » 4 months ago, # ^ |   0 In c++ set or multiset, it is better to use the in-built lower bound function as in the worst case, it has logarithmic complexity. The other one has a quadratic worst case time complexity.
 » 4 months ago, # |   +52 Is tourist's submission for problem E somehow rejudged?
•  » » 4 months ago, # ^ |   +6 Yeah, it was stated that $v_i > 0$, while in fact it was generated (and then validated) that they can be zeroes. It turned out that the second Gennady's submission failed just because of that (and then he wasted the last hour finding out what was wrong).
•  » » » 4 months ago, # ^ |   +8 He came first without having to solve D, so it is all good. Otherwise, the round would have been unrated for him right?
•  » » » » 4 months ago, # ^ |   -8 I guess yes.
•  » » » » 4 months ago, # ^ |   +6 He doesn’t need D and he can be the first.
•  » » » » » 4 months ago, # ^ |   +10 If someone had solved all problems, I'm pretty sure tourist wouldn't have come 1st. If the statement was correct initially, he would have probably spent the last hour on solving D maybe.
•  » » » 4 months ago, # ^ |   +117 What a good way of preventing someone from AKing your contest. (Just kidding. )
 » 4 months ago, # |   +27 There's O(n) solution for div2 D using stack. See my code.But it's hard to explain for me..
•  » » 4 months ago, # ^ |   +5 Wow! That's an amazing solution! Thanks!
•  » » 4 months ago, # ^ |   +5 Great! Thanks for the code
 » 4 months ago, # |   0 Did anyone solve D2C with DP? Is it even possible?
•  » » 4 months ago, # ^ |   +3 I did and got FST :(
•  » » 4 months ago, # ^ |   0 Yes I tried Div2 C with DP, but it failed on test-case 25. But I realise now that it is not possible to solve it using DP. Here is my submission 96694460Atleast in the way I have defined my DP states.
•  » » » 4 months ago, # ^ |   0 Same pretest got us :(
•  » » » 4 months ago, # ^ |   0 I believe my solution has the same idea 96688698, but why do you think it's not possible?
•  » » » » 4 months ago, # ^ |   0 Because in DP solution answer of current state is dependent on previous state and we minimise the difference between minimal and maximal fret. But consider the 25th Test case. 101 146 175 if we subtract 1 96 100 respectively from it we obtain our answer.But DP solution will choose :- 96 96 100 for subtraction, because difference between 5 and 50 is smaller as compared to 100 and 50.
•  » » » » » 4 months ago, # ^ |   +2 Lol, I combined our solutions and it got AC. That's weird. I also added sorts to my solution as you did 96712200
•  » » » » » » 4 months ago, # ^ |   0 Whoa......that is interesting
•  » » 4 months ago, # ^ | ← Rev. 2 →   +5 I have a DP + Greedy solution. Here's my submission.Explanation — First comes the greedy part. By observation, it can be seen that after sorting both the notes and $a_{i}$, it will always be optimal to first choose some elements from the first string, then some from the second, and so on. So, after choosing a particular string $k$ for some $i-th$ note, the $(i+1)-th$ note will be always from a string $j\ge k$. This can be implemented using DP.In my solution, $dp_{i,k,j}$ is the state where the $i-th$ note is being considered to be assigned string $j$ and the $(i-1)-th$ note has been assigned the string $k$. The second.first, second.second and first fields of the states represent the minimum value chosen till now, maximum value chosen till now and their difference respectively.
•  » » » 4 months ago, # ^ |   0 Thanks very much for the code. Codes speak for themselves
•  » » » 4 months ago, # ^ |   0 Can you elaborate this point of yours? — it will always be optimal to first choose some elements from the first string, then some from the second, and so on. So, after choosing a particular string k for some i−th note, the (i+1)−th note will be always from a string j≥k.
 » 4 months ago, # |   -31 I am confused about div2 C.First I got WA then WA on pretest 7 and Then WA on pretest 9. It sucks :(
•  » » 4 months ago, # ^ |   0 hahaha same happened with me too
•  » » 4 months ago, # ^ |   0 same bro
•  » » 4 months ago, # ^ |   0 That's why it is a good problem! I got confused so many times with it. And it took me many more hours after contest to debug.
 » 4 months ago, # |   0 Was able to solve B today but not A :|
•  » » 4 months ago, # ^ |   0 Me too, feels really bad. Another bad thing is stuck on C but then find D so easy after contest
 » 4 months ago, # |   0 Div2C : Each note can be played on a specific range of frets. So out of all such ranges, find the minimum end point (at least one note cannot be played using a higher numbered fret) and the maximum start point (at least one note cannot be played using a lower numbered fret) and output max(0, max_start — min_end). Can anyone point what is wrong with this approach and provide a simple counter test?
•  » » 4 months ago, # ^ |   0 I did the same..
•  » » 4 months ago, # ^ |   +4 Firstly, it would not work on test: Test #11 1 1 1 1 5210 13 AnswerThe best solution is to choose 1 note for 10 and 5 for 13. So, the answer will be 1, but your code gives 0.And also consider this testcase Test #21 1 1 1 1 536 8 10 Answermax_start = 5 (10 — 5), min_end = 5 (6 — 1). Your code will give 0, but you cant get 5 from 8 (only 3 and 7).I hope I understand you correctly.
•  » » » 4 months ago, # ^ |   +1 Thanks! Got it.
•  » » » 4 months ago, # ^ |   +1 k1ps thanks a lot for these test cases : )
 » 4 months ago, # |   0 can someone tell my why I am getting tle https://codeforces.com/contest/1435/submission/96685352 in problem D.
•  » » 4 months ago, # ^ |   0 Your solution is O(N^2).final.size() will become N gradually. Since the outer loop is running 2*N times, complexity is O(N^2).Note: 1+2+3+...N = N(N+1)/2 = O(N^2)
 » 4 months ago, # | ← Rev. 3 →   0 Problem D2C/D1A has a DP tag in it. I would really like someone to help me with the DP solution for D2C/D1A if it is possible. Thanks in advance !
•  » » 4 months ago, # ^ |   0 I have shared my DP + Greedy approach here.
 » 4 months ago, # | ← Rev. 2 →   0 i got TLE with N*M complexity on B, helllooo? what's wrong with this " https://codeforces.com/contest/1435/submission/96705114 " ??? i got accepted with " https://codeforces.com/contest/1435/submission/96705022 "
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 ㅤ
•  » » 4 months ago, # ^ |   0 I got faced the same problem. The same solution is getting AC as well as TLE.
 » 4 months ago, # |   +4 Super easy solutions and implementation for problems Div2 D and Div2 E: Div2 D: 96705496 Div2 E: 96705529
•  » » 4 months ago, # ^ |   +4 D is much easier that the one u linked, just one stack and O(n).
•  » » » 4 months ago, # ^ |   +4 Yes, I see. Complexity can be further reduced to O(N).
•  » » » 4 months ago, # ^ |   +3 Thanks for the solution, mate! Very nice observations
 » 4 months ago, # |   0 can some one explain me the solution of div-2 D...?
•  » » 4 months ago, # ^ |   0 Please someone explain it in clear english ??
 » 4 months ago, # |   0 Why problem-B is getting Time limit exceeded on python. I tried to submitting others code which got correct during the contest but still, it is getting Time limit exceeded. I am using python-3 and also tried using pypy.
 » 4 months ago, # | ← Rev. 2 →   0 In problem Div2D, if we are getting queries of form "— num", we are updating the lower bound , what we need to do when we get query of type "+" ? I didn't understand this line from editorial "and remove any one of them, because we cannot remove any other shuriken"
 » 4 months ago, # | ← Rev. 3 →   0 Question D div2 can be solved in O (n) rather than O(n*log(n))with simple stack and easier implementation, here is an ac solution. https://codeforces.com/contest/1435/submission/96693538(Only the solve1() function is the code for this qn)
•  » » 4 months ago, # ^ |   0 your way to solve this problem is very clear! thank you
 » 4 months ago, # |   +9 4 + - 1 + + - 4 + - 2 - 3I have hacked my own solution for Div 1 B/Div 2 D with the above test case. My submission:- 96706130 Systems tests are so weak Golovanov399 amethyst0 Endagorion AndreySergunin
•  » » 4 months ago, # ^ |   +26 Wow, I'm really surprised by that considering how easy it is to make good YES tests: just make a bunch of small random YES tests and it's really easy to combine them into 1 big test(just increase the numbers in each small test by the sum of $n$ of the small tests before it)
•  » » 4 months ago, # ^ |   -42 each price from 1 to n occurs exactly once
 » 4 months ago, # | ← Rev. 2 →   +52 Div2 D O(n) solution explanation: note that if we have several consecutive -x, they must be in an ascending order (otherwise we say NO and stop instantly) if we have + and a -x after it, we can collapse them (it's not hard to understand that if the answer is YES, we can act this way, just a simple mindfulness exercise) Bingo! We've just solved this problem in O(n) using stack (going in the reversed order, putting goods on a stack and then taking them from the top of the stack and putting on the showcase). The only tricky moment left: you must pay attention to the amount of shurikens to avoid the lack of them (e.g. example 2 from the problem).
 » 4 months ago, # |   0 Someone please tell where I went wrong 96696335
•  » » 4 months ago, # ^ |   0 fixed
 » 4 months ago, # |   +1 Can anyone tell me why my submission for Perform Easily failed pretest 9 : 96698325 ?My logic :The maximal and minimal frets are obviously useful frets i.e. they are used by some fret. So I first sorted the notes and then fixed the "minimal used fret" by using the frets required by the first note by the various strings (the logic being that the smallest note will use the smallest fret from all possible combinations). So out of the 6 frets usable by the first note, one of them should be the "minimal" fret of the optimum set. Then I just binary search for the right hand side by checking if all notes can be played for the specified minimal fret.
•  » » 4 months ago, # ^ |   0 Please anyone please help me out why this submission failed testcase:9 Code#include using namespace std; #define ll long long #define rep(i,a,n) for(ll i=a;i=a;i--) #define f first #define s second #define vll vector #define vvll vector> #define vpii vector> #define vpll vector #define pll pair void solve(){ vectorai(6); for(int i=0;i<6;i++){ cin>>ai[i]; } ll n; cin>>n; vectorbi(n); rep(i,0,n)cin>>bi[i]; set>>st; for(int i=0;imp; setano; ll mini=-1e9; ll cnt=0; // iterate and find till we reach the last of any list for(auto val:st){ mp[val.s.f]++; cnt++; if(mp[val.s.f]==6){ mini=val.f; ano.insert(val.s.f); break; } } ll cal=0; ll maxi=1e9+1; // after that last val find val from each set for(auto val:st){ cal++; if(cal<=cnt){ continue; } ano.insert(val.s.f); if(ano.size()==n){ maxi=val.f; break; } } cout<
 » 4 months ago, # |   +118 Div1 D can be solved via top trees without analyzing the problem.
•  » » 4 months ago, # ^ |   +44 It seems to be difficult to find any information regarding top trees. Do you have any article regarding this?
•  » » » 4 months ago, # ^ |   -56 Please anyone please help me out why this submission failed testcase:9 My Code#include using namespace std; #define ll long long #define rep(i,a,n) for(ll i=a;i=a;i--) #define f first #define s second #define vll vector #define vvll vector> #define vpii vector> #define vpll vector #define pll pair void solve(){ vectorai(6); for(int i=0;i<6;i++){ cin>>ai[i]; } ll n; cin>>n; vectorbi(n); rep(i,0,n)cin>>bi[i]; set>>st; for(int i=0;imp; setano; ll mini=-1e9; ll cnt=0; // iterate and find till we reach the last of any list for(auto val:st){ mp[val.s.f]++; cnt++; if(mp[val.s.f]==6){ mini=val.f; ano.insert(val.s.f); break; } } ll cal=0; ll maxi=1e9+1; // after that last val find val from each set for(auto val:st){ cal++; if(cal<=cnt){ continue; } ano.insert(val.s.f); if(ano.size()==n){ maxi=val.f; break; } } cout<
•  » » » 4 months ago, # ^ |   +36 At a high level, top trees are just link-cut trees but with a BBST of light-children at each vertex, which allows you to do subtree queries. If you don't understand link-cut trees, learn those first.The main paper for the popular splay-tree-based top trees is "Self-Adjusting Top Trees" by Tarjan and Werneck. This also contains references to alternate works by Frederickson and others which have alternative (but more complex) constructions, so you can read those for more background/theory.I submitted a top-tree solution to this problem 96759325, but it's rather complex and not a great way to learn. The quintessential top-tree problem is SONE1 at http://www.lydsy.com/JudgeOnline/problem.php?id=3153, but the site seems to be down right now.
•  » » » 4 months ago, # ^ |   +8 https://negiizhao.blog.uoj.ac/blog/4912Here is a brief introduction of top trees (with some applications I could come up with), but it is in Chinese. I may supply an English version later.
 » 4 months ago, # | ← Rev. 2 →   0 How to solve Div 2D using priority queue ?? I don't understand this line from editorial. Otherwise, for all shurikens that had a lower bound of something less than x we increase it to x, and remove any one of them, why do we have to remove any one of them when we exactly know which one to remove , how to handle the other type of operation "+" type .
•  » » 4 months ago, # ^ |   0 You can check this solution 96682108
•  » » » 4 months ago, # ^ |   0 can you please explain the idea behind your solution .
•  » » 4 months ago, # ^ | ← Rev. 2 →   +11 We can iterate an array in a reverse order. (Example 1) + + - 2 + - 3 + - 1 - 4 we can create an array like this [-1,-1,2,-1,3,-1,1,4] where -1 indicates that the element was inserted(+ sign). Now maintain a min-heap. Iterate backward on this array, we will keep on inserting the elements. We will first insert 4 and 1. Now whenever we land on -1, it means at this point, some shuriken must have been added on the showcase) which can only be either 4 or 1(Since they are sold at the last). So, we can safely remove the minimum element from the heap and add it to the answer. If the heap is found empty at -1 or if we find one of the element in the heap is less than the element to be insert then the array is inconsistent. Also, we can check if number of +'s is equal to n. If it is not equal then the answer is "NO". 96748102
•  » » » 4 months ago, # ^ |   0 Hey, can you elaborate this part we find one of the element in the heap is less than the element to be insert then the array is inconsistent. ?
•  » » » » 4 months ago, # ^ | ← Rev. 3 →   0 The element x which is currently being added indicates the element was sold at this moment and the elements in the heap stores the element already sold out. It also means that at this point, the elements which we have on the showcase are x and all the elements in the heap. By what question says, x has to be the minimum as the customers purchases the minimum element on showcase, but in this case, there is some element in our heap which is less than x and the customer have purchased x instead of that minimum element in the heap. So, it is inconsistent.
•  » » » 4 months ago, # ^ |   0 Thanks for the great explanation. In the problem it says, It's guaranteed that there are exactly n events of the first type, and each price from 1 to n occurs exactly once in the events of the second type. So, you don't have to check if no. of +'s == n
•  » » » 5 weeks ago, # ^ |   0 Thanks for the explanation. Can you please explain how is this safe? we can safely remove the minimum element from the heap and add it to the answer
 » 4 months ago, # |   0 Div2 D problem can be solved in O(n) time using stacks. Code — https://codeforces.com/contest/1435/submission/96718560
 » 4 months ago, # |   0 Why my submission link1 getting TLE but when i changed vector size to (n*m + 5) it get accepted link2 ?
•  » » 4 months ago, # ^ |   +5 This is a common mistake when there are multiple testcases. It may take 500*500*t if you create/initialize 500*500 arrays every time, and that's about $2.5 \times 10^{10}$. The statements said that "Sum of $nm$ over all test cases does not exceed $250000$" so the second submission could pass.
 » 4 months ago, # | ← Rev. 2 →   +10 About Problem 2E/1C: I didn't like this problem at all. It was boring and a detail-finding kind of problem where you know right after reading the problem that you can get to the solution if you are not lazy enough.about the $-1$ part, we can just see that since $a>bc$, for each spell, whether complete or not, contributes positively to the damage. Hence the damage can be arbitrarily large. No need to think about overlapping parts, as is done in the editorial.
 » 4 months ago, # |   +13 Linear time solution of div1B: Consider the queries backwards. Then — x actually means adding shuriken with value x to the showcase, and + means deleting some shuriken. So we can maintain the sorted stack of all shurikens at the moment. On + we will delete the minimum from the stack (it's easy to see that it's always the best option), i.e. top element, and add it to the answer array. On — x we will check that x is less then the value on top of the stack and add it. If we successfully looked through all the queries we will just print the reversed answer array, and otherwise there's no answer. My code 96667703
•  » » 4 months ago, # ^ | ← Rev. 3 →   0 I had a similar idea , which unfortunately didn't work. can you please explain what's the flaw with with the following logic.Initialise cnt to zero . While processing the queries backwards , if we see + we increment the cnt variable by 1. When we see — x , we push x in the stack if x is less than top element otherwise we can pop atmost cnt number of elements from top of the stack such that now either the stack is empty or top of stack is greater than x.while popping from the stack we decrease the cnt by 1.
 » 4 months ago, # |   0 There is something I can not understand in the proof of 1D.Tutorial says "Let AB be a diameter of the tree, and the optimal answer be EF." But it seems that in the given tree, AB is not a diameter.Did I understand what the tutorial wanted to express wrongly? Could anyone explain it for me? Thank you in advance!
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 This is for the sake of brevity. You can see that the edge CB is a little bit longer than AC. Apparently that means that there are some nodes on a path between B and C
•  » » » 4 months ago, # ^ |   0 Ha, got it! Thank you sir!
 » 4 months ago, # |   0 can someone explain to me whats wrong with my algorithm 96738918 in D2C? i know my idea is wrong,but im confused whats wrong with it
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 The answer may not always be abs(b[n]-a[6]-b[1]+a[1]). An example: 1 3 6 8 100 1000 3 1003 1250 1500 According to your code, the answer must be 502, but the correct one is 398.
•  » » » 4 months ago, # ^ | ← Rev. 2 →   0 thanks dude,but i still dun get the idea of the problem C,can u explain more??
 » 4 months ago, # |   0 Even though my rating dropped . I loved the problems and the NARUTO theme . Nice contest!!!
 » 4 months ago, # |   0 Can someone explain A?
•  » » 4 months ago, # ^ | ← Rev. 2 →   +3 Hint1N is even .. So try to make every two element operation sum = 0 Hint 2Suppose think there have two number a and b.1. a=2 & b=3 both positiveThen answer is b=3 & -a=-22. a=2 & b=-3 negThen answer is -b=3 & a=2.3. a=-2 & b=-3then answer is b=-3 & -a=24. a=-2 & b=3then answer is -b=-3 & a=-2 If have any query..please ask.
•  » » » 4 months ago, # ^ |   0 Originally I thought this was meant to be solved with LCM and was having trouble. This approach is really easy to understand thnx!
 » 4 months ago, # | ← Rev. 2 →   0 Can someone please tell me why my solution is wrong for Div2. A... 96650687I printed all 1's if the sum of all elements of original array is 0 and I got wrong answer on 10th Test.
•  » » 4 months ago, # ^ |   0 if the sum ever does happen to be 0, then you're printing 2 valid answers which is why you're getting WA. Just add a return statement.
•  » » » 4 months ago, # ^ |   0 Got it! Thanks
 » 4 months ago, # |   0 Could anyone recommend similar problems to div2C&|D?
•  » » 4 months ago, # ^ |   0 Just search the tags
 » 4 months ago, # |   +1 How to approach the problem E with ternary search( Proof of maximal value being a convex function) ??
 » 4 months ago, # |   +1 Was div.2-D a standard problem? Everybody is saying that the problem is very easy but I could not come up with a simple idea for the problem. I was thinking of a heavy implementation solution.
•  » » 4 months ago, # ^ | ← Rev. 2 →   +1 The idea of problem D:- 1) pair every "+" with below "-" using upper bound. 2) check the order of elements with given order using stackfor better understanding, check out my code
 » 4 months ago, # |   0 I think there's also a O(n) solution for Div2 D using list and it's easier to explain.We consider matching the '-' operations with '+' operations from "- 1" to "- n".After we matched a '+' operation and a '-' operation, we remove them from the operation sequence.So, when considering the operation "- x":If the previous operation is "-", it's impossible to meet the requirements because we have already removed all the numerically smaller "-" operations.Otherwise the previous operation is "+" and we can easily match this "-" operation with it.This can be maintained with a list by recording the position of each "-" operation in the operation sequence.My code
 » 4 months ago, # | ← Rev. 2 →   0 All problems(DIV2) are good and do not need a hard data structure. some observation and you can solve.
•  » » 4 months ago, # ^ |   0 Except the one about ramen
•  » » » 4 months ago, # ^ |   0 I was talking about DIV2
•  » » » » 4 months ago, # ^ |   0 Oh i was doing the original Technocup and there we had that problem about ramen, so i thought it was in div2 too
 » 4 months ago, # | ← Rev. 3 →   +1 Div2 D can be solved going through the reversed query with a set. Every time we meet a "+" we can just say that the woman takes the one with the least price back. So if set's size is 0 the answer is NO. Else erase that element and add to the answer. Else if we meet a "- x" then x must be the less then the least element in the set now. If not, the answer is NO. If yes, insert x. 40 lines of code.p.s. reverse the answer in the end96676641
 » 4 months ago, # |   0 Can anyone tell me why am i getting TLE in div2B? An n^2 solution should have been fine ,right? 96669009
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 I think n^2 should be too slow (2,5e5)^2 is a lot
•  » » » 4 months ago, # ^ |   0 Oh. Can you kindly explain your approach for B then?
•  » » » » 4 months ago, # ^ | ← Rev. 2 →   0 Oh i see the problem. Im sorry thought it was (nm)^2, but after seeing ur code i saw it's nm, which is ok. The problem must be creating a huge array vis, since all other things have a good time complexity.p.s. I got ur sol accepted with changing vis size to n * m + 1
•  » » » » » 4 months ago, # ^ |   0 Thanks a bunch. It worked after using a map instead of vis array
 » 4 months ago, # |   0 can someone tell me why my code is giving wrong output?? my code is 96679812.. this is code for Div2C..
 » 4 months ago, # | ← Rev. 2 →   0 A simple solution to Div2 D using a stack and a priority queue : 96756883
 » 4 months ago, # | ← Rev. 2 →   0 I don't know why I TLE in B :(( somebody help me, please!!!!!!! MY SUBMISSION
•  » » 4 months ago, # ^ |   0 You used memset for 1000000 bytes in while loop for 100000.
•  » » » 4 months ago, # ^ |   0 Oh got it! Thanks <3 <3
 » 4 months ago, # |   0 Div 2 : Problem D It seems hard to explain my idea clearly. I would add some test cases, which might be helpful to understand. Case 1 : From test case 2 , at any moment, if the number of sold items "- x" is greater than the items (+) we already have, then the ans is "NO" Case 2 : Checking The contiguous blocks of "- x" is in the ascending order or not. If the contiguous block of "- x" is not in the ascending order then the ans is "NO".On this point my thinking was, if we have something like this, For n= 4 : (+) x1 (+) x2 (+) x3 (+) x4, where the particular item has been sold just after it has been placed in the showcase. Then for any sequence of xi, ans is possible.Now extending the idea, for example n = 6 : (+) (+) (+) 4 5 6 (+) 1 (+) (+) 2 3. In this case one of the possible solutions could be (6) (5) (4) 4 5 6 (1) 1 (3) (2) 2 3. Here I considered a case where for every block of sold items we have equal numbers of empty places. Items 4th, 5th, 6th are sold means those items have been placed earlier,and also he brought the items choosing the minimum once in each purchase. Hence, items 4th, 5th, 6th could be placed in any order. Now think, if the sequence was (+) (+) (+) 5 4 6 (+) 1 (+) (+) 2 3. Then it means items 5th, 4th, 6th are sold, whereas he brought 5th item whereas item 4th was already there, which is not possible. Hence, the output would be "NO"Case 3 : Now consider the case. For n = 6 : (+) (+) (+) (+) 3 4 6 (+) 1 2 (+) 5. Here for every block of sold items we do not have equal numbers of empty places : (4 empty place) 3 sold items (1 empty place) 2 sold items (1 empty place) 1 sold items Then if we put the items like this (+) (6) (4) (3) 3 4 6 (1) 1 2 (5) 5. 2nd item still needs to be placed. The only remaining place is at the beginning. If we put 2nd item there, then the ans would be "NO" However, similarly if the given sequence was, for n = 6 : (+) (+) (+) (+) 2 3 4 (+) 1 6 (+) 5. Then one possible solution could be (+) (4) (3) (2) 2 3 4 (1) 1 6 (5) 5 . Here we can easily put the 6th item at the beginning, which will maintain the condition. I have used stack. For checking the case 3, I stored the sequence in a vector. My submission 96765725
 » 4 months ago, # |   0 https://codeforces.com/contest/1435/submission/96662042 can someone tell me what I am wrong?? I can't find anything wrong with my code. please...
•  » » 4 months ago, # ^ |   0 the first matrix given gives you the column of the number and the second matrix gives you the row of that number
 » 4 months ago, # |   0 How to solve div2E with binary search?
 » 4 months ago, # |   0 I thought the title of Div2 E was dota2 reference.
 » 4 months ago, # |   0 https://codeforces.com/contest/1435/submission/96788248 What is wrong in my code for shurikens?
 » 4 months ago, # |   0 How could anyone in A saw the row of array like: -a2, a1, -a4, a3, ..., -an, an-1. I mean, no advices on that path weren't given, is anyone sorted out how this works?
•  » » 4 months ago, # ^ |   0 n is even, and you want to make two adjacent elements make a sum of 0.if you have the array a1, a2 — make the sum a1*(-a2) + a2*(a1) = 0.If you do it to every adjacent pair, the total sum will be 0.
•  » » » 4 months ago, # ^ |   0 Yeah, I understand that I need sum = 0, but I hadn't sorted out how to achieve this at the competition, so that's quite bad. That's pretty hard for me to find easiest ways to solve a problem.
•  » » » » 4 months ago, # ^ |   0 Emmm. What you need to do is just practicing more and thinking more. I believe you can solve such problem quickly in a nearly future.
•  » » » » » 4 months ago, # ^ |   0 Thanks. I hope, I will :)
 » 4 months ago, # |   0 can someone tell me why i am getting runtime error https://codeforces.com/contest/1435/submission/96803170
 » 4 months ago, # |   +5 You can solve KOIREP from SPOJ if you have solved Div 2C/1A
 » 4 months ago, # | ← Rev. 2 →   0 Can someone explain me DIV2 E/DIV1 C. Not able to get the way written in the editorial. Can someone get me through that . Thanks in advance.
 » 4 months ago, # |   0 Could someone please explain the solution of the div1E？I'm totally confused with the conclusion “the outcome of the game is defined by the index of the last move and the last difference between the elements”and“ if we fix the last index and gradually decrease the last difference, the grundy value will not decrease. ”
 » 4 months ago, # |   0 Why we used set>> in div2C why can't we just keep track of the value(0-5) in an array of n
 » 4 months ago, # | ← Rev. 2 →   0 Could anyone please tell me some other solutions (except two pointers) of the problem Perform Easily? There are so many problem tags for this problem, so I guess there will be some interesting solutions (such as binary search).
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 I thought of a different approach and used lower bounds and upper bounds but it is failing at test case 9 can you tell me whats wrong with my code? you can go through this I posted this earlier Can Someone tell me why I my code fails at test case 9 96809688 I have used the approach that after sorting the notes the there cannot be a fret smaller than difference between first note and first string then I took two case 1. I found all the frets for every note which is just smaller or equal to the smallest fret. 2. I found all the frets for every note which is just larger or equal to the smallest fret. then I found the Min of differences obtained from these two.
•  » » » 4 months ago, # ^ |   +1 Sorry, I don't understand your approach. However, there is a set of small data that your code can't get the right answer, maybe you can try to discover your mistakes through it.Input1 1 1 96 99 1003101 146 175Output50Your answer74Good luck!
•  » » » » 4 months ago, # ^ |   0 No this case will work but I get it why my code is not running for test case 9
 » 4 months ago, # |   0 Can Someone tell me why I my code fails at test case 9 96809688 I have used the approach that after sorting the notes the there cannot be a fret smaller than difference between first note and first string then I took two case 1. I found all the frets for every note which is just smaller or equal to the smallest fret. 2. I found all the frets for every note which is just larger or equal to the smallest fret. then I found the Min of differences obtained from these two.
 » 4 months ago, # |   0 5 minutes left time to be ready : )
 » 4 months ago, # |   0 Great Round!
 » 4 months ago, # |   0 O(N) is possible for 1413D — Shurikens My solution
 » 4 months ago, # |   0 Thx for this round!