awoo's blog

By awoo, history, 2 years ago, translation,

1389A - LCM Problem

Idea: BledDest

Tutorial
Solution (BledDest)

1389B - Array Walk

Idea: Roms

Tutorial
Solution 1 (pikmike)
Solution 2 (pikmike)

1389C - Good String

Idea: BledDest

Tutorial
Solution (Ne0n25)

1389D - Segment Intersections

Tutorial

1389E - Calendar Ambiguity

Idea: BledDest

Tutorial
Solution (pikmike)

1389F - Bicolored Segments

Idea: Roms

Tutorial
Solution 1 (Ne0n25)
Solution 2 (Ne0n25)

1389G - Directing Edges

Idea: BledDest

Tutorial
Solution (BledDest)

• +127

 » 2 years ago, # |   0 Super Fast!!
•  » » 2 years ago, # ^ | ← Rev. 2 →   +45 Day by day I feel like nobody is getting the humor anymore they are just downvoting every comment.
 » 2 years ago, # | ← Rev. 2 →   0 In the Tutorial of Problem E: I think you mean yd + x instead of xd + y as the index of the x-day of the y-month in the year.
•  » » 2 years ago, # ^ |   0 xd+y means index y-th day of x-th month which is corresponding pair of x-th day of y-th month whose index is yd+x.
 » 2 years ago, # |   0 Can someone tell me why my solution for D fails? I simply decide whether to gain intersection by joining two new segments or expanding others which already cover themselves totally. Counter-examples are especially appreciated. Here's the submission: https://codeforces.com/contest/1389/submission/88410212.
•  » » 2 years ago, # ^ |   0 The test, your solution fails when n = 10, k = 297 l1 = 741, r1 = 741 l2 = 20, r2 = 770 Try to fix it.
•  » » » 2 years ago, # ^ |   +3 Got my error, thank you!
•  » » 2 years ago, # ^ |   0 Is there any way to solve $D$ in $O(1)$.. maybe little case work ($chukles$), but I guess that could be done!
•  » » » 20 months ago, # ^ |   0 Yes. Use some greedy approach to choose how many segments need to be invested. This is my accepted approach https://codeforces.com/contest/1389/submission/106042967
•  » » » » 20 months ago, # ^ |   0 nice, I'll try to solve it too in $O(1)$
 » 2 years ago, # |   0 Waiting for Div 3 :) After so many difficult rounds to increase rating :(
 » 2 years ago, # |   0 What is Event Processing I see it here and there but never find some concise tutorial on it.
•  » » 2 years ago, # ^ |   +8 https://www.geeksforgeeks.org/maximum-number-of-overlapping-intervals/In that algo the beginning and ending of intervals is treates as "events" where the count of them is incremented or decremented.
•  » » » 2 years ago, # ^ |   0 Thanks man!
 » 2 years ago, # |   0 Can anyone explain this logic of problem B (solution 2): if i % 2 == k % 2: tmp = (k — i) // 2
•  » » 2 years ago, # ^ |   0 Can we do the problem B with kadane algorithm in which I will find the max present in the first k position of the array, and then after reaching towards the max, I will choose whether to go beyond or go back ?
•  » » » 2 years ago, # ^ |   0 No way bro!!
 » 2 years ago, # |   0 I think we can just go through 45 combinations for C instead of the 100 given in the tutorial.TLDR:0101010 has both a 101010 and a 010101 type strings in it.DETAILED:Let us say the string we get after removing all other numbers but two be some 011100001001010.....Then for the above we want it to reduce to the form : 01010101... Note, that if the reduced string ends with 1 then the answer is: size(original string)-size(reduced string). Else, the answer is: size(original string)-{size(reduced string)-1}.
 » 2 years ago, # |   +3 Can someone explain the dynamic programming approach to problem E?
•  » » 2 years ago, # ^ |   0 Problem E doesn't use dynamic programming. It's just math and number theory.
 » 2 years ago, # |   +5 Cool contest!Though I've lost some points :(
 » 2 years ago, # | ← Rev. 3 →   0 .
•  » » 2 years ago, # ^ |   0 Firstly.What does these numbers mean? Test containing 4 numbers, but u give 6. It's not 1 test and not 2.SecondlyX can't be negative.In both pairs {l, r} l <= r, so max(r1, r2) >= max(l1, l2) => max(r1, r2) >= min(l1, l2)I hope you understand
•  » » » 2 years ago, # ^ | ← Rev. 2 →   0 .
•  » » » » 2 years ago, # ^ |   0 Sample are incorrect and occasionally it's work (but it shouldn't)
•  » » » » » 2 years ago, # ^ | ← Rev. 2 →   0 .
 » 2 years ago, # | ← Rev. 4 →   0 Why this code is giving TLE? https://ideone.com/wXdX3A
•  » » 2 years ago, # ^ |   0 Lets calculate O(10 * 10 * n + 10 * n) = O(100*n) n = 2 * 10^5. We get O(2 * 10^7). Its about 10 secs on Python and about 0.2-0.3 secs on C++ (I have solve with the same asymptotics) Friendly advice. Use C++ in real competitions.
•  » » » 2 years ago, # ^ |   0 Same code gives AC in c++ but tle in python :| btw thank you.
•  » » » » 2 years ago, # ^ |   0 С++ ~100 times faster than python
•  » » 2 years ago, # ^ |   0 always use fast I/O in Python. normal I/O in python takes a lot of time. https://www.geeksforgeeks.org/python-input-methods-competitive-programming/
 » 2 years ago, # | ← Rev. 10 →   0 Hey @adedalic I have one doubt in D, You have said, In the case of non-intersecting [l1,l2] and [r1,r2], we should at first "invest" some number of steps in each pair to make them intersect. So let's iterate over the number of segments to "invest" cntInv. We should make cntInv⋅(max(l1,l2)−min(r1,r2)) steps to make segments touch. Now, cntInv segments touch so we can use almost the same formulas for them as in the previous case.  But consider case where input is, 5 16 1 4 8 12 In this case should we invest time in joining all pairs or just invest time in joining one pair and complete operations using only that pair. So steps for joining = 4 Steps to make both equal size = 11 , remaining k = 5. So instead of joining all pairs just invest in first joined pair. Remaining steps = 5*2 = 10. Ans = 4 + 11 + 10.  Why should we invest in joining all pairs ? Am I missing something ? Community, Please help !!! EDIT: Now I get it, u have used a for loop (i to n), that checks by joining first i pairs and compares it with minimum each time . Thanks.
•  » » 2 years ago, # ^ |   0 Just iterate over n and find mininum value. ans = INF for i in :n temp = cost of getting K intersection with i segments joined. ans = min(ans,temp) 
•  » » » 2 years ago, # ^ |   0 got it.
 » 2 years ago, # |   0 I wrote a recursive solution using memoization where I checked for each I it's i-1 and i+1 and updated my answer with maximum sum Does my solution not violate the condition given in the question :. " you can't perform two or more moves to the left in a row" Can anyone please explain this that how my function goes for 2 or more left moves in a row if z is>=2 and does not violate this and gets accepted__88418038
 » 2 years ago, # |   +42 Could anyone please explain the dp with segment tree approach for problem F?
•  » » 2 years ago, # ^ |   +13 Here's how I understand it (I did not solve in contest), feel free to correct me if I'm wrong or there is an easier way to think about this. First, let us sort all segments in increasing order of their right endpoints in the list $segs$. Split the segments based on type and sort them in the same fashion in the lists $s[t]$. Let $dp[i][j][t]$ be the answer assuming that the problem is restricted to the first $j$ segments in $segs$, along with the condition that we can only select the first $i$ segments in $s[t]$, and that we have used segment $i$ in $s[t]$ in our solution. Now, to transition to $segs[j+1]$ , assuming $segs[j+1]$ is of type 1 and is the $k+1$'th segment in $s[1]$, we have the following: $dp[i][j+1][2] = dp[i][j][2] + 1$ if there is no touching between segment $k+1$ of $s[1]$ and segment $i$ of $s[2]$, and $dp[i][j+1][2] = dp[i][j][2]$ otherwise. $dp[i][j+1][1] = dp[i][j][1]$ for all $i <= k$, and $dp[k+1][j+1][1]$ is the max of $dp[i][j+1][2]$ for all segments $i$ in $s[2]$ that do not touch with segment $k+1$ in $s[1]$. In other words, the dp table is largely unchanged, except that we try to add $s[1][k+1]$ to everything that we can, which must be optimal. The transition for where the new segment is of type 2 is very similar. Our answer is the maximum over the whole dp table. Now, notice that for any new segment, the segments that do not touch it must be a prefix of the sorted segment list. Thus, these transitions are exactly range add modifications and maximum queries on two segment trees. We can easily find what range to add by using a binary search. Finally, we can eliminate the 2nd dimension with $j$ just by directly modifying the segment trees. See my submission for implementation details.
•  » » » 2 years ago, # ^ |   +8 So, for implementation purpose, we can ignore the $2nd$ dimension of the dp state you mentioned, right? So our dp state will be:$dp(i,j):$ maximum segments if we have only considered till the ith segment of type $j$ and this one is included. And we traverse the segments on the sorted order of their right endpoints.
•  » » » » 2 years ago, # ^ |   +8 Yes, exactly. When I was trying to figure the solution out based on other people's code, I found the 3 dimensions to be easier (as a solution that I could actually come up with), but you have it exactly right.
•  » » » » » 2 years ago, # ^ |   0 Okay, sounds good. Thanks a lot.
 » 2 years ago, # |   0 https://codeforces.com/contest/1389/submission/88444945 Why my code is wrong? Can anyone help? for length = odd, I check for the most frequent character. for length = even, I iterate i,j and the pattern is i,j,i,j, .... ,i, j ( end with j )
•  » » 2 years ago, # ^ |   0 Because you don't need to start with p1=0 & p2=0 Try this example char1: 3 4 9 13 15 char2: 5 7 11 18 Your algorithm will answer 2 ( 3, 5 ) But the correct answer is 6 ( 4,7,9,11,13,18 ) You can see that it doesn't need to always start with p1 = 0. Thanks to myself ! I do this beacuse I can't delete my comment.
 » 2 years ago, # |   0 I am trying to hack 88307545 solution during the hacking phase. By input data was 2 1 25 100 250 The code will output (on local PC and on online compiler): 12 24 25 5 25 5 25 5 25 5 125 250 I write the second test case to make codeforces think "the first line is for the first test case and the others are for the second", but I got unsuccessful attempt. I didn't understand why this happens? Can someone explain to me?Thanks in Advance!
 » 2 years ago, # |   +5 O(1) solution for D 88447760
•  » » 2 years ago, # ^ |   0 +1 88435747 in Kotlin
 » 2 years ago, # |   0 Could somebody please explain the dp approach to B?
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 Dp approach of B will be like that we have two choices at every index either we can move left or right, we will explore the moves and the maximum between these two options will be considered. base cases are:1 . when we don't have any moves left (k==0), in this case, we will return the current index value.2 . when the last move taken were left or we don't have any left moves remained or we are at the starting index of the array, then we have only one option to move right at this index.3.else we have two option to move right and left, and take the maximum among those to the answer. you can check my solution for more clarity, 88714266.
•  » » » 2 years ago, # ^ | ← Rev. 2 →   0 anurag122898 could you please have a look at this? With the same approach, I'm getting two different results just because I'm swapping the two options' sequence. I never knew something like this could make a difference. I'm putting the SS of the comparison between 92397103 and 92402059 for convenience.
•  » » » 19 months ago, # ^ |   0 How did you decide the states to cache? because here index,prev etc also change in every state...
•  » » » » 16 months ago, # ^ | ← Rev. 2 →   0 Even I had the same query VaibhaveS. Could you explain it if you've understood it?
 » 2 years ago, # |   0 Can anyone help me with my code for Problem C ? Getting a Wrong Answer. def func(): cnst = "01233456789" test = input() ans = 1000000000 for i in cnst: for j in cnst: c1 = 0 c2 = 0 for k in range(len(test)): if (i != j and len(test) % 2 == 1 and k == len(test) - 1): continue elif (k%2 == 0): if (test[k] == i): c1 += 1 else: if (test[k] == j): c2 += 1 c111 = c1 + c2 ans = min (len(test) - c111,ans) print (ans) t = int(input()) for i in range(t): func()   
•  » » 2 years ago, # ^ |   0 Your code is checking if i is at even and j is at odd positions, which is wrong. You need to find the longest SUBSEQUENCE of alternating i and j of even length. INPUT: 1 2025 YOUR OUTPUT: 1 CORRECT OUTPUT : 2
•  » » » 2 years ago, # ^ |   0 Can you please explain it a bit more clearly?
 » 2 years ago, # | ← Rev. 5 →   0 Same approach as Editorial still can't be accepted in python3 due to TLE on test 2 , what is this?1389C-Good String 88530443
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 First: Submit in PyPy instead of CPythonSecond: Convert the input to a list of ints once instead of converting it in each iteration.
 » 2 years ago, # |   0 O(1) solution for problem D Segment Intersections.
•  » » 2 years ago, # ^ |   0 If anyone is interested in the explanation then comment.
•  » » » 2 years ago, # ^ |   0 interested
•  » » » » 2 years ago, # ^ |   +3 There can only be three initial configurations for the two intervals:1> One completely overlaps the other.2> There is some partial intersection b/w them.3> They are disjoint.Now, let us tackle them one by one.1>> We will have an initial I which can be calculated easily, thereafter if the current I is smaller than k we can increase the intersection length by one for all n pair of segments in just 1 step and we can do this untill all the pairs have become equal It is obvious that we cannot increase I by more than 1 in a single operation, So this is optimal. Now if still I is less than k we only have one way to increase I by increasing the segment lengths of a pair this will give us a +1 for each 2 step. Also, we can do this indefinitely.2>> Its the same as the first one except that the initial I has to be calculated differently.3>> Initial I=0 , Now how can we increase I? we first have to make a pair of segment meet each other while doing this our I remains the same, lets call this cost as gap. Now as soon as the segments meet we can start having a bunch of +1s . Let us calculate how many +1s we can get in a row, Suppose the worst case when initially both segments are points, then we will have no of +1s equal to the no of +0s (we used to fill the gap) . And in every other case the no of +1s will be greater than +0s , Remember this. Now let us see what choice do we have after using up these two operations , At this point we reach a situation analogous to one explained earlier in 1>> where we will be getting +1 for every 2 steps. We have all the tools ready . Let us expand this for n. I claim that in an optimal final solution we will have a situation where x of the n segments are completely overlapping and at most 1 pair of segments which has been extended by a 2 step operation or has a partial overlap , Also x will be maximum. Suppose a case where x is not the maximum then we have a case where x-1 segments have complete overlap and 1 segment with a partial overlap or an extended 2 step operation segment. We can simply rule out the first case because we can never compensate for the loss of one completely overlapping segment with a partial one. As for the other case the no of 2 step operation will be greater than the complete overlap of the segments as compensation. This is where we use the fact we proved earlier, that we can get a complete overlapping with total no of steps less than or equal to twice the size of the segments, which will always be at least as efficient as the +2 step operation and hence we can carve out a completely overlapping segment from this segment without loosing optimality (for the lack of better word).
•  » » » » » 2 years ago, # ^ |   +3 Thanks, man for the effort! stating 3rd case more mathematically -suppose we have to make 'rem' out of 'n' segs, let 'ex' be the amount to 'activate' the seg (= max(l1,l2)-min(r1,r2)), and 'tot' (= max(r2,r1)-min(l1,l2)) we can get by doing 1 inc per 1 move operation (+1/1 ops) after 'activation'.Its oBvious that we have to 'activate' the 1st seg and do +1/1 ops If still we have something remaining (let it be 'x')now if 'x' >= 'tot' then its better to activate another seg and get tot why? because 'tot'+'ex' <= 'tot'*2 as 'ex' <= 'tot'if 'x' < 'tot' then we decide on the basis of which is bigger, 'x'+'ex' or 'x'*2 so how is this o(1) -> after the 1st activation, we can activate segs until reqd amount < 'tot' and then we decide between 'activation'+'+1/1 ops' and '+1/2 ops' for the last one.
•  » » » » » » 2 years ago, # ^ |   0 Yeah at one point I thought the way I described it isn't going to help anyone, happy to see that you got it... P.S. yours seems much formal and clear.
•  » » » » » » » 2 years ago, # ^ |   0 Hello Can you check my submission of D problem..I tried the O(1) solution.I cannot find any mistake in it.I tried with over 100 cases. https://codeforces.com/contest/1389/submission/88646237
•  » » » » » » » » 2 years ago, # ^ |   0 Found my mistake I didnt consider the case separately in disjoint where k is less than r2 -l1
•  » » » » » » 2 years ago, # ^ | ← Rev. 2 →   0 __
 » 2 years ago, # |   0 This is in no way a criticism, but I am indeed slightly surprised to find 3 different languages in the implementation section.
 » 2 years ago, # |   0 Can anyone please explain the second solution code of problem B?
 » 2 years ago, # | ← Rev. 3 →   -8
•  » » 2 years ago, # ^ | ← Rev. 3 →   0 Try to write the code in the code snippet. Yours is very unreadable and takes a lot of space in the comments for(int i = 0: i < n; i++) { likeThis(); } EDIT: For your problem, a string is good if it has all same characters in case or even length, or same parity indexed characters are same in case of odd length.
 » 2 years ago, # |   0 A rather random question though, for the explanation for question B, it says any position between 1 and k-2(t) + 1 will be visited. But if k == 4 and t == 2, supposedly i can visit position 1 and 2, but the formulas will suggest I only visit between 1 and {4 — 2(2) + 1}, which means visit between 1 and 1?
•  » » 2 years ago, # ^ |   0 I get what it is trying to say now, thanks!
 » 2 years ago, # |   0 in the div2 b problem whats the intution for this"it is optimal to choose exactly the same i for all the pairs (right, left). And that i should be such that ai+1+ai is maximum possible."
 » 2 years ago, # |   0 Can someone please explain the dp approach for B?
 » 2 years ago, # |   0 Can we do the problem B with kadane algorithm in which I will find the max present in the first k position of the array, and then after reaching towards the max, I will choose whether to go beyond or go back ?
 » 2 years ago, # |   0 Really nice editorial for Problem B really liked it
 » 2 years ago, # |   0 Can anyone provide the Test case 2 — 19th input for task B. Most people had this error "expected 218 found 216". If not exact same test case, then maybe something similar?
•  » » 2 years ago, # ^ |   0 I don't know the exact test case but you should also handle for the case k=0 explicitly because my solution got accepted after I make a case for k==0.
•  » » » 2 years ago, # ^ |   0 In that case the answer would be just the 0th element which I've handled. Thanks for the help!
 » 2 years ago, # |   0 Man, I am getting confused in problem A editorial. It says if 2l>r then its not possible else its l and 2*l. Then how can in the example l=1 and r=1337 produce an output of 6 and 7. Can anyone please explain?
 » 2 years ago, # |   0 in problem B, the tutorial assumes that the left steps will always be accompanied by right steps as they have been taken in pairs but it may so happen that the last step will be a left step, so in that case the tutorial fails and so does the solution in the following test case18 11 4 11 19 18 19 19 5 14 15 17 4 10 9 8 17 9 2 15 10
•  » » 2 years ago, # ^ |   0 The tutorial assumes left steps are always preceded by a right step; Hence the order (right,left).
•  » » » 2 years ago, # ^ |   0 oh thanks... i had misunderstood it
•  » » » » 2 years ago, # ^ |   -10 hey can you help me in this i'm trying it since last 7 hours  def solve(arr,z): maxm=0 for i in range(len(arr)-1): maxm=max(maxm,arr[i]+arr[i+1]) z=min(z,5) i=1 ans=0 moves=len(arr) flag=0 for i in range(len(arr)): ans+=arr[i] moves-=1 if i>=1 and (arr[i]+arr[i-1])==maxm: break f=True #(left , right ) while z!=0 and moves!=0: if f: ans+=arr[i-1] z-=1 moves-=1 f=False if f==False and moves!=0: ans+=arr[i] moves-=1 f=True if z==0 and moves!=0: print(arr[i+1:i+moves+1]) ans+=sum(arr[i+1:i+moves+1]) return ansfor _ in range(int(input())): n,k,z=map(int,input().split()) arr=list(map(int,input().split()))[:k+1] print(solve(arr,z)) 
 » 2 years ago, # | ← Rev. 2 →   0 The idea of G is quite similar to this problem Museum TourVery interesting: Undirected Graph -> find Bridges -> Tree -> Can be solved using DPDirected Graph -> find SCC -> DAG -> Can be solved using DP
 » 2 years ago, # |   0 The *2100 Prob. D seems much more difficult than the *2200 Prob. E.
 » 2 years ago, # | ← Rev. 9 →   -10 I'm trying the array walk since 7 hours still not able to get it can someone please help me with this code where i'm making mistake seriously i just wanted to solve it at any cost please help!!!!  def solve(arr,z): maxm=0 #store maximum(sum of ajecent pair ) for i in range(len(arr)-1): maxm=max(maxm,arr[i]+arr[i+1]) z=min(z,5)#z cannot be greater than 2 i=1 ans=0 moves=len(arr) flag=0 for i in range(len(arr)): ans+=arr[i] moves-=1 if i>=1 and (arr[i]+arr[i-1])==maxm: #ossilating condition break f=True while z!=0 and moves!=0: if f: ans+=arr[i-1] z-=1 moves-=1 f=False if f==False and moves!=0: ans+=arr[i] moves-=1 f=True if z==0 and moves!=0: print(arr[i+1:i+moves+1]) ans+=sum(arr[i+1:i+moves+1]) return ans for _ in range(int(input())): n,k,z=map(int,input().split()) arr=list(map(int,input().split()))[:k+1] print(solve(arr,z)) 
 » 2 years ago, # |   0 can anyone explain the dp approach to problem B?
•  » » 2 years ago, # ^ |   0 there are things on youtube which explains the dp approach of B, you just type codeforces educational round 92 then you will get what you want :)
 » 2 years ago, # |   0 For the Good String question: I read the tutorial and worked on it myself for practice. Below is my solution in C++. It passes the test 1. But for test 2. my output is wrong for some cases. Could you please let me know where is wrong? Thanks a lot!using namespace std;int main(int argc, char** argv) { int i, j, cases, x, y, temp, tempX, tempY, ans, maxAns = -1; string s; cin >> cases; for(i = 0; i < cases; i++) { cin >> s; for(x = 0; x <= 9; x++) for(y = 0; y <= 9; y++) { ans = 0; tempX = x; tempY = y; for(j = 0; j < s.length(); j++) { if((s[j] - '0') == tempX){ ans++; temp = tempX; tempX = tempY; tempY = temp; } } if((ans % 2 == 1) && (x != y)) ans--; if (ans > maxAns) maxAns = ans; } cout << (s.length() - maxAns) << endl; } return 0;}
 » 2 years ago, # |   0 In problem B editorial "You have t pairs of moves (right, left) to insert somewhere inside the sequence of k-2t moves to the right."Why k — 2*t ?
 » 2 years ago, # |   0 Not expecting DP in second problem.
»
2 years ago, # |
Rev. 3   0

88352400 for probblem C , can someone please tell me what's wrong in my code?

Spoiler
 » 2 years ago, # |   0 Hey, Can somebody please help in problem B, I am getting TLE. I have used DP.Submission Link
 » 2 years ago, # |   +8 I have tried to make editorial for questions A-E . please have a look. Language :- Hindihttps://www.youtube.com/watch?v=S_-BaaH7P80&list=PLrT256hfFv5X1GNpiqEh5njN_WJmQu8Q6
•  » » 2 years ago, # ^ |   +6 Great Tutorial!!
 » 2 years ago, # |   0 Why in Problem G we have to find biconnected components? Any counterexamples to SCC?
 » 2 years ago, # | ← Rev. 2 →   0 In Problem B, Solution 1. Why are they adding arr[i+1] to the maximum pair sum. I mean we can go till pos, so we can add the pairs till pos but because of arr[i+1] pair , the last iteration will also add arr[pos+1]. Also, if we need to add that, why they not added (arr[i+1]) into the 's' variable.
 » 2 years ago, # |   0 Can anyone explain me what the line "Also, you can't perform two or more moves to the left in a row" in the problem 1389B-Array Walk, is meant to be? Can anyone elaborate this line, please?
 » 2 years ago, # |   0 I used a dynamic programming approach to problem $B$, but I get an output of $18$ instead of $19$ for the inputs 5 4 1 1 5 4 3 2 Here is my code cin >> t; while(t--) { cin >> n >> k >> z; for(int i = 0; i < n; i++) cin >> a[i]; memset(dp, -INF, sizeof(dp)); dp[1][0] = a[0] + a[1]; for(int i = 2; i <= k; i++) { dp[i][0] = dp[i - 1][0] + a[i]; for(int j = 1; 2 * j <= i && j <= z; j++) { dp[i][j] = max(dp[i - 1][j - 1] + a[i - j], dp[i - 1][j] + a[i - j]); } } int ans = 0; for(int i = 0; i <= z; i++) ans = max(ans, dp[k][i]); cout << ans << "\n"; } I don't know why this is happening, since for every possible number of total moves and left moves, we could have previously had one less left move or the same number of left moves. This is the main logic in my idea. Any hints or help would be appreciated.
 » 20 months ago, # |   0 In problem C, if we are using brute force, for all 100 combinations it is O(n) then it becomes overall O(100*n) which is O(10^7) and with that we have 1000 test cases also, so how does it even work? At first i didn't do it because i thought maybe it will cause TLE, but I realized that they have done the same thing.
 » 19 months ago, # | ← Rev. 2 →   0 can someone please share dp approach to problem C.. i'm thinking hard but still can't able to do using dp i also searched on YT but found only brute force method someone please help....
 » 17 months ago, # |   0 I think there is an error in the tutorial for [1389A — LCM Problem], it should be "If 2l > r, then there is no pair".
 » 12 months ago, # | ← Rev. 2 →   0 Problem B : Array Walk — I solved it with DP using memoization technique. Now, when I declare map, ll> (ll means long long int) to cache the values where the vector contains three elements, it results in TLE but when I declare map, ll>, ll>, the solution gets accepted. In both ways I am storing 3 elements as keys in map, first way using vector and second way using pair but first way results in TLE and second gets accepted, all code remains exactly same. Can someone please explain why this is happening ?
 » 3 months ago, # |   0 https://codeforces.com/contest/1389/submission/162863152 what is wrong in my code, pls help