strange list codeforces

Edit: Oof: I misremembered and got dominating set/vertex cover backwards. Not the answer you're looking for? My friends list seems to be fine. For example, $$$f(321) = 123$$$, $$$f(120) = 21$$$, $$$f(1000000) = 1$$$, $$$f(111) = 111$$$. Description of the test cases follows. So what you are suggesting, is to find the largest square that divides every number in the range, then replace every number in the array by itself, divided by that largest square divisor, and we will be left by the product of primes (with exponent equal to 1), which we are looking for. A few weeks ago, I started noticing that none of my friends on the friends list were green anymore, which was surprising, until I realized that I was still friends with some green and grey coders, but for some reason, they don't show up on the list anymore. 8000 = (4*4) * (10*10) * 5; k=5 is the constant for 8000. What am I missing? A path can contain at most $$$c$$$ light edges, and each vertex has at most $$$1$$$ heavy outgoing edge. And, how does the example with persons A and B profe that a greedy solution works/exists? 2) A, B (0) 2021.04.12: Codeforces #1512D Corrupted Array (0) 2021.04.11: Codeforces #1504A Dj Vu (0) 2021.04.05: Codeforces #1471A Strange Partition (0) 2021.04.05: Codeforces Round #712 (Div. Otherwise, if $$$q$$$ is not divisible by $$$x$$$, the robot shuts down. 2) 4: 90: Jzzhu and Cities: Codeforces: Codeforces Round #257 (Div. EDIT: Since this comment is downvoted I'd like you to tell me why they aren't similar? You have given an array $$$a$$$ of length $$$n$$$ and an integer $$$x$$$ to a brand new robot. Strange List Ideological analysis: The first queue is the result of Test5 reported, so I can only find the law. Can someone help figure out why my solution for A is failing on pretest 3, I followed the same approach as in the editorial ? This integer is given without leading zeroes. Explanations for the two first test cases of the example: Virtual contest is a way to take part in past contest, as close as possible to participation on time. Virtual contest is a way to take part in past contest, as close as possible to participation on time. While) and case, Building a domain name and server personal website based on GitHub and Hexo, Encapsulating APIs that interact with the background (axios), Use C ++ # if, # Elif, # else and #ENDIF instructions, P4281 [AHOI2008] emergency collection / gathering (LCA approach), When debugging OpenNE: Attempted relative import in non-package, Qt5.4 input Chinese compiler does not pass under win7, Python asynchronous way to request multiple interfaces at the same time. By clicking Post Your Answer, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct. That element is your answer. You essentially wrote that LCM(x,y)^2 is a perfect square. Visiting a Friend time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pig is visiting a friend. Igor is in 11th grade. 3) Personal Simple Questions, Codeforces Round #552 (Div. I tried precomputing all perfect squares in range [1..1e6]. if $$$n = 4$$$, then for every integer $$$x$$$ such that $$$1 \le x \le n$$$, $$$\dfrac{x}{f(f(x))} = 1$$$; if $$$n = 37$$$, then for some integers $$$x$$$ such that $$$1 \le x \le n$$$, $$$\dfrac{x}{f(f(x))} = 1$$$ (for example, if $$$x = 23$$$, $$$f(f(x)) = 23$$$,$$$\dfrac{x}{f(f(x))} = 1$$$); and for other values of $$$x$$$, $$$\dfrac{x}{f(f(x))} = 10$$$ (for example, if $$$x = 30$$$, $$$f(f(x)) = 3$$$, $$$\dfrac{x}{f(f(x))} = 10$$$). Thanks a lot! can you explain the editorial ? In both cases we use maximum of (300 + 1e5 / 300 + 300) < 1000 questions. Thanks for contributing an answer to Stack Overflow! The problem statement has recently been changed. Then we can use path compression for the heavy edges, and use that to answer each query in time $$$O(c \log n + c^2)$$$. To learn more, see our tips on writing great answers. How about numbers 15 & 135. The only programming contests Web 2.0 platform. Initially, $$$a_{i-1} \le a_i \le a_{i+1} \le a_{i+2}$$$, since $$$a_{i-1} = a_i = a_{i+1} = a_{i+2}$$$ $$$\forall i$$$ such that $$$p \not\in \{ i-1, i, i+1, i+2 \} $$$. 1 C), no binary search is needed. $$$p_1 = 5$$$, $$$p_2 = 3$$$, $$$p_3 = 6$$$, $$$p_4 = 1$$$, where $$$p_1 < p_3$$$ and $$$p_2 > p_4$$$ satisfies the inequality, so one of $$$[a,b,c,d]$$$ tuples is $$$[1,2,3,4]$$$. Title link This Round is done very poor. If we assign lowest price to people with lower c values , then for people with higher c value we need to spend large amount of money either as gift or cash. Strange List | Codeforces Round #694 (Div. How to help my stubborn colleague learn new ways of coding? G. Removal . Can anyone tell me, what is wrong with this solution for problem Strange List? 1470A - Strange Birthday Party I overcomplecated that problem, 103444635. If you've seen these problems, a virtual contest is not for you - solve these problems in the archive. Igor already knows which numbers the teacher will give him. Gary2005 and codephilic , It's not necessary that impostor will be present in ['i'-root(n),'i'] , because with each query the game is also played , thus the impostor could be also present at position less than 'i'-root(n) since we have asked 2*root(n) queries . After you have done that you will get number Ai written in a form x^p * c, where p is some integer(the number of times you can divide Ai until it is no longer divisible by x) and c is that last number that is no longer divisible by x(so if x = 2 and current number is 12 you can write it as 2^2 * 3, basically meaning 12 is divisible by x two times, since exponent is 2). 2), Invitation to SmallForces Monthly Contest #3, [GYM] HIAST Collegiate Programming Contest 2023, EPIC Institute of Technology, 2023-2024 Enrollment Campaign, How to use Centroid Decomposition to solve IOI 2011 RACE. Solution - https://codeforces.com/contest/1471/s. We claim that after the 0th second or the first iteration, the set with even number of elements will always form a perfect square. "Let's sort guest in order of descending value $$$K_i$$$." You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.. . On Jul/27/2023 17:35 (Moscow time) Educational Codeforces Round 152 (Rated for Div. This gift is gone so repeat the process from the next gift. In the first test case, it is optimal to apply the third operation. The complexity would be $$$ A/1^2 + A/2^2 + A/3^2 + = O(A) $$$. #sksama #codeforces #codechef #div2 #solution #explanation #tuf #gfgTelegram- https://bit.ly/30jGLHZ USE CODE - SKSAMAGFG for FLAT 10% off on all Geeks For . At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters. If you're interested in working on a project for the banking sector, then you've come to the right place. I advise you to read the editorial solution for this problem and try to solve it that way. Me too. Maybe I considered it to be not sorted because it would be more interesting problem if C was not sorted. if n < 600 just check every position for solution keeping in mind that position of answer is the one which still has k cards and following person has less than k cards. The first line contains a single integer $$$n$$$ ($$$4 \leq n \leq 5000$$$) the length of permutation $$$p$$$. Now lets explain how to count how many times you add which number. That is exactly why I was also confused by the editorial. I meant this. The only programming contests Web 2.0 platform, Educational Codeforces Round 99 (Rated for Div. Which, in that case, says nothing about x and y themselves, when the goal is to determine whether they are adjacent. check this maybe will be more clear for you. Codeforces Round #710 (Div. I also proved that the period is always smaller than $$$k^{\frac{n}{2}}$$$, but it's still far away from the solution. your first comment was "what you wrote is not true for any x and y" . Every one second, the array A will change: A DFS do not understand why the wording of collapse, chaos and violence, because the meaning of the title is not very good. Let me demonstrate what confuses me. Replacing endl by '\n' can also optimize your code. He studies in a special school and his grade can be equal to any positive decimal fraction. I was not even close to this approach before. Still a tautology. $$${C_K}_i$$$ was sorted? Cannot retrieve contributors at this time. Brilliant. Hi, sorry I still don't understand. Then, we can divide that perfect square by GCD(a,b)^2 and we will receive another perfect square, denoted by X^2. How can I fix problem with excessive memory? Due to that silly mistake, it cost me too much time at debugging. You should use the function exactly once in the beginning. Else, we colour it black. Is there a plugin to hide the rating column from the problemset page? The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) the number of test cases. https://codeforces.com/contest/1471/problem/B. Sounds great, but you failed system tests so I assume that approach is wrong? If you have no gift then you are forced to give money to everyone. we have to check whether [ (x*y) / gcd(x,y)2 ] is perfect square, Note k = x*y => (k^2/gcd(x,y)^2) = (k/gcd(x,y))^2 and thats a perfect square. It is guaranteed that the sum of values $$$n$$$ over all test cases does not exceed $$$10^5$$$. 70 lines (39 sloc) 1.08 KB !XD, The editorial is seriously fast. 3) Editorial 1506A - Strange Table - Programmer All This algorithm can only guarantee that no adjacent black nodes (teachers) will get the same color. C++ Java Python3 C# #include <bits/stdc++.h> using namespace std; int main () { int lower_bound = 2; int upper_bound = 10; for (int i = lower_bound; i <= upper_bound; i++) { cout << i << endl; int response; I wrote "what you wrote IS TRUE for any x and y". The questions were nice today, enjoyed it. The editorial mentions, that LCM(a,b) = a*b/GCD(a,b). I misunderstood the algorythm, we do not choose all which are connected to a white one, but only a single one. I want more rounds like this, Nice round and fast editorial :) The solutions are pretty elegant to me as well lol, If we simply color the graph with two colors, how is it garanteed that no two adjacent vertex get the same color? OverflowAI: Where Community & AI Come Together, codeforces problem named Strange List the "MEMORY LIMIT EXCEEDED", Behind the scenes with the folks building OverflowAI (Ep. Raw Blame. Edit: Ok, now I got it. I don't know if proof of the correctness of the greedy approach is simple, I for sure, cannot formalize it. Let say that you have n = 10^5, x = 2 and ai = 2^25, you can see now that when you pass the array the first time you will add 2 * 10^5 numbers to the end and they will all be 2^24, if you continue doing that at the end you will have 2^25 * 10^5 size of array filled with 1. Two students can connect to each other (get the same color) as long as there is a teacher connecting both of them. 103476026 My Accepted Submission After the contest. If that element is less than $$$k$$$, just keep incrementing your index until you hit an element equal to $$$k$$$. 2) B. I earlier said it was AlogA but then I remembered that sum of inverse of squares = pi^2/6. EDIT 3: C'mon boys, what's wrong with you? CodeForces. 3) Editorial 1506A - Strange Table, Programmer All, we have been working hard to make a technical sharing website that all programmers love. In this problem the vertices of the graph are pairs $$$(i,j)$$$, where $$$i \le c$$$ is the remaining money to pay for reverses, and $$$j < n$$$ is the position in the permutation. Strange Functions time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Let's define a function f(x) f ( x) ( x x is a positive integer) as follows: write all digits of the decimal representation of x x backwards, then get rid of the leading zeroes. Then, let's say we are given x and y. After that you have to find minimum value of p(also first one if there are more minimums) for all given inputs, lets say it's at some position K. After finding the minimum value note that all integers before that K will be added to the sum p(K) + 2 times, and after position K(including K) will be added p(K) + 1 times. And the numbers a and b are adjacent if and only if that number is a perfect square, so (a*b)/GCD(a,b)^2 = X^2 for some integer X. [Tutorial] Floors, ceilings and inequalities for beginners (with some programming tips), Educational Codeforces Round 152 Editorial. Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide, The future of collective knowledge sharing, @Md Ayyan Fahim .then why don't you accept my answer please :), I am very new here so i dont know actually how to accept the answer, @MdAyyanFahim Just click on Right tick on my answer near upvote or downvote arrows and it'll be accepted, New! Sorting by $$$K_i$$$ is the same as sorting by $$$C_{K_i}$$$, because $$$C$$$ is sorted ascending, as the staments states clearly. We can partition the edges into heavy edges ($$$(i,j) \to (i,j+1)$$$) and light edges (all of the other edges). Fun solution for Div2 E / Div1 C. Ask 300 random questions. A. Tokitsukaze and Strange Inequality time limit per test 1.5 seconds memory limit per test 256 megabytes input standard input output standard output Tokitsukaze has a permutation p p of length n n. In the second test case the array initially contains integers $$$[4, 6, 8, 2]$$$, and $$$x=2$$$. The resulting array in this case looks like $$$ [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1]$$$. The intended solution is not to do what is written in task but to think of a faster way. Can anyone explain the even case? That's about 3 * 10^12 numbers, and obviously to slow to fit in 1 second, and also to big to fit the memory limit. it.F is the 'k' value I talked about earlier. Can't 2 perfectly square numbers be "adjacent" to themselves? Connect and share knowledge within a single location that is structured and easy to search. Well that is equal to (a*b/GCD(a,b))/GCD(a,b) = (a*b)/GCD(a,b)^2. The first input line contains a single integer $$$t$$$ ($$$1 \leq t \leq 100$$$) the number of test cases. You just put x everywhere but when you divide with x second time the second parameter should be x^2. Something along those lines is also written in the editorial. In our case (4 * 4) = 16, it is a perfect square. Thus we keep a map of occurrences of lists of factors which occur an odd number of times. In this article, we'll talk about a detailed banking project in Python so you can get started with ease. "Now let's prove that the number of cards that players $$$p+1,p+2,,n,1,p1$$$ have is not increasing. What does the following mean in the editorial for Div. Subscribe. What is LCM(a,b)/GCD(a,b) then? Essentially, by giving the cheapest gift to the person with the greatest k_i, you will be always as well off as in any other situation. at the start of your main. Strange Partition | B. $$$b_i \le b_{i+1}$$$. Editorials. F. Strange Triples. B. 1470B you wrote __ "b- the total number of elements with a class of odd size or with the class equal to 1". In Question A For my code, in test 4 it is giving, wrong output format Expected integer, but "1e+010" found. The only line for each test case contains two integers $$$a$$$ and $$$b$$$ ($$$1 \le a < b \le 10^6$$$). If you've seen these problems, a virtual contest is not for you - solve these problems in the archive. Making statements based on opinion; back them up with references or personal experience. (5*5*5) can never generate a perfect square. 1) & Codeforces Round #259 . 3)G. Strange Beauty, CodeForces Round # 710 (Div. We can do binary search in ['i'-(n-3)/2],'i'] if n is odd , ['i'-(n-2)/2,'i'] if n is even (left limit will wrap around if it goes less than 1).Here 'i' is any position with value >k. It probably is simple though. Like if the graph is a cycle of 5 nodes in the form of a pentagon then however you put colors two neighboring white nodes will appear? But then finding the largest perfect square for any number should still take root(n) complexity right? What the robot does is the following: it iterates over the elements of the array, let the current element be q. 6th Oct, 2022 Read Time 8 Mins In this article The banking sector has many applications for programming and IT solutions. cin.tie(0); cout.tie(0); also helps. After. If you still don't know how to do that, you can click here to have a look at the AC submission. Really nice work, What a great problems with interesting solutions! Thanks for clarification . Note that: black and white are not symmetric! 1 + Div. Like once I read a problem statement I am sometimes stuck on how to proceed. 16 can be divided 4 times X, 2 can only except X1, 4 can be remov [CodeForces 830C] Strange complexity reduction, codeforces873C Strange Game On Matrix, codeforces 842D. 2). Like if the graph is a cycle of 5 nodes in the form of a pentagon then however you put colors two neighboring white nodes will appear? I precompute the lowest prime factor for each $$$a_i$$$ (similiar to Sieve of Eratosthenes), then factorization can be done in $$$O(log_{a_i})$$$ by repeatedly dividing $$$a_i$$$ by its lowest prime factor (I think drayc's method is faster though). 1 C Strange Shuffle? Split it! Please guide me the way to tackle it. You can solve problems unofficially. I even solved like that and got correct answer for each 0 query, because I thought there was one case only. Educational Codeforces Round 143 (Rated for Div. Do I need to change the code in implementation? Can you explain the process. Because of that, you are getting TLE. Can someone prove or disprove the following claim about Div1C: After exactly $$$n$$$ questions, the table looks exactly the same as in the beginning?