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By awoo, history, 3 years ago, translation,

1598A - Computer Game

Idea: BledDest

Tutorial
Solution (BledDest)

1598B - Groups

Idea: fcspartakm

Tutorial
Solution (BledDest)

1598C - Delete Two Elements

Idea: fcspartakm

Tutorial
Solution (Neon)

1598D - Training Session

Idea: BledDest

Tutorial
Solution (Neon)

1598E - Staircases

Idea: BledDest

Tutorial
Solution (awoo)

1598F - RBS

Idea: BledDest

Tutorial
Solution (BledDest)

1598G - The Sum of Good Numbers

Idea: BledDest

Tutorial
Solution (Neon)
• +114

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 » 3 years ago, # | ← Rev. 4 →   +11 The easiest solution for C(in my opinion): 131432277P.S. sum * (n-2) = (sum — a[i] — a[j]) * n => we can find for every a[i] all a[j] in O(n log n) using binary search
•  » » 3 years ago, # ^ |   +9 If you simplify that equation, you would get (a[i] + a[j]) = (2 * sum) / n, so basically this is just a 2-sum problem where the target value is (2 * sum) / n,
•  » » » 3 years ago, # ^ |   0 Yep, u r right (:
•  » » » » 3 years ago, # ^ |   0 Can you please check my soln. I have used 2-sum approach but it is giving me WA on test 2. 131563369
•  » » » » » 3 years ago, # ^ |   0 ++I also tried the same way and getting wa on 78th in 2tc..hbarp can you pls tell me why it's wa.solution: 131653903
•  » » » » » » 3 years ago, # ^ |   0 I havent looked at the correctness of your algorithm but ans = ans + c * d; line has int overflow for sure.
•  » » » » » » » 3 years ago, # ^ | ← Rev. 2 →   0 XD. Thanks, for mentioning this. now, after using it long it still showing wa on that tc. Can you please tell me why is it so?
•  » » » » » » » » 3 years ago, # ^ | ← Rev. 2 →   0 Changing ans = ans + c * d; to ans = 1LL*ans + c * d; doesn't fixes anything. ans is already long long int. Over flow is in multiplication of c*d, change this to ans = ans + c*1LL*d.Also ll val = (n * (n - 1)) / 2; has int overflow. Just having val of type long long int doesn't do anything. n and n-1 are still of int type. Change this to n*1LL*(n-1)/2. Also, next time just uses C++ 64 bit and long long everywhere.Also, if (st.size() == 1) is incorrect fix. You will still get WA on Spoiler1 8 1 2 2 2 2 2 2 3 
 » 3 years ago, # | ← Rev. 3 →   +20 sorry i forgot that x is a good number :-_
 » 3 years ago, # |   +7
 » 3 years ago, # |   0 IDk why but my solution for C gives TLE on Test 17 even though it's the exact same solution as the tester's : https://codeforces.com/contest/1598/submission/131542518
•  » » 3 years ago, # ^ |   0 Don't know why, but using unordered_map gives TLE in this question I have the same problem... use map instead this will solve problem.
•  » » 3 years ago, # ^ |   +1 Maybe because you're using unordered_map? Link to Another CF Blog.
•  » » » 3 years ago, # ^ |   +3 Thnxx a lot!!
 » 3 years ago, # |   0 Why using unordered_map gives TLE in this question... searching elements using hash map should be faster than the map which takes log(n) time to search instead.131514464Also I don't get it... how by adding just these 2 lines which where on a blog, mp.reserve(1024);mp.max_load_factor(0.25);solves the problem of TLE and my soln gets accepted.131524838Then I tried a custom hash function rather than the built-in hash function. Which also got accepted. Also the same solution gets accepted in python using dictionaries which also uses hashmap.Does this mean that the C++ built-in hash function is not good ?131520347I tried to find but didn't get some concrete answers when to use map or unordered_map ?Or using custom hash is better not ?Please if anyone can help it in.
•  » » 3 years ago, # ^ |   0 There are many hacks on unordered_map and I suppose we should avoid using it in Codeforces contests. Here a blog written by neal showing how it works.
•  » » 3 years ago, # ^ |   0 It's just because c++'s hash function is deterministic so people can hack on this. max_load_factor() basically increase the number of buckets which will break the malicious hacks.
•  » » 3 years ago, # ^ |   0 Note to self: ALWAYS use custom hash function when dealing with unordered map no matter how unnecesary in a codeforces contest
•  » » » 3 years ago, # ^ |   +16 The funny thing in this contest is sjcsjc 's template had custom_hash but still he got hacked for using c++ default hash. 131404435 After that he dropped from 4th rank to 87th rank among rated users.
•  » » » » 3 years ago, # ^ |   +1 I am a noob.
•  » » » » » 3 years ago, # ^ |   +16 He is a noob.
•  » » 3 years ago, # ^ |   +18 mp.reserve(1024); mp.max_load_factor(0.25); only changes the test that's required to hack the solution, it doesn't make it unhackable.
 » 3 years ago, # |   0 Nice solution for D!
 » 3 years ago, # |   +13 I suppose that the time complexity of F should be $O(n2^n + A\log A)$ instead of $O(2^n + A\log A)$. (:Could someone tell me if there are some mistakes?
•  » » 3 years ago, # ^ |   0 Can you tell me what is A in this time complexity? I understand where $N$ and $2^N$ comes from but I just can't find anything related to $A$.
•  » » 3 years ago, # ^ |   0 You are right, it's $n\cdot 2^n$ instead of $2^n$.
 » 3 years ago, # |   +3 Any other solution for problem D?
 » 3 years ago, # |   0 In Problem D, I don't understand why the only bad triples are (xa,ya),(xb,ya),(xa,by). Isn't there a possibility of having three same topics, such (1, 1), (1, 2), (1, 3)? I've been looking at the explanation for so long but still can't figure it out :(
•  » » 3 years ago, # ^ |   0 but their difficulties are different so they aren't bad triples.The problem statement says "at least one of the conditions are met"
 » 3 years ago, # | ← Rev. 2 →   0 For D, can someone explain to me why there can't be a triple of the form [(Xa, Ya) , (Xb , Yb) , (Xa, Yb)] ? EDIT: I figured it out, it's actually the same thing except that the third pair is the central one, instead of the first.
 » 3 years ago, # |   0 For 1598C:No, you don't have to use map in such a hacky way (or in another word, full of patches).131572801======================================Instead of keeping a map of all elements, we keep a map for number of first i items ([0, i-1])So the core logic can be much more easier:in C#: for (int i = 0; i < n; i++){ // update the answer if (cnt.ContainsKey(target - data[i])) answer += cnt[target - data[i]]; // update the cnt map, for (i + 1) to use. if (!cnt.ContainsKey(data[i])) cnt.Add(data[i], 0); cnt[data[i]]++; } or in C++: for (int i = 0; i < n; i++){ // update the answer answer += cnt[target - data[i]]; // update the cnt map, for (i + 1) to use. cnt[data[i]]++; } 
 » 3 years ago, # | ← Rev. 12 →   0 i tried to add photo for my comment but the comment become empty any one can help me
 » 3 years ago, # |   0 In problem D can anyone please explain why we won't be overcounting in the editorial solution?
•  » » 3 years ago, # ^ |   0 problems and diff. are unique pair, no overlap when multiplication as in editorial
 » 3 years ago, # |   +8 Alternative solution for 1598D - Training Session. I did see restriction that all problems pairs of parameters are different, but I had no idea how to use this fact. My solution doesn't rely on this fact at all. Solution turned out to be much harder so I wasted a lot of time to implement it during round.Here is main idea. Think of how to calculate all triples with different topics. There is easy waySort all problems by their topics in non-decreasing order. You should pick three problems in strictly increasing order. Now you can group them by topics. To count all triples you can fix topic of second problem in increasing order, and count how many ways you can pick problem with this fixed topic, task with topic below, and task with topic above. All problems before the fixed group will have topic below, so their number is just zero-based index of first problem within the fixed group. All problems after the fixed group will have topic above, so their number is n minus zero-based index of last problem within the fixed group minus one. And number of problems with the fixed topic is just length of fixed group. Multiply those and get all triples.Apply the same idea to problem's difficulties. We count something twice. What to do? WellI don't know. I did try to think in this way, but there are too many of possible situations. So I dropped this idea. Instead, think in other way. Suppose instead we calculated all triples with different difficulties. What are we missing after it? We missingTriples with different topics but some of difficulties coincide. I'll stress out: topics should differ, but difficulties coincide. IdeaLet's forget for a moment about different topics. Focus on difficulties. Try list the cases. Idea about which cases do we need to listBecause of different topics, we still want to use groups of fixed topic. Think of what fixed topic would be. First or second or third. CasesI tried to fix second topic (middle one). Here is what I've got. Look at them, mark them by their property. Focus on those we should include as they wasn't yet counted. 1 2 0 1 2 1 1 2 2 2 2 1 2 2 2 2 2 3 1 2 3 3 2 1 3 2 2 3 2 3 3 2 4  Cases detailedI put letter v as a check mark sign of triple we do need to include. 1 2 0 1 2 1 v 1 2 2 v 2 2 1 v 2 2 2 v 2 2 3 v 1 2 3 3 2 1 3 2 2 v 3 2 3 v 3 2 4 How to count them? Cases even more detailedNotice some of them just has same first and last difficulty. I'll put mark = 1 2 0 1 2 1 v = 1 2 2 v 2 2 1 v 2 2 2 v = 2 2 3 v 1 2 3 3 2 1 3 2 2 v 3 2 3 v = 3 2 4 How to count other v that we still missing? Cases even more detailed than beforeNotice we left with those which coincide first and second difficulty, I'll mark them l, and second and third, I'll mark them r. 1 2 0 1 2 1 v = 1 2 2 v r 2 2 1 v l 2 2 2 v = l r 2 2 3 v l 1 2 3 3 2 1 3 2 2 v r 3 2 3 v = 3 2 4 Now we counted 2 2 2 three times, we should subtract it twice. I claim we can use this strategy to count all triples we need. SolutionCount all triples with different difficulties using simple method described in the beginning. Then, sort problems by topics, group them by their topic. Solve for fixed topic. And within fixed topic we will iterate over problems. Let's say problem we iterate has difficulty d. I'll list how to calc things that related to corresponding marks: l is just number of tasks with difficulty d before the fixed group multiplied by number of problems after the group. r is just number of tasks with difficulty d after the fixed group multiplied by number of problems before group. = is sum of squares for each diffivulty v the number of problems with difficulty v before the fixed group multiplied by the number of problems with difficulty v after the fixed group. to subtract 2 2 2 we just need to subtract twice the number of tasks with difficulty d before and after the fixed group. We will maintain this using associative arrays left[d] and right[d] (map in C++ or dict in Python) which will maintain how many problems with difficulty d before and after the fixed group correspondingly. And we will maintain integer s which will store sum of multiplication. So l will use left[d], r will use right[d], and = is just s. To count 2 2 2 we just subtract twice left[d] multiplied by right[d]. Time complexity is $O(n \log n)$.In my code s is named sums.C++ 131451686Python 131455867
 » 3 years ago, # |   +8 Detailed description how to implement 1598E - Staircases in $O((n + m + q) \log (n+m))$ time and space. SolutionSimilar to editorial, define base staircases. The only trick I need to describe is how easily to index all stuff. Because we will have $O(n+m)$ base staircases, and we need to be able pick the one we need to operate at that moment. Idea is to store within staircase indexes of blocked cells in set in C++, then we can use lower_bound method to find next blocked cell or previous blocked cell when we need. Requirement for indexing is simple: we need their indices go one after another, so that length of empty cells would be next blocked cell minus previous blocked cell minus one. We don't actually have any other requirement for single staircase, because we'll use trick to initially install blocked blocks around whole table. So far as our new / old blocked cell is located between those indexes of blocked cells outside of table — we're fine.And here is neat idea to index within single staircase. Notice that no matter which base staircase we choose it will have exactly two cells on each row (not taking into account restriction to being inside of table). Let's say left one of those cell is 0, and right one is 1. Now, it's easy to see, that index of left cell $= row * 2$ and index of right cell $= row * 2 + 1$ will satisfy all our requirements. All cells within staircase will go one after another, and we have very easy formulas for left cells and right cells. You can think of it like formula for apartment numbers if there are exactly 2 apartments on each floor.Now, we need to make indices for our base staircases. We also don't need particular starting point, all we need that they also go one after another. Notice that each base staircase can be uniquely defined by set of their left cells within rows. Also note: they form a diagonal. There is easy way to index diagonal, just $x - y$, here $x$ is row number. And this is exactly what we need.Notice, when we block cell, in one base staircase we block left one, and in other base staircase we block right one. And the one which we block right cell is next base staircase. Next diagonal (and base staircase) would be $x - y + 1$. So, each time we block cell we can implement it as block index $2*x$ in staircase $x - y$ and block index $2*x + 1$ in staircase $x - y + 1$. You could use indices of staircases $x - y - 1$ and $x - y$ correspondingly, this only change index of starting base staircase. Blocking cell removal implemented similarly. Also, checking is there already blocking cell can be also done like this.All we left to do is to figure out how many stairs there are initially after installing all blocking cells around the table. To do that, we can just walk over all base staircases, and find those with at least two blocking cells (there shouldn't be any with 3 or more blocking cells), and count number of staircases within as usual.131469659
 » 3 years ago, # |   0 for que B https://codeforces.com/contest/1598/submission/131729850 can u check this once i don't which test i am missing so pls
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 if(r>=n/2&&q>=n/2&&s==0){cout<<"YES"<
 » 3 years ago, # |   +17 A slightly different solution of 1598F - RBS. SolutionThis solution will work in $O(n\cdot2^n + A)$ where $A$ is total length of brackets sequences.Notice, that for any fixed mask of used brackets sequences, there is fixed balance. Because balance is sum, and no matter how do you rearrange sequence, it will still have same balance in the end.Notice, that for any prefix of used backets sequences, if there is better arrangement for this set of brackets sequence, you can change into it, and get better result. Thus any prefix should be optimal for corresponding mask.Thus we have idea of following dp[mask] — number of RBS prefixes in best arrangement of sequences from bit mask mask. But idea is slightly different. Let's say dp[mask] is this thing if and only if it still potentially may have more RBS prefixes, otherwise None in Python or minus infinity in C++ or Python as well.In this case, any potential answer is either dp[mask] as is, or dp[mask] + single sequence of brackets after which we can't have any more RBS prefixes.Suppose we somehow maintain total balance for mask: it could be additional dp for balance of masks, or precalc, or other approaches. Then, when we append sequence to current mask, we need to somehow find out how many times we reach 0 balance to count additional RBS and also we need to find out do we reach negative balance at any moment. To do that, it's pretty straightforward. Just precalc balance of each sequence, and minimum, and how many times it reach minimum. To count how many times we reach 0 we use number of times we reach minimum. Because, if we reach 0 when it's not minimum, then we will have negative balance and won't update dp[] because we don't store sequences which failed to continue. So when we reach 0 it should be minimum within this sequence of brackets. Also this shows us how to check does it fail for continuation.Now, tricky part. In the way above we can get maximum over dp[mask], but there are also cases when best answer is some dp[mask] + sequence of brackets after which we won't have any additional RBS prefixes. For this case we already know what balance of mask of sequences. Let's call this balance of mask b. We only need to know how many additional RBS prefixes we will get when we add single sequence of brackets. And, this is just how many times balance of this single sequence reach -b, because at that point mask has balance b, so to reach 0 it should add -b. But it could fail to be RBS before reaching 0, for example, fourth time. So, we actually need to know how many times this single sequence of brackets reach balance -b before reaching balance which is -b-1 (less by one). And this quantity we can precalculate in linear time of $A$, in other words, in linear time of total length of brackets sequences. Just make array for each sequence, and in its index b store how many times balance reach -b before reaching -b-1. Balance never grows more than length of brackets sequence, so total space of this precalculation is total length of brackets sequences $O(A)$.So, for each sequence of brackets we precalculate: how many times it reach balance -b before reaching -b-1. Also, for each mask we maintain or precalculate total balance of sequence within mask. And write dp[mask] = maximum RBS prefixes for some arrangement of mask which still may have additional RBS-prefixes. Otherwise None in Python or minus infinity in C++ or Python as well. Then, answer is either maximum dp[mask] or some dp[mask] + sequence of brackets after which we won't have any additional RBS prefixes.131544153
 » 3 years ago, # |   0 in problem C why i am getting tle.. Pls tell[problem:1598C] https://codeforces.com/contest/1598/submission/131512968
•  » » 2 years ago, # ^ |   0 This is due to the use of unordered_map instead of map. Try map and you will get AC.
•  » » » 2 years ago, # ^ | ← Rev. 2 →   0 even , i used unordered map earlier it also gave me tle but when i am using map it is not giving tle . can anyone explain why is it so ? please.
•  » » » » 23 months ago, # ^ | ← Rev. 4 →   0 I think it may be because of special anti-hash testcases.Unordered map relies on hashing to determine where the value of a key is stored. When there are hash collisions, the the hash function makes the collided keys share the same position. This is done with a linked list, which needs to be linearly searched if you are looking for a key that has collided with other keys.Essentially, you can design testcases that abuse this workaround, causing unordered_map access complexity to approach O(n)
 » 3 years ago, # |   0 132438878 can anybody help me with this I'm getting tle even though the time complexity of my son is same as that of the solution :)
•  » » 3 years ago, # ^ |   0 ^in problem c
•  » » » 3 years ago, # ^ |   0 This may upset you, but unordered_map uses more time than normal map. Confer to this link web : https://codeforces.com/blog/entry/62393?#comment-464775
•  » » » » 3 years ago, # ^ |   0 thnx man
 » 2 years ago, # |   0 My solution for F fails at test 32. Can someone please look at the submission and point out the error?I would be happy to explain what my code does if it is not clear. Thanks.
 » 2 years ago, # |   0 A Dp approach to problem E in $O(NM + Q*min(N,M))$ SolutionLets calculate the number of staircases of both types at $cell (i,j)$ using Dynamic Programming$L[i][j] =$ number of staircases starting at $cell (i,j)$ of type $1$$D[i][j] =$ number of staircases starting at $cell (i,j)$ of type $2$Notice that at $cell (i,j)$, all staircases of type $1$, lead to all staircases of type $2$ in $cell(i,j+1)$Also note that at $cell (i,j)$, all staircases of type $2$, lead to all staircases of type $1$ in $cell(i+1,j)$This leads to the recurrence formula: $L[i][j] = 1+D[i][j+1], D[i][j] = 1+L[i+1][j]$ (if $cell (i,j)$ is free)$L[i][j] = D[i][j] = 0$ (if $cell (i,j)$ is blocked)The total number of staircases = sum of all $L[i][j]$ and $D[i][j]$ minus count of all free $1*1$ cells. This is so because both types of staircases are $1*1$ squares so we have to subtract due to overcounting.As for queries, we can just update $cell (i,j)$ to new status and recalculate all dp values but this is too slow since there are $N*M$ values to update. You can observe that if one cell is updated, not all dp values change. The only dp values that change are the ones that can reach $cell (i,j)$ through a valid staircase of either type. The number of such cells is at most $min(n,m)*2$ for either staircase type. So all we need to do is update only those cells, which is $O(min(n,m))$ per query. To do that, we can update current cell then run a dfs from current cell to all cells that can be visited either by going up or down and update the dp values for those cells. The sum of all $L[i][j]$ and $D[i][j]$ can be stored in a sum variable and updated accordingly. The same thing can be done for the number of free cells.Time Complexity: $O(NM + Q*min(N,M))$code
•  » » 2 years ago, # ^ |   0 Good job bro!
 » 2 years ago, # |   0 For problem D, is there anyway to find the number of triples that are different both in topic and in difficulty?
 » 23 months ago, # |   0 Problem A can be solved using DFS like this: 161393885.
 » 5 months ago, # |   0 hey I have a doubt in ques C sorry if its too silly.. should not the ans+=min(freq of a[j] , freq of (2*sum/n a[j]). ie minimum of the counts should be added right? because if needed sum is 10 and freq of 7 is 4 and freq of 3 is 2 then pair with sum 10 should be only 2 right?
•  » » 5 months ago, # ^ |   0 the number of pair should 4 times 2 equals 8. You have four different ways to choose a 7 and two different ways to choose a 3.
•  » » » 5 months ago, # ^ |   0 ohh got it! thank you
 » 5 weeks ago, # | ← Rev. 3 →   0 1598C - Delete Two Elements1598C — Delete Two Elements [python3]Thanks for the tutorial!Here is the translation to python3: import os, sys, getpass, platform import math, random, decimal, queue, heapq, bisect, itertools, functools, collections, string, operator, timeit, pprint from queue import Queue, PriorityQueue, LifoQueue from itertools import accumulate, chain, combinations, combinations_with_replacement, compress, count, cycle, dropwhile, filterfalse, groupby, islice, permutations, product, repeat, starmap, takewhile, tee, zip_longest from functools import cmp_to_key, lru_cache, partial, partialmethod, reduce, singledispatch, total_ordering from random import choice, choices, shuffle, sample, random, randint, randrange, uniform, seed, getstate, setstate, getrandbits from string import ascii_letters, ascii_lowercase, ascii_uppercase, digits, hexdigits, octdigits, punctuation, printable, whitespace from heapq import heapify, heappush, heappop, heapreplace, nlargest, nsmallest from bisect import bisect, bisect_left from collections import defaultdict, OrderedDict, deque, Counter from math import gcd, factorial, isqrt, comb, perm, prod cond = getpass.getuser() == '4a' if cond: sys.stdin = open(os.path.join(os.getcwd(), 'a.txt'), 'r') I = lambda: [int(a) for l in sys.stdin for a in l.strip().split()] S = lambda: [a for l in sys.stdin for a in l.strip().split()] IM = lambda: [[int(a) for a in l.split()] for l in sys.stdin] SM = lambda: [[a for a in l.split()] for l in sys.stdin] D = lambda k=1: {i: list(map(int, input().split())) for i in range(1, 1 + int(input()) * k)} DS = lambda: {i: [(int(x[0]), x[1]) for _ in range(int(input())) for x in [input().split()]] for i in range(int(input()))} d8 = ((1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1)) d4 = ((1, 0), (0, 1), (-1, 0), (0, -1)) az, AZ = ascii_lowercase, ascii_uppercase mod, inf = 1_000_000_007, float('inf') CNR = lambda n, r: factorial(n) // (factorial(r) * factorial(n - r)) PNR = lambda n, r: factorial(n) // factorial(r) A = IM() if cond: print(A) def solution(A): for i in range(1, len(A)): if i % 2 == 1: size = A[i][0] else: arr = A[i] res = 0 counts = Counter(arr) sum_arr = sum(arr) if 2 * sum_arr % size != 0: print(0) continue target = 2 * sum_arr / size for i in range(size): first = arr[i] second = target - first if second in counts and counts[second] > 0: res += counts[second] if first == second: res -= 1 print(res // 2) solution(A)