### pikmike's blog

By pikmike, history, 9 months ago, translation, , 1207A - There Are Two Types Of Burgers

Idea: BledDest

Tutorial
Solution (Roms)

1207B - Square Filling

Idea: BledDest

Tutorial
Solution (BledDest)

1207C - Gas Pipeline

Tutorial

1207D - Number Of Permutations

Idea: Roms

Tutorial
Solution (Roms)

1207E - XOR Guessing

Tutorial
Solution (BledDest)

1207F - Remainder Problem

Idea: BledDest

Tutorial
Solution (BledDest)

1207G - Indie Album

Tutorial
Solution (PikMike)  Comments (64)
 » 9 months ago, # | ← Rev. 2 →   [DELETED]
 » In solution for C) editorial says that: If s[pos]=1 then we have to stay on the height 2, but in the sample test 1 8 2 5 00110010 It is optimal to go to height 1, after index 3, where s = 1, and s = 0 (as shown in the problem figure). What am I missing here?
•  » » The "s" array in the editorial corresponds to intervals (so s corresponds to the height of the interval (3, 4)), but the dp array "d" corresponds to the height at specific x-values. In other words, because s = 1, the road needs to be at height 2 at x=4, meaning that d doesn't exist. s = 0 instead means that d and d are both OK.
•  » » It depends on what's the value of a and b. whether to get down at the gap of two crossroads or not, depends on how many piller you need to add. case 1: If you get down, then it takes 2 more pipes. which adds 2*a more values case 2: if you don't, then it takes k more pillers. which adds k*b more values take the min of case1 and case 2. That's all.
 » I didn't get the 2nd query of F ? What sum is supposed to calculate ?
•  » » sum of all those elements whose indices give remainder y when divided by x.
 » During contest, after solving first 5 problems, spent 1 hour working on problem F looking for a solution which was faster than O(n^(3/2)) because I assumed, incorrectly, that O(n^(3/2)) cannot pass... I even designed an O(n^(4/3)) algorithm to solve a simpler version of the problem where all the numbers are fixed from the beginning and never modified. (The solution is similar to the solution in the editorial except that the value of K is n^(2/3) instead of sqrt(n)). I then spent all the time trying to improve my algorithm to support modification to the sequence...Honestly, I also felt that problem E was much easier than problem C and D.
•  » » 9 months ago, # ^ | ← Rev. 2 →   I think 500000 is too large that my friend sxd666 got FST only because he used long long instead of int.
 » Can anyone explain more elaborately :For queries with small x (x≤K), we may notice two things: there are only O(K2) possible queries; each number in the array affects only K possible queries.
•  » » k=4 1 2 3 4 5 6 7 8 then 1 2 3 4 is only possible for <=k and each can have atmost ai-1 remainder, hence total approximately k^2 queries.
 » IN ques.D for counting of cnt1,given is cnt1=c1!*c2*!..cn! my doubt is while calculating c1! ,it is not necessary that the corrsponding elements in the another column will be all different and give different permutations..so we must divide it by those factorials..
•  » » Even if you consider the second column, even two sets of identical pairs will need to be considered twice so there is no need to divide.
•  » » 5 months ago, # ^ | ← Rev. 2 →   I was thinking the same thing. The third sample case is as follows 3 1 1 1 1 2 3 The given answer is 4. However, there are only 3 ways to arrange the above pairs (3!/2!). What am I misunderstanding?
•  » » » Permutations->1,2,32,1,3.So only two bad sequences. Hence 4 good sequences.
•  » » » » 4 months ago, # ^ | ← Rev. 2 →   I thought 1,2,3 and 2,1,3 shouldn't be counted twice, since they are identical sequences, but the permutations are distinct, so they should be counted separately. Got it, thanks!
 » 9 months ago, # | ← Rev. 2 →   Can someone explain the question C please?
 » An online solution to problem G :We build the suffix automaton for the trie which is given, and then we build a segment tree for every node (endpos set) . For a query, we do the heavy-chain decomposition for the trie, and then we query the segment tree sum.Total time complexity is $\mathrm O(n\log^2n)$.This method use so much memory that it will get MLE on test 21 (https://codeforces.com/contest/1207/submission/59296517), but you can get AC by replacing the array with std::map.Sorry for my poor English.
•  » » Why your code links to submission by different user?
 » 9 months ago, # | ← Rev. 3 →   The editorial solution for C, else { d[pos + 1] = minOf(d[pos + 1], d[pos] + a + 2 * b) } should it be else { d[pos + 1] = minOf(d[pos + 1], d[pos] + a + 2 * b); d[pos + 1] = minOf(d[pos + 1], d[pos] + 2 *(a + b)); } why the solution ignores the 0 to 1 transition when the s[i] == '1'?
•  » » 9 months ago, # ^ | ← Rev. 2 →   When s[i] = '0', there are 4 possibilities, you can reach 'pos+1', that is two transitions each from height 1 and 2. When s[i] = '1', there is only one 1 possibility, you can reach 'pos+1', that is using one gas pipeline and 2 pillars. Pictorial illustration
•  » » » but When s[i] = '1', why it can not be height 1 to height 2 transition? which uses 2 pipes and 2 pillar
•  » » » » 9 months ago, # ^ | ← Rev. 2 →   Because you can only have zig-zagged line in the interval (i, i+1) when s[i] = '0', when s[i] = '1', you cannot break the line, it must be linear coming from the left side. Hence you can only modify dp[i+1].
•  » » » » » ok, I got it. Thank you.
 » 9 months ago, # | ← Rev. 2 →   I am getting wrong Answer on Test 13 in Problem 1207C Gas Pipeline. My logic is that i am using dp[j] which is the cost to bring pipeline to height 1 at position j and dp[j] for cost to bring pipeline to height 2 at j. if Character at j is 1 dp[j] is -1 which means that it is impossible to bring pipeline to height 1. I am using MAX_VALUE in place of -1 in dp[j-1] if dp[j-1] is -1 which means it has a very high cost to build the pipeline from there.Link to my Solution..https://codeforces.com/contest/1207/submission/59357734
 » 9 months ago, # | ← Rev. 2 →   Can someone tell me why my submission is giving WA for question D. Thanks in advancehttps://codeforces.com/contest/1207/submission/59385418
•  » »  cout << ((tot - (cnt1 + cnt2 - cnt12))%MOD); } else{ cout << ((tot - (cnt1 + cnt2))%MOD); Here's the mistake. you are subtracting something from modded value. In some case value of tot will be less than (cnt1 + cnt2 — cnt12) or (cnt1 + cnt2) that's why your final answer is landing to negative value.replace your lines with below lines.  cout << (((tot - (cnt1 + cnt2 - cnt12))%MOD)+MOD)%MOD; } else{ cout << (((tot - (cnt1 + cnt2))%MOD)+MOD)%MOD; 
•  » » » Oh right Thanks a lot . After one hour of struggling it was the MOD :p
•  » » » » 9 months ago, # ^ | ← Rev. 2 →   Hey, I understood the approach to calculate c1,c2. Can you elaborate on how to calculate c12.It would be much helpful. Thank you :)
•  » » » » » c12 is the case when both the first and second elements should be sorted in order. So just sort the pair and check if the new sequence is sorted according to both first and second element. Then just check for the repeating pairs and subtract it from c1 and c2.
•  » » » » » » Could You please elaborate a bit..Thankyou!!
 » F is not worth 2100.
•  » » 9 months ago, # ^ | ← Rev. 3 →   What do you mean? For each problem you get 1 point for solving it.UPDATE: OK, I see. Tags with numbers of points have been added.
 » Could anyone give me a greedy solution for problem C. I came up with this approach in the contest but I can't implement it correctly. Thanks in advance!
•  » » 9 months ago, # ^ | ← Rev. 2 →
•  » » » thank u
•  » »
 » 9 months ago, # | ← Rev. 2 →   In the editorial of Problem F it states that: "Note that, as in most problems related to sqrt-heuristics, it may be optimal to choose the constant that is not exactly sqrt(N), but something similar to it (but most solutions should pass without tuning the constant)"Could anyone please explain why it may not be optimal to choose exactly sqrt(N)?
•  » » 9 months ago, # ^ | ← Rev. 2 →   The two parts of the algorithm have O(K) and O(N/K) complexities but the constant factor will not necessarily be the same (e.g. we do a modulo and use more memory for the small x part and always update K elements, while the naive algorithm for large x just sums N/x values which is only at most N/K... many things to consider).Pushing K on one side or the other of sqrt(N) will increase the time spent on one part and reduce the other, so because of the constant factors the optimal sum of complexities might be for 0.7sqrt(N) or 3.5sqrt(N) or whatever. Here sqrt(N) is 750 but experimentaly i got the lowest execution time with K around 400.Do people really test this in competition though ? Or just guess and round a bit up or down ?
 » 9 months ago, # | ← Rev. 3 →   Never mind. Solved it.
 » 9 months ago, # | ← Rev. 2 →   [deleted]
 » I've used greedy approach to solve problem C. It fails on test case 13. Please tell me where it goes wrong. Code
 » can someone help me in problem D,i got the concept how to solve it but i am not getting how to calculate cnt12
 » Please someone explain the problem E !!I couldn't understand from the editorial.
•  » » 1) In each query generate numbers in the way to check some bits in any $i$ case (if we are sure that $k_{th}$ bit is '1' in choosed number and '0' in answer, then $x$ has $k_{th}$ bit also '1', otherwise it is '0' — it follows from the properties of XOR).2) Let's fix first $a$ bits as '1' and choose any combinations in another $14 - a$ bits. We have to generate $100$ different combinations, so $2^{14 - a}$ must be better than $100$ => $14 - a = 7, a = 7$ ($2^7 = 128 > 100$).So we fix first $7$ bits as $1111111$ and increase another, getting the following sequence: $11111110000000_2$, $11111110000001_2$, $11111110000010_2$ and etc.3) Using the same algorithm generate another $100$ numbers, but visa versa: $00000001111111_2$, $00000011111111_2$, $00000010111111_2$ and etc.4) From first query answer we check first $7$ bits using XOR, from another query answer — last $7$ bits and finally get complete answer.
•  » » » Oh so, now I get it. Thanks a lot !!
 » Task F is quite easy to solve and code, but statement is quite hard to understand :DAt first I thought that it is necessary to calculate the sum of all elements $a_i = k \cdot x + y$, not indexes)
 » In G instead of adding on path to the root and extracting value from vertex we can do: add to a vertex and calculate sum in a subtree (it can be easily verified to be equivalent)
 » 9 months ago, # | ← Rev. 5 →   Another solution for problem G:There are no more than $S$ = $2 \times sqrt(4 \times 10^5)$ different lengths in the set of all patterns, say $L_1, L_2, ..., L_S$.When DFS to some trie node $v$, among all different patterns with length $L_i$, there is at most one occurrence ending at node $v$.Therefore, maintaining number of occurrences of all different patterns can be done in $O(S)$ whenever DFS moves to next vertex.
•  » » So you do an ordinary Aho-Corasick but with the text inside which we search being a trie rather than just a plain text?I think we can even bound S by sqrt(2*4*10^5) but is it enough? Your total compexity is O(S*T). It can be as much as 350e6 operations. Have you checked if this solution passes the tests?
•  » » » 4 months ago, # ^ | ← Rev. 4 →   Sorry I forget to reply for a very long time. :PYes I did, right after the contest. Here is the link:https://codeforces.com/contest/1207/submission/59360068https://codeforces.com/contest/1207/submission/59322378(I got two and now I don't know why!)I think 350e6 is acceptable considering TL is 3 seconds.
•  » » » » Thanks anyway!
•  » » I get your meaning. But the total complexity is $S * totLen * O(hash)$, which is $4e8 * O(hash)$I suspect it will get $TLE$ verdict though.Did you implent the method you mentioned? I'm just curious.Thank you
•  » » » 4 months ago, # ^ | ← Rev. 4 →   Well, I don't get why there is an $O(hash)$ in the complexity.If I remember well, it's $O(S \times totLen)$. (Maybe I'm wrong because it's very long ago :P) Here's the link of my implementation:https://codeforces.com/contest/1207/submission/59360068 https://codeforces.com/contest/1207/submission/59322378(I got two and now I don't know why.)
•  » » » » Oh, I get your method, thanks!!Previoulsy, I thought you'll use hash to record the answer, but indeed you still use Aho-Corasick. Besides, your method is simplier than using a data structure to calculate the answer.
 » 9 months ago, # | ← Rev. 2 →   Can someone explain how to calculate c12 more elaborately in Div2D?
 » 9 months ago, # | ← Rev. 2 →   For EI have an approach but don't know whether it's feasible or not.Let us denote numbers of $1^{st}$ query by $a_i$ and numbers of $2^{nd}$ query by $b_j$.So if we select numbers in such a way that xor of any pair ($a_i$,$b_j$) is unique. Now let us denote the output to two queries by $N$ and $M$ respectively then we can say that $N$ = $a_i \oplus x$ and $M$ = $b_j \oplus x$.If we now do $N \oplus M$ then we get $a_i \oplus b_j$ .Since we selected such numbers that every pair has unique xor we know value of $a_i$ and $b_j$ and once we get this we can get $x$ by $N \oplus a_i$ or $M \oplus b_j$.So my question is there any way to select such numbers that their xor is unique.BledDest
 » 6 months ago, # | ← Rev. 2 →   1207E - XOR-угадайкаI solved this problem in a different way than editorial, so i'm sharing it. It's pretty simple too.For the first query, We simply give numbers from 1 to 100.now we the judge will return me ($Ai$ xor $X$)we xor all number from 1 to 100 with this number.for the number $Ai$ it will become ($Ai$) xor ($Ai$ xor $X$) = $X$so the required number is in the list of numbers, let's call this list suspects.Now we give query numbers from $1 * 163$ to $100 * 163$The judge will return some value lets call it $Y$. For each value in the list suspect we check its xor value from ($1$ x $163$) to ($100$ x $163$), if this equals $y$, then that value of suspect is our answer. Solved without any kind of bitmasks.P.S i choose 163 because its the smalles number who on multiplication with 100 will be less than $2^14 -1$