Hello Guys I have divided problems into posts according to difficulty. Hope u all don't mind this. ;) This is the part 2 of two posts of unofficial editorials for November Long Challenge. For editorials of problems VILTRIBE, CLRL, PERPALIN and CHEFHPAL, click here. This post has editorials for problems SEGPROD and CSUBQ. Problem SEGPRODProblem Difficulty : MediumPrerequisites:Modular Multiplicative Inverse (euclid's method), Simple maths, and patience.Problem ExplanationGiven an array and a number P (not necessarily prime), Answer range product queries modulo P. (Not as simple as it looks). SolutionThe first thing i want you all understood is the euclid method of finding modular multiplicative inverse, vital to my approach, about which you may read from wikipedia and find the algorithm here. The naive brute force (not at all successful even for 20 points, due to integer overflow) would be to create a prefix product array (similar to prefix sum array) and for every query L to R, output prefixProd(R) / prefixProd(L1) and output modulo P. That's correct, but incapable of giving us 100 points. So, we move to modular multiplicative inverse (MMI) for help!!. We will to the same thing. Instead of prefix product, we will make an array storing prefix modular product (mod P ofcourse). Now, to answer queries, we are going to answer queries just as prefixModProduct(R)/prefixModProduct(L1). This is same as (prefoxModProduct(R) * ModInverse(L1))%P. If you have followed the logic upto here and understood MMI from geeksforgeeks page, you have earned 20 points. Hurray. The reason of only 20 points is that Modular multiplicative inverse of a number A exists only when A and P are coprime (gcd(A,P)==1). For first subtask, it was given that P is prime and all numbers are smaller than P. From that, it's obvious that gcd(A, P) is one for every element, so we get 20 points. Now, for full solution, we will use the same technique, but with a slight change in plan. Now, we will first find prime factors of P (U'll understand why) and handle numbers in a different way, which will be explained by an example. Suppose P = 6 and given array is 1 2 3 4 5 6 7 8 9 10 Now, Prime factors of 6 are 2 and 3. Now, we will handle powers of 2,3 separately and rest numbers separately. Create an array factors which will have value {2,3} for this example. create a 2d array factorPowSum[NumberOfFactors][N] factorPowSum array will look like below (explained how to make this below that) factor 0 1 2 3 4 5 6 7 8 9 10 11 2..... 0 0 1 1 3 3 4 4 7 7 8 8 (dots to adjust position, spaces are truncated automatically) 3..... 0 0 0 1 1 1 2 2 2 4 4 4 And we will think(it's necessary) we are given given array 1 1 1 1 5 1 7 1 1 5 11 (all numbers are divided by 2 and 3). Now, you will see that all numbers are coprime to 6. Now we can use the algorithm to solve the queries for reduced array. Let's assign factor 2 to index 0 and factor 3 to index 1. Creation of factPowSum array Just assign factPowSum{factorIndex}{0} = 0 and for 0<i<=N factPowSum{factorIndex}{i} = factPowSum{factorIndex}{i1} + Power of factor divided from ith number(1based indexing). For example, from 8, 3rd pow of 2 was divided, factorPowSum{0}{3} = factorPowSum{0}{4} + 3. Hope i made the array clear. Now, consider that we are given only powers of factors of P. Continuing same example, The same array we were working with now becomes (I know this is lengthy as hell, but I tried making it as easy to understand as possible) 1 2 3 4 1 6 1 8 9 2 1 (Only powers of 2 and 3 are considered from given array). Now, if someone ask us range query on this array, we can answer each query within O(number of factors of P) time. This is simple, just refer to factPowSum array and give it a try. Read further only after a try. let ans = 1. for every factor of P, ans = ans*pow(factor, factPowSum{R+1}factPowSum{L}). Now, combining above two subproblems, we get answer of query as let ans = (prefixModProd(R+1)*MMI(L) * product(pow(factor, factPowSum[R+1]factPowSum[L])) )%P. // The queries in question are 0indexed You are about to lose 100 points if you do this. Shocked!!. Be sure to take modulo after every multiplication so as to lose your AC to Integer overflow, use long long int/long only. Now you all deserve 100 points in this problem, but you will get TLE because test cases were too tight. Calculating powers with fast modulo exp even takes O(logN) time, which is too much for 10^7 queries. So we calculate powers of factors of P before answering queries and store in an array. I am not going to explain this but it's easily understandable from my code. In case you need help, feel free to ask. Added Proof:lemma: P can have at most 10 different prime factors. Proof: Try multiplying first eleven prime numbers. The product will be greater than 10^9. but as P <= 10^9, it follows that maximum number of distinct prime factors P can have is 10. Here's a link to my Code. (Took nearly 30 submissions for 100 points :D) Problem CSUBQProblem difficulty:MediumPrerequisites:Segment Tree, Too much patience unless u r a segment tree expert (msg me if u are, want a talk.)Problem ExplanationAn array of Size N and M queries, Tow Integers L and R (L<=R), handle following queries. 1. update ith value to X. 2. Find number of subarrays betweeen L and R, whose maximum values lie between L and R. (both inclusive) SolutionOne basic formula. (very basic) let function c denote c*(c+1)/2 (The number of all the possible contigious subarrays from an array of length c.) 1+2+3+4+5 ... (N1) + N = N(N+1)/2 Now, take L = 1, R = 10 and given array 2 0 11 3 0 (example given in problem) (I'm directly taking an array instead of a series of updates to explain it). Required subarrays between 1 to 5 are {1, 1}, {1,2},{4,4} and {4,5} An interesting thing to observe is that ans = c(2)c(1)+c(2)c(1) (I'll explain how i came up with this.) Two Facts: 1. Any element > R is never included in any subarray (like element 3). 2. Any number of elements smaller than L can be included in subarray as long as there is atleast one single element between L and R inclusive. So, we are going to get 25 points first using this fact. See this solution for details. in this solution, inc means consecutive elements till now which are smaller or equal to R, and exc means all consecutive elements < L till current position. Whenever we reach an element >= L, subtract excluded subarray count ( c(exc) ) from count and set exc = 0 and inc++ and whenever u get an element > R, add c(inc), subtract c(exc) and set both to 0. otherwise inc++, exc++. Ans is count + c(inc)c(exc). Got 25 points. This solution runs in O(QN) time. I hope this logic is clear. Because if it isn't, be sure to read again once, twice, 10 times, 10^2 times, 2^ 10 times (I like this quote :P). Now, Getting 52 points isn't difficult if You have heard about square root decomposition. I have submitted a solution using this technique too. :P. (I submitted too many solutions for this) I used in this solution apart from the counting technique we discussed above and one thing mentioned below Cheers, now we got 52 points with this ease, but no longer. Problem setters aren't going to gift 52 points with this ease. Now, think here, what information about a segment L to I and a segment from I+1 to R we need to find out all required information. Refer to my solution (One more solution :P) alongwith to understand what's going on. Let me help u. You need 8 variables (too much u feel i guess), as follows: 1. count => The count of subarrays already included in range, except those at borders. 2. incL => Number of elements from left to leftmost element which is > R. (0 if there's no element > R in range) 3. incR => Number of elements from right to rightmost element which is > R. (0 if there's no element > R in range) 4. excL => Number of elements from left to leftmost element which is >= L. (0 if there's no element >= L in range) 5. excR => Number of elements from right to rightmost element which is >= L. (0 if there's no element >= L in range) 6. size => size of range 7. found1 => boolean, which tells whether range contains atleast one element > R. 8. found2 => boolean, which tells whether range contains atleast one element >= L. Any ideas how we are going to merge ranges?, it's simple enough. Consider array 2 0 1 11 3 0 5 Answer of this Problems is c(3)c(1)+c(3)c(1) = 61+61 = 10 Think of this problem in two parts. for elements > R and for elements >= L. For first part, consider any element > R as dividing element. for above example, 11 is dividing the total ranges into two subranges of length 3 each. For second part, consider any element >= L as dividing element. For above example, 2,1,11,3 and 5 are dividing elements, resulting in 2 ranges of length 1. ( {1,1} and {5,5} ). If you understood this part, You can divide the given problem into two parts, the one dealing with elements to be included, and the one dealing with elements to be excluded. Here comes the Main part of solution,The Segment tree. Now, as you may see, i have used an iterative segment tree,(I messed up recursive one and also, iterative is slightly faster if u take the trouble to understand it) about which you may read here. I first of all, created a class S (struct equivalent in java) which hold all this info about each segment of tree. Now, the most important thing is to create a merge function, which will merge two segments into a larger one correctly. You may see from my code, i have dealt with inclusion and exclusion separately, making my code simpler. For first part, i check if both sides have element > R (our dividing element) (using boolean found1). if both has, include Left of large range will be outL of left segment and include right will be includeRight of right segment. count will be increase by c(left.incR + right.incL) because they are no longer boundary elements of segment. In case only one segment has dividing element, incL and incR of output range is assigned accordingly. Same way for large. And Here we go. We have solved the problem. Bet you didn't realize that. All that is left to implement it. And Now, nothing is going to stop you from 100 points except a TLE in one file. (Not Sure, because i had use the following tip to save time.) Link to My code. PS: This was my first editorial using segment trees. So, i am sorry if any error might have crept in (without my knowledge, of course) Do give your review for this editorial. As always, i wholeheartedly invite your suggestions and thank for your response to my previous editorials. PS:Sorry for delay, was held up in something important. Delay gift will be posted as soon as I learn the technique of problem Polynomial. :D Hope you don't mind delay in delay gift. :D asked 14 Nov '17, 00:00
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I would like to describe the fastest solution for SEGPROD that I came up in O(N log N + Q). Let's use SQRTdecomposition and divide array in sqrt(N) buckets of size sqrt(N), for each bucket let's calculate product for each prefix and for each suffix. And for each pair of buckets x <= y let's calculate a product of all elements between them in buckets F[x] * F[x+1] * ... * F[y], where F[i]  product of elements in ith bucket. All these operation can be done in 2*N+sqrt(N)*sqrt(N) = 3*N ticks. Let's answer for queries, suppose that l and r not in the same bucket, hence answer can be calculated in O(1) <> use some precalculated suffix production in lth bucket multiply it by some precalculated prefix production in rthe bucket and multiply all of it by precalculated F[idx(l)+1] * F[idx(l)+2] * .. * F[idx(r)1], where idx(x) denotes number of bucket where x is placed. But what if l and r in the same bucket? We can use SQRT decomposition one more time!! Let's split every bucket into sqrt(sqrt(N)) buckets, and for every bucketbucket will precalculate the product on the prefix, on the suffix and product for every pair of bucketbuckets inside the bucket, now we have 6*N operations, let's split every bucketbucket into sqrt(sqrt(sqrt(N))) buckets and for every pair of bucketbucketbucket use the same procedure as above. Then if query is in the same bucketbucketbucket(, rl <= sqrt(sqrt(sqrt(N)) ~ 5, and we can find the production simply. T(N)=3*N+sqrt(N)*T(sqrt(N)) for precalculating and answer in O(1) answered 15 Nov '17, 00:18
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One of the most unique and brilliant solution i have every heard, repeated application of sqrt decomposition to get query time O(1). I would be glad to see ur submission. Thanks for sharing ur approach friend. PS:if u r willing, i am recieving request for editorial of POLY, but wasn't able to solve that one. If u r willing to share ur approach with me, that would be really helpful. Be assured, due credit will be given to u. If u urself want to write editorial, that would also be welcome. In case u r busy, no problem mate!!!
(15 Nov '17, 00:32)
Thanks for this approach, I'm sure I can't implement this. Looking forward to viewing your submission :D Once again, thanks a lot buddy :)
(15 Nov '17, 00:35)
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I might be wrong, but i guess this solution is going to get TLE because of "precalculated F[l+1] * F[l+2] * .. * F[r1]" as you mentioned. Bounds of this problem were so tight even to refuse O(QlogN) solutions to pass even after many optimisations. And In case we try precomputation like F[L+1][R1] is going to give us MLE/TLE or both. Can you please explain that part??
(16 Nov '17, 00:18)
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Here is my very first AC submission  https://www.codechef.com/viewsolution/16282523 (there are a lot of optimizations, the most crucial is adding fast inputoutput(0.5 seconds and building this SQRTdecomposition tree layerbylayer, in that case, there will no many memoryjumps). I didn't want to write it, cause solution couldn't be understandable. The biggest problem is enumeration of buckets in the 2nd, 3rd, .., imagine that you write segment tree, but not for 2 childrens, for arbitrary K > 2. About POLY, do you know what is the convexhull trick?
(16 Nov '17, 01:42)
I've replaced that line, let we have sqrt(N) buckets. For every pair let's calculate production between them, totally there will be sqrt(N) * sqrt(N) = N such productions. And if we have a query for L and R from different buckets, let K = [sqrt(N)], idx(L) = [L / K], idx(R) = [R / K], let's take the production F[idx(L)+1]F[idx(L)+2]..*F[idx(R)1] if idx(L) != idx(R)
(16 Nov '17, 01:53)
I have already tried writing a segtree for more than two children in past, successfully written it too, but didn't feel that it was worth the effort at that time, to say the truth. And yes, i know about convex hull trick, but wasn't able to generalise it for nonlinear functions (got 30 points). I too have built sqrt buckets layer by layer to avoid memory complications. (upto x layers .. x i define myself using a formula on N) avoiding creating extra layers, but with magic = 4 instead of 8. Having a nice time with ur solution. Your solution is understandsble, thanks for writing it that way.
(16 Nov '17, 13:50)
The only thing that's had caused me a confusion was ur MUL array. I expected it to give MLE, but i was wrong there, it didn't. :) Now I'm crystal clear about implementation. Will soon implement it, possibly tonight. Thanks again for coming up with such a unique solution. PS: Nice use of iterative Segtree to reduce iteration count. :P
(16 Nov '17, 13:56)
thanks for sharing such a great solution :) i understand it but unable to implement it! I have first created sqrt(N) blocks and then for each blocks another sqrt(N) ...(as mentioned) but facing problem on how to map them to appropriate index during query. can you please explain a little bit how did you map indices in the inner levels ?
(22 Nov '17, 19:36)
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I also solved CSUBQ using seg tree. But my approach was a little bit different. It is easy to see that: Here is a link to my solution. I use recursive seg tree and did not need any optimizations. I also added an additional variable in the node that stores size of node to make code simpler. answered 14 Nov '17, 01:18
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Exactly what I did, except that I used 2 segment trees one for elements >= L and one for elements > R.
(14 Nov '17, 01:25)
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Thanks to both of u for sharing ur approach... Though ur approach was nice, But to tell the truth, this is my first ever segtree non classical problem which i managed to solve. 2 segtree would be too much for first timers...
(14 Nov '17, 15:24)
@taran_1407 you don't necessarily have to make 2 segment trees. Since the working of the 2 segment trees is exactly the same, I made a class for segment trees and used 2 objects of that class as mentioned above. It made the code look clean and simple. Here's the code.
(14 Nov '17, 20:13)
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Mate, i too used one segtree only. Thanks for sharing. :)
(14 Nov '17, 20:15)
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I too used segment tree, but my approach is kind of a standard one. Just save positions of maximum left and the right element which is < R and merge the segments.
(17 Nov '17, 01:09)

You can also solve SEGPROD by Sparse Tables. Initially, in C++, it was giving TLE in some cases. So, I submitted it in PYPY. Here is the link to the accepted submission. But after the contest, I saw that many contestants have solved it in C++ using sparse tables after optimizing the code. answered 14 Nov '17, 00:46
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I appreciate your approach but as a great coder said, everyone likes his own code even if other's code is too beautiful to be disliked. :) Thanks for shring your approach though.
(14 Nov '17, 00:48)
You are answering each query in (log {R  L + 1}) time using sparse table ? That is log n in the worst case. I tried using sparse table too, in C++. But even the first subtask did not pass. :/
(14 Nov '17, 00:49)
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His solution passed. Maybe it's time to learn efficient implementation of Sparse table.
(14 Nov '17, 00:51)
But, there are 2e7 queries in the worst case and answering each in log n time would take (2e7 * log n) computations which was failing clearly in C++. Not sure how it passed in python. I dont know if 'range product queries' can be answered in O(1) per query using Sparse Table. In that case it would pass.
(14 Nov '17, 00:56)
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I am amazed too.
(14 Nov '17, 01:05)
@jaideeppyne Yes, my solution is O(NLogN + QLogN). But I've submitted it in PYPY, it's time limit is same as that of Python i.e. 5x that of C++. However, PYPY is significantly faster than Python. So, in the end, my solution passed just within the time limits.
(14 Nov '17, 01:20)
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I read the comments and came to know that many people were doing the same thing as suggested in the editorial,but still getting TLE. I was having the same problem during the contest. And finally the only thing which solved my problem was 1based indexing. Because of 1based indexing the ifelse condition in query part (for checking left > 0 or not) can be avoided and hence the total time will decrease by some factor within the query. And there you go AC. Hope it helps :) For reference check by submissions : answered 14 Nov '17, 18:01
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Honestly, I feel that setter and tester were extremely wrong in setting the time limit. I wonder how much time setter's code takes to run, cause if its something like 3.83.9seconds, then time limit is wrongly put. Saying because, languages like java etc. got accepted (time bonus), while c++ got tle.
(14 Nov '17, 18:24)
Really @vijju123?? Is this the case?? Are the solutions with same approach but different languages are getting different verdicts?? This is a matter of serious concern as this would cause unfair advantage to some. I would really hate this if something of this sort happens. I guess this need strict action. Perhaps it should be made responsibility of problem tester to solve the problem with same approach using c++, python(PYPY or CPYTH) and Java, for these three languages are most popular ones. Hope this matter won't go unnoticed.
(14 Nov '17, 19:13)
Setter's solution works in 2.5 sec holy shit :O O_O
(15 Nov '17, 20:05)
Oh by god.... @vijju123 Well, I hope you convey this suggestion to admin. There should be proper testing in these three languages at least. No one should ever get undue advantage just because of a particular language. Hope once again, This matter won't go unnoticed.
(15 Nov '17, 20:40)

CSUBQ, I took 89 variables as well. You cannot say SEGPROD required patience, try it in C++. My code was taking ~4.5 seconds on local machine for worst case I produced, and its a constant TLE for 20 hrs straight on that SINGLE TEST FILE OF FINAL SUB TASK. Then got AC after a hard day XD answered 14 Nov '17, 00:09
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I know, the 9th file. But learnt something knew from Both these problems. That trust problem testers to make your programs work in worst complexity cases. :D (just joking) Learnt about Nonclassical application of Segment Tree, First time about MI. Like the problem set but cursed the test cases :D PS: Delay in delay gift post :D
(14 Nov '17, 00:14)
We cannot say the problems had weak cases this time I guess :p
MI?
:(((((((((((((((((((((((((((
(14 Nov '17, 00:25)
I had to find someone ready to explain me their approach (I haven't solved the problem myself yet, nor know how to). I found someone few minutes ago... I cannot really ask that person to explain right now, not at 12:30 am. I will post as soon as i get their approach
(14 Nov '17, 00:30)
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Yes, I understand. Anyone will be bewildered if he has to explain one of hard problems of long at 122am at night :p
(14 Nov '17, 00:34)
I know you would be surprised, but that person was willing to, but i myself had to refuse. I can't trouble that person at this time even if that person is willing to help me. So, hopefully only a small delay. :)
(14 Nov '17, 00:36)
Thats quite polite of you :D
(14 Nov '17, 00:42)
And generous of that person. :)
(14 Nov '17, 00:44)
@vijju123, Unluckily, That person got some unavoidable reasons, keeping that person busy, because of which I cannot do an editorial on POLY unless anyone else is willing to help me. Sorry for inconvenience.
(14 Nov '17, 20:05)
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I solved the SEGPROD problem using SQRT Decomposition and segment tree. Precomputed (each buckets product from range i to j, also precompute prefix and sufix of each bucket) out of query loop and answered each query in constant time. This approach had one flaw, that is, if L and R are both in same bucket, so I used segment tree for this case, and luckily it worked out. Here is the link to my solution  code answered 14 Nov '17, 03:45

@prakhar17252 I think u should precompute the modular multiplicative inverse for all elements of answer array. Since size of array is only 10^6 so complexity 10^6(logN) However in worst case of urs,complexity is O(210^7logN) which exceeds the Time limit Hope your first subtask gets passed with this answered 14 Nov '17, 08:29

just see that minute difference in my code to get rid of TLE in task 9.took me 8 hours of continuous effort to find it. TLE https://www.codechef.com/viewsolution/16259177 AC https://www.codechef.com/viewsolution/16260183 just replaced x with s new variable and it worked!! answered 14 Nov '17, 21:54
Happens, mate. You are not the first one for this, nor you will be the last. I too wasted nearly 12 submissions (WA Segment tree ones) on CSUBQ because of a variable mistake. Just changed variable and got AC.
(15 Nov '17, 20:44)

I solved CSUBQ(after contest) using 2 Fenwick trees cuz i am no expert with segment tree and Fenwick seemed easy to implement with need of just handling the merge or break segment conditions with if else blocks. answered 16 Nov '17, 01:42
Great... Its just that i haven't worked with fenwick tree, because segtree usually serve the purpose.
(16 Nov '17, 14:26)

There is no link associated for editorials of first four questions on Click Here.
Oops!!, forgot. Just updating right now
@taran_1407 I have asked a question for SEGPROD. It will be really helpful if you could help me debug my code and telling me where I am wrong.
No need for square root decomposition to get 52 points (CSUBQ). :P
https://www.codechef.com/viewsolution/16211947
Nice use of binary search. :) @eugait
@taran_1407 There is no binary search.
@eugalt, I'm not much in c++, but i thought lowerbound function uses binary search. Correct me if I'm wrong. :)
@taran_1407 There is lower_bound() function in the algorithms library which uses binary search on an iterator range. Here it's a map/set method with the same name.
Oops. My mistake :D