### ch_egor's blog

By ch_egor, 6 weeks ago, translation, ,

Thanks for the participation!

1313A - Fast Food Restaurant was authored by Endagorion and prepared by ch_egor

1313B - Different Rules was authored by meshanya and prepared by DebNatkh

1313C2 - Skyscrapers (hard version) was authored by meshanya and prepared by Sehnsucht

1313D - Happy New Year was authored and prepared by voidmax

1313E - Concatenation with intersection was authored and prepared by isaf27

• +80

 » 6 weeks ago, # |   +92 How Can I read Such a big editorial for B. havoc editorial to read...
 » 6 weeks ago, # |   -8 In the B problem, I think we just need to compare a + a with b + c
 » 6 weeks ago, # |   +52 Solution for B looks like some neural network ;)
 » 6 weeks ago, # |   +34 Contest has become mathforces with B
 » 6 weeks ago, # |   +17 All comments are related to B :D
 » 6 weeks ago, # |   +7 Is there any easier understanding of problem B?
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +24 For the minimum possible overall place: For a number $a$ in round 1 and another number $b$ in round 2, we should make sure $a + b \geq x + y + 1$. $1$ should find $x + y$, $2$ should find $x + y - 1$... And if $x + y > n$, $1$ should find the smallest number in round 2 which can be selected.So, we can see that the final answer will be $x + y - n + 1$, cause we can not arrange number $[1, x + y - n]$ to make them greater than $x + y$, and don't forget the ans must in $[1,n]$. For the maximum possible overall place: In round 1, for each number $a$ between $1$ and $x-1$(it can be empty), we can find a number $b$ in round 2, and make $a + b \leq x + y$.In round 2, for each number $b$ between $1$ and $y-1$(it can be empty), we can find a number $a$ in round 1, and make $a + b \leq x + y$.We cannot find any other pairs to make $a + b \leq x + y$. Thus, the answer is $x - 1 + y - 1 + 1$, and don't forget the ans must in $[1,n]$.Let $a = x - 1, b = y + 1$, while $a$ is decreased by $1$, we can add $1$ for $b$. If $b$ is greater than $n$, we can select any number remaining, cause at this time, $a + b < x + y$. To proof the second one, we can just swap the round 1 and round 2.It seems like it's a kind of greedy, but the editorial can proof the answer strictly. And also sorry for my poor English. Wish it can help you. :DUPD: swap((a,b),(x,y)), sorry for the mistake.
•  » » » 6 weeks ago, # ^ |   +13 what is a & b ?
•  » » » 5 weeks ago, # ^ |   0 thanks
•  » » » 5 weeks ago, # ^ |   0 thx, it helped alot
•  » » » 5 weeks ago, # ^ |   0 For the first case (minimum possible overall place) did you find a+b>=x+y+1 because it will give the count of all the combinations of final ranks which are more than x+y sum's rank? Also, I don't understand what you are saying in the next line ( the x+y>n condition)
•  » » » » 5 weeks ago, # ^ |   0 Of course, we can't make sure we'll always find such $a$ and $b$ that $a + b \geq x + y + 1$. And at that time, the answer will be increased by 1, because whichever we select the $b$, we can't make $a + b >= x + y + 1$, in other world, the person who get the $a$ place in round 1 will finally placed before Nikolay.
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +23 Not sure if it is easier, but I solved it with the following intuition: let's rephrase this task in a different way: you have a square grid of size n. Each participant with scores (xi, yi) will mark one cell with same coordinates. There should be only 1 marked cell on each vertical and horizontal line. Total score of a participant is xi + yi and participants with a same score share same diagonal (from top-right to bottom-down), participants with lower score are on diagonals on the left, and participants with higher score are on diagonals on the right.Now the task is the following — having Nikolay's mark on (x, y) how can we put other marks in order to have max or min number of marks on a right side from Nikolay's diagonal. You can check three cases depending on a position of Nikolay's diagonal from main diagonal x + y ? n + 1, where ? on of <,=,>
•  » » » 5 weeks ago, # ^ |   0 awesome.
•  » » » 5 weeks ago, # ^ |   0 Awesome. Thanks for sharing your intuition.
•  » » » 5 weeks ago, # ^ |   0 Could You please explain "You can check three cases depending on a position of Nikolay's diagonal from main diagonal x + y ? n + 1, where ? on of <,=,>"..How did you determine the position.
•  » » 6 weeks ago, # ^ |   -8 Fire woman is beautiful!~
 » 6 weeks ago, # |   -16 can anyone write the bruteforce solution for problem A.
•  » » 6 weeks ago, # ^ | ← Rev. 4 →   0 The problem clearly states- * All visitors should receive different sets of dishes and each visitor should receive no more than one portion. So, if a b c is the input: Only seven sets are possible they are as follows a,b,c,ab,bc,ca,abc. and their permutations. Here comes the greedy portion. For input: 1 2 2 3 a=2 b=2 c=3 Visitor 1 takes a, therefor new values are: a=1 b=2 c=3 Visitor 2 takes b, therefor new values are: a=1 b=1 c=2 Visitor 3 takes c, therefor new values are: a=1 b=1 c=2 Visitor 4 takes ab, therefor new values are: a=0 b=0 c=2 OUTPUT: a,b,c,ab 4 which is not optimal as (ca and bc) could have served two visitors and (cc) don't satisfy the conditions. For an optimal answer, we first sort all three number in reverse then map values to a, b and c, which are a=3 b=2 c=2 now automatically ab and ac will be chosen. OUTPUT: a,b,c,ab,ac 5
•  » » » 6 weeks ago, # ^ |   0 thanks sir
•  » » 6 weeks ago, # ^ |   0 You can check my solution https://codeforces.com/contest/1313/submission/71659186
•  » » » 6 weeks ago, # ^ |   0 thankyou
•  » » 6 weeks ago, # ^ |   0 you just if else about 7 case. First you need to sort element to non-decreasing. I call it a[0], a[1], a[2]. a[0] for smallest, a[2] for largest. First 3 case is, a[0] have at least 1, a[1] have at least 1, a[2] have at least 1. For next 3 case is, a[2] and a[0], a[2] and a[1], a[1] and a[0]. For the last case is a[0], a[1] and a[2]. Each food is have at least 1 amount in each if else statement. You should use variable to count how many case above true, each case true you should add 1 to count variable. Then output count as the answer.
•  » » » 6 weeks ago, # ^ |   0 thank you
•  » » 6 weeks ago, # ^ |   0 This is how I approached it — https://codeforces.com/contest/1313/submission/71659645
•  » » » 6 weeks ago, # ^ |   0 thank you
•  » » 6 weeks ago, # ^ |   0 t=int(input()) for i in range(t): count=0 n,m,r=sorted(map(int,input().split()))[::-1] if n>=1: count+=1 n-=1 if m>=1: count+=1 m-=1 if r>=1: count+=1 r-=1 if n>=1 and m>=1: count+=1 n-=1 m-=1 if m>=1 and r>=1: count += 1 m-= 1 r-= 1 if r>=1 and n>=1: count += 1 r-= 1 n-= 1 if n>=1 and m>=1 and r>=1: count+=1 n-=1 m-=1 r-=1 print(count)
•  » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 @sanjeevkumar113f did ur soln. gave AC?
•  » » 5 weeks ago, # ^ |   0 You can use Greedy method  /// Sort a > b > c if (a < b) swap(a, b); if (a < c) swap(a, c); if (b < c) swap(b, c); int res = 0; /// Take 1 disk if (c) res++, c--; if (b) res++, b--; if (a) res++, a--; /// Take 2 disks if (a && b) res++, a--, b--; if (a && c) res++, a--, c--; if (b && c) res++, b--, c--; /// Take 3 disks if (a && b && c) res++, a--, b--, c--; cout << res << endl; 
•  » » » 5 weeks ago, # ^ |   0 thanks
•  » » 5 weeks ago, # ^ |   +8 Notice that you should take (a-b) and (a-c) before take (b-c) else you will get WA
•  » » » 5 weeks ago, # ^ |   0 i have sorted them in increasing order i consider them m ,n r this solution is accepted during contestbefore sorting i was getting error what you have noticed
 » 6 weeks ago, # |   +12 B looks so hard in the editorial:)
 » 6 weeks ago, # |   +56 I would like to understand D. Why we use bit mask of size k?What does it mean if "we can find free bit and create match to this segment" to create the dp?
•  » » 6 weeks ago, # ^ |   +2 Let’s imagine this situation. You gave every segment some number 1..k, such there are no intersecting segments with the same number. In this case you can easy maintain dp.When we try to add new segment, we take some remaining color to match it to this segment (you have mask of free colors). So we came back to previous situation.
•  » » 6 weeks ago, # ^ |   +31 Suppose we sort and process a start/end event queue over the segments. At each start event, we will assign that segment the smallest integer not in use by another active segment. Since there is no index with more than $K$ segments covering it, during this process we will assign segments at most $K$ unique values.These values correspond to an index of a bit in a bitmask; this bit is on if we currently have this segment selected. Because each index is covered by at most $K$ segments, a bitmask $[0, 2^K)$ can uniquely identify any subset of segments which are crossing some index.Then, you can write a take/leave DP on the segments, and score ranges you cover based on the parity of your bitmask. code
•  » » » 5 weeks ago, # ^ | ← Rev. 5 →   0 The language of the question is a bit confusing. Does this line — "It is also known that if all spells are used, each child will receive at most $k$ candies." mean that the intervals that are supplied already satisfy this criteria or does it mean that our selection of intervals should be such that this criteria is satisfied (i.e. at most $k$ intervals pass through one point)? because I think only in the latter case this algorithm would work.
•  » » » 5 weeks ago, # ^ |   +8 thanks a lot
 » 6 weeks ago, # |   0 Can someone please help me find error in my logic with question B.my submissionFor minimum case, I started making pairs such that their score is one greater than our score. Then after that, some scores in first contest remain which are greater than a, and some in second contest which are greater than b. we take min of these to give additional people with score greater than us.sg1 = min(n-1-x, y-1)sg2 = min(n-y-1, x-1)og = min(n-x-sg1, n-y-sg2)mn = n - sg1 -sg2 - ogSimilarly I do for max, but for this case, we search for equal score. After that we search for scores less than a in first contest, and scores less than b in second contest. We add min of these to give additional people that may come at or before us in position.gr1 = min(n-x, y-1) gr2 = min(n-y, x-1)ol = min(x-1-gr2, y-1-gr1)mx = gr1+gr2+ol+1
•  » » 6 weeks ago, # ^ |   0 The answer should in range [1,n].
•  » » » 6 weeks ago, # ^ |   +3 my god I feel silly now. I only had to change sg1 and sg2 so that they dont go negative! Thanks mate. The problem was that when a, b were last then sg1 and sg2 should have been zero, but I overlooked that issue!
 » 6 weeks ago, # | ← Rev. 2 →   +18 Problem B For worst rank: Round 1 ......x........ Round 2 ..........y.... The total score needs to be made as greater as possible. So, Step 1 — We can couple entries like (x+1,y-1), (x+2,y-2)..... and (x-1,y+1), (x-2,y+2)..... Step 2 — After this step we will be left with some left out numbers is the first few ranks of few rounds, so we'll couple them together. The reason we are going only for the starting ranks is that coupling of these elements will result in score less than (x+y). For example: Round 1: 1 2 3 4 5(x) 6 7 8 9 10 Round 2: 1 2 3 4 5 6 7 8(y) 9 10 In step 1, we'll couple (6,7),(7,6),(8,5),(9,4),(10,3) and (4,9), (3,10) After step 1, we'll look for the left out ranks among the starting ranks of rounds 1 and 2 --> 1,2 (round 1) and 1,2 (round 2). So, we'll couple them. Answer = 1(including itself) + 5(same score part 1) + 2(same score part 2) + 2(score less than x+y) For best rank: Similar to the worst rank case. Here we will try to make score greater than (x+y). Sum = (x+y+1) will yield the most number of possible results, so we'll aim for sum to be x+y+1 (similar to what we did for sum=x+y in the previous case). Edge case: min(x-1,n-x-1) may produce negative numbers, so we'll need to take max with 0. This will give us the maximum possible number of scores that are greater than (x+y). So we'll subtract this from N. Code for reference: // worst rank part1 = min(x-1,n-y) part2 = min(y-1,n-x) w = part1 + part2 + min(x-1-part1,y-1-part2) worst = 1 + w // best rank part1=min(y-1,n-x-1) part2=min(x-1,n-y-1) part1=max(part1,0) part2=max(part2,0) best = N — (part1 + part2 + min(n-x-part1,n-y-part2)
•  » » 6 weeks ago, # ^ |   0 I did exactly same as this, but stupidly I didnt do the part1 = max(part1, 0) for best rank. Bye bye expert, on the bright side I can give div 3 ;-)
 » 6 weeks ago, # |   0 Problem B. I forgot about the borders and place max(1,min(n,x+y-n+1)) wrote max(1,x+y-n+1). and finally I didn't solve the problem during the time of the Contest
•  » » 5 weeks ago, # ^ |   0 I know that feel, bro
 » 6 weeks ago, # |   +3 Why cant I view other people's code ? :/
 » 6 weeks ago, # |   0 What is the approach of finding "nearest" numbers to the right and to the left that are smaller than the current one?
•  » » 6 weeks ago, # ^ |   0 Maintain a monotonic stack
•  » » 6 weeks ago, # ^ |   0
 » 6 weeks ago, # |   0 Can someone tell me what's the problem in this code? https://codeforces.com/contest/1313/submission/71672652
 » 6 weeks ago, # |   +3 How to search the "nearest" in c2 (second solution)?
•  » » 5 weeks ago, # ^ |   +1 Pls, check mine, there are comments: https://codeforces.com/contest/1313/submission/71813248stack approach from here helped me: Next smaller element
•  » » » 5 weeks ago, # ^ |   0 Thanks man
•  » » » 5 weeks ago, # ^ |   0 This helped me a lot. I never needed nor thought about how to precompute this information. Thank you!
 » 6 weeks ago, # | ← Rev. 2 →   0 Problem C. in editorial(second solution). please help. can you this code O(n*n) optimize to O(n log n) or O(n). I could not find j faster. for (int i = 1; i <= n; i++) { int j = 0; for (int k = 1; k < i; k++) { if (m[k] <= m[i]) { j = k; } } l[i] = l[j] + (i - j) * m[i]; } 
•  » » 5 weeks ago, # ^ |   0 you can use segment trees or fenwick trees for log n search of minimum element in a range
•  » » 5 weeks ago, # ^ |   0 Use a strictly ascending stack (one for L and one for R) — O(n) as every element will only be popped once.
•  » » 5 weeks ago, # ^ |   0 Only I can't understand the answer of problem C?
 » 6 weeks ago, # | ← Rev. 2 →   0 Can anyone explain D. What do those bits represent. For ex. what would the state represent, and what does dp calculate (I guess it gives the maximum score for that particular state but I am not sure). I am not getting any ideas. Also, I am not completely convinced -> because for the case when k = 1, this won't work.
 » 6 weeks ago, # |   0 can someone please share the code of the 1st solution mentioned for c2 problem? it is quite difficult to understand just with the editorial.
 » 6 weeks ago, # |   +2 What if in C2 we have more than one min element??
•  » » 6 weeks ago, # ^ |   0 It does not matter, in case it has more than 1 min elements, choose the first one and divide the array into two parts — one that is left to it and one that is right to it. Calculate ans recursively for them
•  » » » 5 weeks ago, # ^ | ← Rev. 2 →   0 S.Jindal I am doing exactly the way you are telling but still I am not able to get the maximum possible answer...can you help my submission 71992921
 » 6 weeks ago, # |   +3 Please explain solution of problem 'D' in detail
•  » » 5 weeks ago, # ^ |   +6 Well detailed and explained solution of problem 'D' Dp states explained71761226 Hope it helps :)
•  » » » 5 weeks ago, # ^ |   +10 good job bro
•  » » » 2 weeks ago, # ^ |   0 How can we memoize using dp[id][mask] ? The same mask can denote a different set of segments right ?
•  » » » » 2 weeks ago, # ^ |   0 A mask can represent different set of segments but a mask at a particular 'id' represents a unique set of segments.(This is due to the fact that atmost only 8 segments can be active)
•  » » » » » 2 weeks ago, # ^ |   0 Is the fact that only 8 segments can be active the only reason why for an id, a mask corresponds to a unique segment set ? I didn't understand how the code is ensuring that a mask for an id is a unique set of segments.
•  » » » 11 days ago, # ^ |   0 for a segment, why is l, r+1 added and not l, r ?
 » 6 weeks ago, # | ← Rev. 5 →   +5 I am getting Runtime error "Exit code is -1073741571" on problem C2 with C++1471730114while the same code is ACCEPTED with C++17 71727975What could be causing the difference?
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   0 I looked further into this and surprised to know that stack size limitation is the culprit. Default "8192" KB (in linux) is too low. Setting it to "100000" worked for me, and looks as if a higher value is set for c++17. Not sure what could be the reason for setting this as low for c++14
 » 6 weeks ago, # |   +5 In the recursive approach for C2 how to make choice after finding minimum i.e, if we have to recurse on the right or left part?
•  » » 6 weeks ago, # ^ |   +3 You have to recurse on both the parts and calculate their respective scores. Depending upon which side has better score, you need to set the opposite side to the min value.
•  » » » 6 weeks ago, # ^ |   0 thanks, but what if there are multiple minima ? do we have to consider any one of those or
•  » » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 Yes just consider the first such minima (from left) and break the array at the point.
•  » » » » » 6 weeks ago, # ^ |   0 thanks! so for finding minima(RMQ), we have to use segment tree/ sqrt decomposition. What is the O(N) approach?
•  » » » » » » 6 weeks ago, # ^ |   0 Sparse table.
•  » » » » » » » 6 weeks ago, # ^ |   0 sparse table supports query operation in O(1) time but O(N Log N) preprocessing time.
•  » » » » » » » » 6 weeks ago, # ^ |   0 And segment tree preprocessing is O(N logN)? But with sparse table you can answer RMG query in O(1).
•  » » » » » » » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 Segment Tree construction is O(N), because there are ~2*N nodes in the tree and each node needs constant time.Query/Update takes O(Log N) time
•  » » » » » » » » » 5 weeks ago, # ^ |   0 And solution with segment tree is O(N logN) and with sparse table is O(N).
•  » » » » » » » » » 5 weeks ago, # ^ |   0 We look to main solution complexity in this we always say that preprocessing take O(NlogN), and solution take O(N).
•  » » » » » » » » » 5 weeks ago, # ^ |   0 Hi in (C1)easy version problem i have found the minimum element and then found the sum of elements on the left side of minimum and similarly on the right hand side . If the sum on the right side is less than that of left side then then i have assigned minimum value to all elements to the right side and vice versa. can u pls tell why this approach is not working
 » 6 weeks ago, # | ← Rev. 2 →   0 Why second solution work(for problem C2), can someone explain?
 » 6 weeks ago, # |   0 Anyone solve problem B with binary search? Cause this problem has this tag.Please show me.
 » 5 weeks ago, # |   0 Thanks for the tutorial <3
 » 5 weeks ago, # |   +1 what stand between me and the rating add,is the math
 » 5 weeks ago, # |   0 I tried a recursive approach for C1, getting WA can anyone tell me whats going on !!My code
 » 5 weeks ago, # |   0 I wrote a segment tree & divide and conquer solution for C2, based on the first idea in this tutorial of problem C2. But I get TLE here: https://codeforces.com/contest/1313/submission/71738916. I tried many times at my local machine, using a variant to count the time complex and found the divide and conquer procedure is O(N), and my segment tree is O(17) = O(logN). So I used ST table to reduce complex from O(N logN) to O(N) here: https://codeforces.com/contest/1313/submission/71741474, get Accepted.
 » 5 weeks ago, # | ← Rev. 2 →   0 .
 » 5 weeks ago, # |   +8 Can anyone just explain D more clearly ? I'm too stupid to understand the given editorial.
 » 5 weeks ago, # |   -8 How to solve the problem C2 recursively in first solution,isn't it got a comlexity 2^n?
•  » » 5 weeks ago, # ^ |   0 Learn about sparse table you will get the answer
 » 5 weeks ago, # |   0 Help needed in problem c2.My approach: for selecting the pick index I calculated the following for each index.sum=dpl[i]+dpr[i]-a[i] here dpl[i] is sum from 1..i making non decreasing segment and dpr[i] is same from n..i for which index I get the maximum sum I selected that index to be the peak index. And then just calculated the final result. Can someone please tell me why this is wrong or any test case? my code: 71701033
 » 5 weeks ago, # | ← Rev. 2 →   0 In (B) some proofs may become much more symmetrical if you notice that you can make the substitutions $x\leq y$ Change $x,y$ so that (case 1) $x=y$, or (case 2) $x+1=y$. Also an observation is that you don't have to solve a completely new problem for the minimum after solving the maximum problem; specifically if $x+y = c$, then the worst possible for $x,y$ corresponds to the best possible for some imaginary opponent with $x'+y' = c+1$; one just needs to subtract the number of "ties" in the $x',y'$ case, and add in a $+1$ because $x,y$ beats itself.
 » 5 weeks ago, # |   0 In C2 editorial second soln, what's an optimal way to find "nearest" numbers to the right and to the left that are smaller than the current one.
•  » » 5 weeks ago, # ^ |   0
 » 5 weeks ago, # |   0 Can anyone help me with problem Skyscrapers (Hard Version), I am getting memory limit exceeded for my solution I used Sparse Table for finding range minimum query and did a recursive method My submission
 » 2 weeks ago, # | ← Rev. 2 →   0 There are death of resoures to understand DPBitmasking concept . I am feeling helpless. please someone share enough the resource to get this concept easily. This would also enable me to solve problem D at myown
 » 2 weeks ago, # | ← Rev. 2 →   0 I struggled a lot with problem C / C2 (skyscrapers). Some comments said to use a monotone stack, which was a big hint, however I have only done a few problems with this approach and I didn't know what the stack actually represents. Finally I came up with it:For now we'll focus our attention on nondecreasing skylines. For every index $i$, we will identify the optimal skyline from index $1$ to $i$ inclusive. For $i=1$, clearly we use $a_1=m_1$. More generally, for $i=1$ to $n$, we push $i$ onto the stack $s$ after popping all elements $j$ such that $m[j]>m[i]$. Finally we push $i$. The stack right now describes the optimal skyline ending at $i$; Specifically, if $s$ contains $i=i_k > i_{k-1} > \cdots > i_1$, then we make assignments as follows: indices $(i_{k-1},i_k]$ get height $m[i_k]$, indices $(i_{k-2},i_{k-1}]$ get height $m[i_{k-1}]$, $\ldots$, and indices $(0,i_1]$ get height $m[i_1]$. It is a little exercise to see that this is a valid assignment and optimal.Let's call $maxIncreasingSkyline[i]$ the sum of the heights of this optimal skyline. It can be computed in amortized $O(1)$ at each step while doing the popping/pushing.One can do the same procedure for nonincreasing skylines, by iterating from $n$ to $1$, and similarly obtain $maxDecreasingSkyline[i]$, which represents the optimal skyline that "begins" at index $n$ and ends at $i$.Now the actual optimal skyline will be one of 3 possibilities: a nondecreasing skyline (whose value is $maxIncreasingSkyline[n]$), a nonincreasing skyline (whose value is $maxDecreasingSkyline[1]$), or a combination of the two, in other words, there is some $j$ such that $maxIncreasingSkyline[j]+maxDecreasingSkyline[j+1]$ is maximal. The latter can be computed in $O(n)$ time since we have already prepared the two arrays. Then one does another linear scan to reconstruct the assignments $a[\cdot]$ from the stack.My submission is here but hopefully the explanation above is adequate.