By Anadi, history, 3 years ago,

Tutorial
Solution

Tutorial
Solution

Tutorial
Solution

Tutorial
Solution

Tutorial
Solution

Author: Rzepa

Tutorial
Solution 1
Solution 2

#### 1043G — Speckled Band

Tutorial
Solution 1
Solution 2

Author: isaf27

• +113

 » 3 years ago, # |   +1 Auto comment: topic has been updated by Anadi (previous revision, new revision, compare).
 » 3 years ago, # |   +13 Superfast editorials!!!
 » 3 years ago, # |   0 Auto comment: topic has been updated by Anadi (previous revision, new revision, compare).
 » 3 years ago, # |   0 There's an appearance issue in C tutorial
 » 3 years ago, # |   +47 Wow this G solution...
 » 3 years ago, # |   +24 This solution of Problem G is saying, "You need an editorial for me? Meh, just take a look at me, I don't need one."
 » 3 years ago, # |   +55 Author had solution for G, but this tutorial is too small to contain it.
 » 3 years ago, # |   0 Auto comment: topic has been updated by Anadi (previous revision, new revision, compare).
 » 3 years ago, # |   0 For F, you may have considered this right end part of formula: , right now it's confusing.
•  » » 3 years ago, # ^ |   0 Thank you. Fixed.
 » 3 years ago, # |   0 Auto comment: topic has been updated by Anadi (previous revision, new revision, compare).
 » 3 years ago, # |   +5 There is one correct solution for G as scoreboard showes. Who is this one? Or it is error?
•  » » 3 years ago, # ^ |   +5 I think LHiC solved it after contest.
•  » » » 3 years ago, # ^ |   0 Thx. I think it is wrong behavior to show count of solutions that are not showing at table.
 » 3 years ago, # |   0 Auto comment: topic has been updated by Anadi (previous revision, new revision, compare).
 » 3 years ago, # | ← Rev. 2 →   0 What am I missing for problem D, 31th test case : 10 37 4 10 8 3 6 2 9 5 17 4 10 8 3 6 2 9 5 12 9 5 1 7 4 10 8 3 6We have subarrays {2,9,5,1} and {4,10,8,3,6}, so I'm calculating number 26. (4*3/2+5*4/2)+length_of_array(when subbaray is one number)
•  » » 3 years ago, # ^ |   0 Well, if you observe more closely, you'll see that the second group contains the 7, too (it appears on all of the 3 permutations!), so you get 6 elements for the second group! Answer: 4 * 5 / 2 + 6 * 7 / 2 = 10 + 21 = 31 elements.(note that we can include the single elements when we are calculating the answer for each group, by adding one more number. Please note that if a group is single (only 1 element), the answer is 1, because we include only that element and note that 1 * 2 / 2 = 1. This makes the calculations easier! :) )
 » 3 years ago, # | ← Rev. 2 →   +5 In problem F:"Our answer is the smallest i such that dp[i][1] is non-zero. Since dp[i][j] can be quite big we should compute it modulo some big prime."Can we be sure that if dp[i][j] % p is 0 then dp[i][j] must be 0 as well?
•  » » 3 years ago, # ^ |   +3 As you only need to check seven numbers probability of collision is extremely low.
•  » » 3 years ago, # ^ |   0 You can take 2 DP arrays with different big primes, to reduce the chances of collision.
 » 3 years ago, # |   +19 Why is the distance in a suffix array not greater than ?
•  » » 3 years ago, # ^ |   +20 As i is defined as the maximum index, it means that s[i..r] has no suffix that is also a prefix. Thus the suffixes between l and i in the suffix array (they share at least the same prefix as l and i) must have indexes such that the distance between them is . An thus there are at most such indexes.
 » 3 years ago, # |   +13 In G's solution, isn't aabc == aab and bcaa == baa or did I misunderstand something?
•  » » 3 years ago, # ^ |   +8 it should probably be abac & baca
 » 3 years ago, # |   0 tex formatting is broken in G
 » 3 years ago, # |   0 what is a border?
•  » » 3 years ago, # ^ |   +10 Equal prefix and suffix of the string
 » 3 years ago, # | ← Rev. 3 →   0 I am interested in how everybody solved D.Personally, I did some prefix hashing and then binary search on maximum length substring starting at value k for a mnlogn solution.Some ones I've heard:1) Some type of Map solution (a lot of people did this)2) O(nm) DSU solution (rotavirus mentioned)3) Rolling hash (anyone else did prefix hash... ?)I'm especially interested in DSU explanation, but also I've seen some variation in how people have used map, and want to hear how you guys approached that. For example, I saw radoslav11 make some suffix automaton, which unfortunately got TLE on 36, but with some constant optimization it could work.Please tell me about your solutions for this problem!
•  » » 3 years ago, # ^ |   0 i made a dsu of size n, then for each pair of adjacent elements in the first permutation, i checked whether these elements are adjacent in all other permutations, if yes, i united them. So, if two numbers are in the same disjoint set, then they form a valid subarray, so i counted the number of pairs in each disjoint set and printed the sum of them.
•  » » 3 years ago, # ^ |   +4 I had a vector consisting of all pairs of adjacent numbers in the permutation and used binary search to find the count of occurance of a pair instead of using a map for keeping track of the count. I prefer doing this instead of using a map generally when the time limits are strict. My Submission
•  » » 3 years ago, # ^ |   +13 I used a revolutionary new data structure. For some people it can take years to master, but once you figure out how to use "array", you are set for life.
•  » » 3 years ago, # ^ |   +7 I just keep track of the position of each element in each permutation for a linear-time solution. See the code.The only observation needed is that if the first permutation's substring starting at index i is valid until index j, then the substring starting at i + 1 is valid until at least j as well.
•  » » 3 years ago, # ^ |   0 what is difference between rolling hashing and prefix hashing? what's prefix hashing btw?
•  » » » 3 years ago, # ^ |   0 Basically you store the prefix sum of a[i]*BASE^i so that: h[i] = h[i-1] + a[i]*BASE^iSo when you want to get hash code for substring (L, R) in O(1) you can do it like this(h[R]-h[L-1]) * invPow[L]invPow[i] represents BASE^(-i)
•  » » » 3 years ago, # ^ |   +1 For rolling hash you just keep a window (actually a number to represent the window) of size k and you "pop" and "push" to slide the window.For example, you maintain k = 3 at i: a[i]*BASE^2, a[i+1]*BASE^1, a[i+2]*BASE^0You would subtract a[i]*BASE^(k-1), multiply everything by BASE, then add a[i+k] (BASE^0 is just 1)You could also maintain it in reverse and "divide" with modular inverses, but that's just more work.
 » 3 years ago, # |   +23 For F, I got accepted by a randomized algorithm with some greedy at the beginning (45016248). Greedy part: remove some numbers on the array such that there is no pair on it which the one is divisible by the other one. (e.g. ). Randomization part: Attempt many (e.g. 1, 000, 000) sessions to randomly pick some elements from the array until its GCD is 1 or restart if the number of elements picked is more than the best one.
•  » » 3 years ago, # ^ |   0 What is the expected runtime?
•  » » » 3 years ago, # ^ |   0 The expected runtime is just a big constant for the randomization part, around O(1, 000, 000 × 50) but can change 50 to 7 or 8 as it is the maximum possible answer.
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   0 Actually, neal showed below that the expected runtime of one iteration is not 7.
•  » » » » » 3 years ago, # ^ | ← Rev. 2 →   0 We cap the 7, if after picking 7 numbers doesn't give a GCD of 1 then skip and restart the iteration.This number btw can be lower, as we keep track the progress of iterations.
•  » » » » » » 3 years ago, # ^ |   0 Thanks, then it is deterministic runtime :)
•  » » 3 years ago, # ^ | ← Rev. 2 →   +49 Here's a test case that your algorithm will have trouble on:[10, 21, 6 * p for all primes p that fit]The answer is 2 because you can pick 10 and 21. However every other pair of numbers in this array will have a GCD greater than 1. So you only have a roughly chance of picking the right pair on every attempt you make, and n can get fairly large, over 5,000.
•  » » » 3 years ago, # ^ |   +3 Ah, that's correct; I only did calculation for case of instead and found out of success. The success percentage of your case is just 4% for my program. I guess I got lucky then, ><
 » 3 years ago, # |   +3 I think that in the last paragraph of tutorial of problem E it should be Similarly we have that yi  +  xj  <  xi  +  yj if xi  -  yi  >  xj  -  yj.
 » 3 years ago, # |   +1 can some admin link this to the problems, right now only the announcement is linked, thanks
•  » » 3 years ago, # ^ |   0 It looks like instead of linking the tutorial, the announcement was unlinked!Please fix, thanks.
 » 3 years ago, # |   +1 There are no contests this week. :(
 » 3 years ago, # |   0 How to Solve C ?
•  » » 3 years ago, # ^ |   0 I add 'b' to input string at last. Then if i see 'ba' or 'ab' in string then revers at that point. You can try some examples to better understanding of this approach.
•  » » » 3 years ago, # ^ |   0 Thanks
•  » » 3 years ago, # ^ |   0 Read the problem statement wrong for C. Here's that problem. The solution to my wrong assumption also works since its a subset XD. I wrote a blog abt it check it out here
•  » » » 3 years ago, # ^ |   0 Thanks
 » 3 years ago, # | ← Rev. 2 →   0 The answer is 3 if the string s is baac, bcaa or aabcCases bcaa and aabc don't make sense. It probably should be baca and abac.
 » 3 years ago, # |   0 I solved Problem C during contest. But I wanted to know DP approach for solving it. Can anyone share DP approach to solve Problem C ?
 » 3 years ago, # |   0 can some one explain c with example and telling all the cases ?
 » 3 years ago, # |   +19 Pure stl challenge for problem A :) int n; cin >> n; vector a; copy(istream_iterator(cin), istream_iterator(), back_inserter(a)); cout << max(*max_element(a.begin(), a.end()), 2 * accumulate(a.begin(), a.end(), 0) / n + 1); 
 » 3 years ago, # | ← Rev. 2 →   +12 I solved C with another interesting approach. Let's have a recursive function sol(int pos, bool asc).pos is the position (or prefix) till which we need the answer, and asc tells us whether to make this prefix ascending or descending.The final answer is sol(n, true). Let's call the position with the rightmost 'a' (or the minimum character in the prefix) inside the current prefix as idx. (rightmost max character in case of asc == false)Now, to get the optimal answer, we need to flip at position idx.But, to make sure that this flip will be optimal, we have to make the prefix before idx in descending order, so we call sol(idx - 1, !asc) which will make the prefix in the optimal descending order, and then let recursion do its job. Base case would be when we reach the prefix of size 1, where it doesn't matter whether to flip or not.My submission — 45069274.UPD: It seems I have solved this problem for a regular string (without the restriction of containing only as and bs) :DI need to read problems more carefully!
•  » » 8 months ago, # ^ | ← Rev. 2 →   +8 I know i am late, but if you are still reading this.. i don't understand this part "But, to make sure that this flip will be optimal, we have to make the prefix before idx in descending order" Can you elaborate? thanks.UPD: i got it, when we flip the prefix 0..idx, we want the 0..idx-1 to be in biggest to smallest order so that when we split upto idx...it is the smallest possible order.
 » 3 years ago, # |   +25 Another solution for F:First compute for each i from 1 to MAX how many multiples of i are there. Then iterate over all input numbers, find it prime factors, iterate over all subsets of its prime divisors, get number of numbers divisible by all those prime numbers from precomputed table. Use it to calculate for each subset how many numbers divisible by exactly those prime factors are there. Now use basic dp to find out what is minimal number of numbers in set, whose elements gcd is divisible by particular subset of those primes. Start with dp[product of all] = 1, then iterate over smaller and smaller subsets and for each subset m such, that there is at least one number divisible by exactly those set to dp[current_set] + 1To optimize it a bit we can first replace each number by product of it's distinct prime factors and then remove duplicates and numbers divisible by some smaller number from input.Solution complexity is , where ψ(x) is number of distinct prime divisors.Implementation: 45104292
 » 3 years ago, # |   0 Very cool and fast editoral, thank you!
 » 3 years ago, # |   0 my solution for D looks asymptotically correct but it gives TLE at testcase 36 can someone help 45130338
 » 3 years ago, # |   0 Thanks!
 » 3 years ago, # | ← Rev. 2 →   0 In problem D :-> my solution is giving tle in O(M*n), i tried to take input in one go through string which makes it fast but giving wrong answer can anybody tell me where i am wrong?Here is my submission by simply taking the input one by one: First SubmissionBy taking string as input: Second submmissionUPDATE: I GOT MY FLAW!! thanx if anyone GETTING TLE USING O(M*N) TRY TO TAKE INPUT USING GETLINE IN ONE GO,IT WILL REMOVE THE TLE PROBLEM!!
 » 3 years ago, # |   0 Will someone please explain problem F a bit more clearly..
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 You can remove the duplicates in the array and get the same result. Why? Because gcd(a,a) = a. So now let us force every element to be unique. If we greedily maximize how many element we can have it is 1, 2, 2*3, 2*3*5, ...2*3* 5*7*11*13 = 7 elementsThis means we only care about making gcd(arr) = 1 with 7 elements at max.So we can reduce O(n*maxA) DP to O(7 * maxA) statesWe we want to know for a number j, how many ways can we use i elements to make gcd(the i elements) = kj. This is can be calculate with combination modulo some number. We can precalculate some value mult[j] which says "how many values are multiple of j" to achieve this.But we want to exclude answers where k > 1, so we subtract those. This loop is actually (AlogA) because O(n + n/2 + n/3 ... 1) = O(nlogn)So we get O(7 * maxA * logA) complexity
 » 3 years ago, # |   +11 Another approach for problem F: Do a binary search first. Notice that: It's hard to determine whether a subset of size k and with its gcd equals to 1 exists directly, but calculating the number of such subsets modulo a big prime is easy: simply use the inclusion-exclusion principle(with the mobius function). Use a good prime and you'll get accepted. :P
•  » » 3 years ago, # ^ |   0 Your approach sounds interesting. Can you explain a little more how you use inclusion-exclusion? And mobius function?
•  » » » 3 years ago, # ^ | ← Rev. 3 →   0 The number of ways to select k coprime elements from a multiset is sigma(mu(x) * comb(number_of_elements_that_is_a_multiple_of_x, k) for 1 <= x <= MAXA).You can precalculate the factorials and the inverses of the factorials of each positive integer <= n to compute binomial coefficients and a sieve to compute mu(x) and number_of_elements_that_is_a_multiple_of_x for each x.Time complexity: O(n + MAXA log MAXA)
 » 3 years ago, # |   0 can anybody explain the logic behind problem D?
 » 3 years ago, # |   0 45341495my solution for D is WA on #7, can someone tell me what i'm doing wrong?i'm counting for every 'i' the size of the largest array [i, i+1, ...] that is present on every array, then i sum all these values. my code returns correct results for all other testcases with small size
 » 3 years ago, # |   0 isaf27 About problem G , why the distance between l and i in suffix array <= sqrt（n）?and what is the second solution to check if there exists a border for [l,r]
 » 3 years ago, # |   0 i have read some other solution for G and it's look like they don't find any tandem repetitions. Is there a easier solution for G? I think this solution is too hard to implement in contest. Tks you
 » 3 years ago, # |   0 I don't know if this is my first time to solve D by my own.Well my approach is a little different from others. I stored first array as it is. While for other arrays I stored the position of each element (eg. [1, 4, 2, 3] will be stored as [1, 3, 4, 2]). We will also maintain an array p of m size for storing current positions of m arrays.Now we traverse the elements of first array. If for all remaining arrays, the position in the array for current element in first array is 1 greater than the current array position i.e. if p[i] + 1 = array[i][array[1][p[1]]] then subarray size will be incremented. In case the condition is not true and if size of common subarray is k till now then ans will be incremented to k × (k + 1) / 2 and k will 1.Complexity is O(mn).Link to code
 » 3 years ago, # |   0 Could someone explain the alternative solution of Problem F(Solution 2) to me?I confuse at bitmask-part in get_edge function.
 » 2 years ago, # | ← Rev. 2 →   +3 in there anybody who can explain bit-mask solution for problem F ?
 » 2 years ago, # |   0 In G, the "finding a border" part can be done "easily" without a suffix array. The observation is that if a substring with length $L$ doesn't start with a tandem, then for any $K$, it contains at most $2L/K$ occurrences of its prefix with length $K$ -- otherwise, two of them are sufficiently close to create a tandem. I fix $K \approx \sqrt{N}$, precompute hashes of all $O(N)$ substrings with length exactly $K$, and for each query substring, check for all borders with length $\le K$, look at occurrences of its prefix with length $K$ inside it and check for each of these occurrences if the border's end half starts there. With $O(1)$ substring comparison using hashes, this is $O(\sqrt{N})$ as well.On the other hand, I used a suffix array approach instead of Main-Lorentz, since I didn't know it.
 » 3 months ago, # | ← Rev. 2 →   0 For F, can someone please give me some intuition(or proof) behind the first line, that it not contain more than 7 elements...
 » 2 months ago, # |   0 For solution D, why do we substract i from each reach[i]?