There is O(n) solution. Notice that in this faster way we will not visit all the $$$\binom{k-h+29}{29}$$$ possible choices $$$s_1,s_2,\dots, s_{k-h}$$$ because we are assuming that exactly half of them belong to $$$A$$$ and exactly half of them belong to $$$B$$$. Any counting problem, like counting pairs of elements/counting subarrays satisfying some property: If any common technique, like fixing the L pointer or 2pointer approach, does not work, try to divide and conquer. After that, we start processing the request. clue: BITSETS, Maximum Manhattan distance between 2 points: convert every point $$$(x,y)$$$ to $$$(x',y') = (x-y,x+y)$$$. The time complexity is O(n*logn). If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive. image source: https://youtu.be/HkGdJod75Po, Easiest div2 contest so far, yet I got slain by mere problem D :(. compareTriplets has the following parameter (s): int a [3]: Alice's challenge rating int b [3]: Bob's challenge rating Return int [2]: Alice's score is in the first position, and Bob's score is in the second. The answer is $$$\lceil \frac{s}{2} \rceil$$$. Submit codes. For each request p_i, c_i, we perform a binary search in the corresponding Fenwick tree in order to find the leftmost position with p_i 1's before it. Otherwise, we iterate over the points in $$$A_{k-1}$$$ and we consider as candidate elements for $$$A_k$$$ the points that can be obtained by changing the value of one coordinate by $$$k$$$. What to do if all the elements are negative? This depends on the time all calls are going to end at. The way to build tree and find the position of K-th letter is quite simple if you under stand line-segment-tree :). One and Two Codeforces Full solution in c++ || Codeforces Contest Solution Code In Seconds 1.19K subscribers Subscribe 821 views 1 month ago codeforces, codeforces for beginners,. It may be easy to come up with a solution sometimes. 328. The following choice works: One should repeat the whole process until a solution is found. If $$$k\le 6$$$, we simply iterate over all points with $$$|a|+|b|+|c|+|d|\le 1 + 2 + \cdots + k$$$. My submission: 193040606. I tried to solve this problem in the hard way, but I don't know why it failed on the middle of the test 2. Why this tutorial hs considered m as n-1/2 how we got.. pls someone clear the doubt. If you've seen these problems, a virtual contest is not for you - solve these problems in the archive. One of the problems will be divided into two subtasks. For even you can just do n / 2, n / 2 For odd you will find answer (n + 1)/2 + x, (n 1)/2 x Where x = 4, 49, 499, 4999, , 8, 89, 899, 8999, Can anyone explain the pairing part of C? 2) One and two question solution, coded in python Then choose the best point take O(n). Here in this video we have discussed the approach to solve\" A. It's easy to show that those two numbers always occur as adjacent numbers in $$$sorted(A[])$$$, The number of distinct gcd prefixed/suffixed at an index in an array will never exceed $$$log(A_{max})$$$. I didn't participated in the contest, but I solved after . We can now describe the solution. Virtual contest is a way to take part in past contest, as close as possible to participation on time. If the list of absolute differences is the same for both strings, they are funny. It can be implemented in linear time in the size of the two groups we have to iterate over, which have size $$$\binom{(k-h)/2+15}{15}\approx 5\cdot 10^5$$$. either its moving right towards the ith element (which is closest for it), or left (to some other closer element on its left as compared to the ith element, we dont care what that element is, only the direction matters to us). B. 2), CerealCodes is excited to announce our Summer 2023 Programming Contest!. Note the recurrence in the official solution: Solving it with brute-force costs $O(n^2)$ time. This contest counts for GP30 scores. Indices are numbered from left to right from $$$1$$$ to the length of the string. In E, what does it mean to "use coordinate compression on $$$p_i$$$" where $$$p$$$ is the prefix sum array of $$$a$$$ ? first you will solve easy question in div 2 and than go for div 1, The only programming contests Web 2.0 platform, Editorial of Codeforces Round 889 (Div. Print an answer for each test case in the input in order of their appearance. We describe a randomized solution that solves the problem for $$$m$$$ up to $$$10^{11}$$$ (and, with some additional care, may be able to solve also $$$m$$$ up to $$$10^{12}$$$). This has consecutive sums from $$$12$$$ to $$$18$$$. Every thing is Nice, But how do you know that we can skip the whole series if the number is not present at (l+r)/2 i.e. Any One has code for problem B, I tried binary search it's not working https://codeforces.com/contest/1788/submission/192954046. Find all the points such that the expected number of necessary moves is wrong. The component with node $$$1$$$ contains a cycle. Will studying number theory help me with this problem? I found the error in my code. Lately I receive a message that my submission 192896600 is same as 192935411.I cheked it and find that his solution to the problem is different from mine.I guess may be comments at the end of the code is the reason why two submissions is thougt as same. The choice of $$$a_1, a_2,\dots, a_{h}$$$. In particular, this trick guarantees $$$|A_k|\le 5000$$$ for all $$$0\le k\le 62$$$. It should be: dp[i][j] = min(dp[i-1][j-1], max(dp[i-1][j],left[i])+length[i]), not, dp[i][j] = min(dp[i-1][j], max(dp[i-1][j-1],left[i])+length[i]). https://codeforces.com/contest/1788/submission/192977088, You can check his code: https://codeforces.com/contest/1788/submission/192954046. Each final pair will be $$$2n + 1 + x$$$ for $$$x$$$ in some consecutive range, that adds to $$$0$$$. Author: TheScrasse Preparation: akifpatel, dario2994. Can you clarify a bit more about what do you mean by, you chose the j-th index before you and the j-th index had a direction left/right. I can not explain why it works for this problem. Both Lemmas can be easily proved with a bit of math. Once you find that, break the loop. There is way to find palindromes is O(n) using Z-function-like algorithm. The first line of the input contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) the number of test cases in the input. i thought a lot but couldnt find any such test cases? In fact, since a Div 1 contest is usually (always?) Can anyone please explain the condition of excluding the unnecessary dots from the subsets in problem D ? One and Two Codeforces Round #851 (Div. Prefix Bitwise Or/And can take a maximum of $$$log(N)$$$ values. So, you can assume you always use the card on the top. Find out if an integer $$$k$$$ exists so that the following conditions are met. A reasonable interpretation is given by considering an $$$(n+1) \times (m+1)$$$ grid, where the $$$i$$$-th row initially contains a block in column $$$S_i$$$. The first line of each answer should contain $$$r$$$ ($$$0 \le r \le |s|$$$) the required minimum number of positions to be removed, where $$$|s|$$$ is the length of the given line. The point values will be 100-200-300-400-500-500-550-600. Can we deduce a relation between $$$L$$$ and $$$L'$$$? https://codeforces.com/contest/1788/submission/192961698. Assume by contradiction that $$$|a|+|b|+|c|+|d-k|> 1 + 2 + \cdots + k-1$$$. Note : this is one of the approach ,there can be multiple optimized approaches. Suppose you take a specific subset of $$$\geq 2$$$ dots. We iterate over all possible $$$s_1, s_2, \dots, s_{(k-h)/2}\in A$$$ and over all possible $$$s_{(k-h)/2+1},\dots, s_{k-h}\in B$$$ and check whether the sum of one choice from the first group and one choice from the second group yields the result. One and Two \" of codeforces round 851 in hindi.Please like, Subscribe and share channel for more amazing content :) Also press the icon to get notified quickly. What I cannot understand is, why is no one recommending just GIVING UP. I'm guessing that I'm not understanding that part correctly, where does the constant $$$c$$$ comes from? In some sense, we are saying that the inductive step works also in dimension $$$4$$$, but the base cases don't. If $$$|a|+|b|+|c|+|d-k|\le 1 + 2 + \cdots + k-1$$$, then it must be $$$(a, b, c, d-k)\in A_{k-1}$$$ because if $$$(a, b, c, d-k$$$ were reachable with $$$k-1$$$ moves then $$$(a, b, c, d)$$$ were reachable with $$$k$$$ and we know that this is not true. A. I had a similar sol but couldn't implement it-tons of debugging :(. Every "connected component" of blocks (except the last one) represents an element in the set. Problem Name Online Judge Year Contest Difficulty Level; 1: Spreadsheet: Codeforces: Codeforces Beta Round #1: 1: 2: Before an Exam: Codeforces: Codeforces Beta Round #4 (Div. Each test case consists of one non-empty string $$$s$$$. (Like points) And do you have links to the games before? Problem B can also be solved with just randomly choosing $$$x$$$ until you find good one. If you are thinking that maybe the dp[i][j][left] is not a valid state because j has no closer element on its left w.r.t to i. Its length does not exceed $$$1.5\cdot10^5$$$. I think the editorial for C kinda glosses over inspiration for the construction. What is your question? Is Div.1 more difficult or Div.2? Instead of counting 2's I just maintain prefix and suffix array of multiplication My submission. First of all, notice that it is then possible to determine in $$$O(1)$$$ whether a point belongs to $$$A_k$$$ or not. Then that state will never be calculated in our answer. Whenever You want to maximize/minimize bitwise properties among some elements, consider iterating from the last bit and checking its possibility. How many points can you get? As a Gray I don't recommend it unless you know there won't be any collisions given the constraints of the problem or your submission could be hacked or it will just give WA, actually I didn't think about that when I answered you, it works for this problem because the constraints are small, same approach i used it failed on tc2 also https://codeforces.com/contest/1788/submission/192882231. 2.5K Companies Given an integer array nums, return all the triplets [nums [i], nums [j], nums [k]] such that i != j, i != k, and j != k, and nums [i] + nums [j] + nums [k] == 0. Thank you. Slightly longer: we start out with 26 arrays, all filled with 1's and having a length of k*repeats[c], where repeats is an array containing the number of occurrences of each character in the initial string. If you know the whole cycle (of length $$$x$$$), you can win in $$$n-x$$$ queries by asking for each node if it ends up in the cycle after $$$n$$$ moves. ,. First preprocess all Add queries for O(n), and then answer all Sum queries. Index of two/2 th 2 is the output in that case. Suppose we processed the first i calls and j of those were ignored, the best way is to minimize the last minute of conversation (i.e. It means I need to decrease left by (last_odd 2) and increase right by (last_odd 2) + 1.Then again I increase left by 2 to iterate over even numbers and decrease right by 1 to make step=1 in sums of pairs.To make it all work we need to choose x at the beginning. A. Thanks for the fast tutorial! Doing that will decrease the chance of binary search getting stuck or getting WA due to precision. In fact, since a Div 1 contest is usually (always?) So, you know the total number of (cards + points) that you get. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. E. Black and White Tree. 2) Editorial awoo Educational Codeforces Round 145 [Rated for Div. We expect to see many different solutions for this problem. For the second test case, there is no $$$k$$$ that satisfies $$$a_1 \cdot a_2 \cdot \ldots \cdot a_k = a_{k+1} \cdot a_{k+2} \cdot \ldots \cdot a_n$$$. Is that the way to get the initial values of $$$p_i$$$? How to do fractional cascading on an iterative segment tree? Simple solution to vk cup Qualification Round 1 and 2. Codeforces: As Simple as One and Two Mike the Coder 15.5K subscribers Subscribe 337 views 3 years ago Codeforces Problems Hi guys, My name is Michael Lin and this is my programming. The only programming contests Web 2.0 platform. One and Two Codeforces Round #851 (Div. One of the problems will be interactive, so please read . I get it now. 1 + Div. 2 Only) C, D and F are great problems. Suppose that you unlock $$$x$$$ cards in total. So final number formed from a is a1=220 and from b is b1= 211 (i am assuming you may know how to form a number from given digits.). Definition: For $$$k\ge 0$$$, let $$$A_k$$$ be the set of points $$$(a, b, c, d)$$$ such that $$$|a|+|b|+|c|+|d|\le 1 + 2 + \cdots + k$$$ and $$$a+b+c+d$$$ has the same parity of $$$1 + 2 + \cdots + k$$$ but $$$(a, b, c, d)$$$ is not reachable with exactly $$$k$$$ moves. A tree is an undirected connected graph with no cycles. What is its complexity? Each of the following t t lines contains 3 3 integers a a, b b, c c ( 1 a, b, c 108 1 a, b, c 10 8 ). It's fast, small, cross-platform and powerful. Codeforces. Thanks to this observation, if one is able to efficiently find $$$A_k$$$ for all interesting values of $$$k$$$, then solving the problem is (comparatively) easy. 2) A. Construct a Rectangle. in the multiplication, only 2 matters. Suppose that you know, for all previous i's and j's, the least time at which all calls end. So, you can calculate $$$dp_{i,j} =$$$ expected number of moves of the block $$$x$$$ before it reaches the block $$$x+1$$$, if the block $$$x$$$ is in position $$$i$$$ and the block $$$x+1$$$ is in position $$$j$$$. dp[i][j] =dp[i][j] = min(dp[i-1][j-1], max(dp[i-1][j],left[i])+length[i]). In this challenge, you will determine whether a string is funny or not. More insights on this can be understood well by studying the xor basis technique. For example, you can try making $$$a_{i+1}$$$ bigger using a positive element. How to do fractional cascading on an iterative segment tree? [1, S - 1], clearly this makes sum S one of even sum. Why does question B take a brute force approach to the pre-test 1 error, can it provide the data of test 1? Divide the digit in two smaller digits:$$$a = \lfloor\frac{d}{2}\rfloor$$$$$$b = d - a$$$. Just maybe you'll have something to ask. Then, check for each node if it's "sufficiently close" to $$$C$$$, by asking $$$(i, k, C)$$$. C. Monsters And Spells. The issue is that we need some flexibility in the process as we may need to repeat it many times, this flexibility is provided by the addition of some additional random elements which don't change the magnitude of the values $$$f(0), f(1), \dots, f(29)$$$ but that modify them as much as possible (if we added a large number it would not affect many $$$f(s)$$$ and thus it would not be very useful). If all the elements are negative, you can win in $$$n-1$$$ moves. Yes, Div 1 A/B are harder than Div 2 A, B. And let 2 combine with themselves. You're given the sequence a1,a2,,an. $$$y_1$$$ and $$$y_2$$$ are defined similarly in the negative case. 6. Otherwise, run a loop until you find the number of two/2 th 2. If you change $$$p_{r_k}$$$, $$$p_i$$$ would also change. The bottleneck is making the big positive element. The j'th element has its direction fixed ie. If you still can't get it, feel free to point out what you are confused :). The backpack problem to find $$$s_1, s_2,\dots, s_{k-h}$$$. there is difference that's why they are i think still going on. In problem C you provided one of the possible pairings but how can someone prove it? In the second test case, the cell in row 2 2 and column 1 1 is already black. Find the nodes in the cycle in the component with node $$$1$$$. There are many other edge cases (like if you have consecutive 9s) that this kind of approach would not be suitable. https://codeforces.com/contest/1788/submission/192980925. The problem statement has recently been changed. Can you "double" the number of nodes in the cycle? they are supposed to be easier but apart from 1 contest, I have felt they are at par with Div2. Approximately the first $$$\log_2(m)$$$ values are set equal to $$$1$$$. Subscribe 707 views 8 months ago CodeForces Solutions Problem Link: https://codeforces.com/contest/1740/p. 12708008. It can be shown that $$$X$$$ takes $$$log(X)$$$ distinct values until it reaches to $$$0$$$. Brief introduction of myself: Chinese, 14yo. $$$a+b$$$ and $$$1 + 2 + \cdots + k$$$ shall have the same parity. My final answer is the sum of dp[i][[j][right] for all pairs of i and j. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. how we can do better? [Tutorial] Floors, ceilings and inequalities for beginners (with some programming tips), There is no element chosen between the j'th and the i'th element. (I did not formally prove that myself but maybe this illustration on paper will help to get the idea in that direction. Array only consists of 1 and 2, you can notice that multiplying by 1 does nothing, so we count the number of 2's in the array, to divide the array into two equal parts, there must be an equal number of twos on each part, using a second variable we can loop from N to 2 and check if the array value is 2, if it is, we add 1 . 1035. To determine whether a string is funny, create a copy of the string in reverse e.g. I was dead stuck on the problem and knew I was missing something for sure but after reading you comment here, solved it in the next 10 mins. B 158B - Taxi Choose the 3 and 1 as much as possible. So, it's enough to check intervals with $$$l = 1$$$, i.e., find the smallest $$$x$$$ that does not divide $$$n$$$. Either you go for a greedy or for a backpack. It is supported only ICPC mode for virtual contests. Swapping the shown pairs produces. What are the requirements for participating in the competition? Blocks $$$x, x+1$$$ both move with probability $$$1/2$$$, unless block $$$x+1$$$ has reached position $$$m+1$$$. It is supported only ICPC mode for virtual contests. Input: The first line contains single integer n (1 n 5*10^5) the size of the array a. We will hold AtCoder Grand Contest 063. Two simple necessary conditions are: It turns out that this two conditions are also sufficient! $$$a_1 \cdot a_2 \cdot \ldots \cdot a_k = a_{k+1} \cdot a_{k+2} \cdot \ldots \cdot a_n$$$. And it seems that suffix array can work well in finding all palindromes , but i'm not good at it :( . I have implemented exactly same approach as mentioned in the editorial. 2, C Hmm now I could not find myself any $$$n$$$ for which the submission would not work. Therefore, the cell in row 1 1 and column 4 4 will become black. FOR digit 4 , a={2} b={2} 4/2=2 (case of even digit). Obviously the sum of min number of every size in two sets is the first answer. after that I randomly tested bunch of ways to solve the match, one worked for all(and kinda proofed that it always work for n odd (like the editorial with the m thing), so I coded it . This also makes it trivial to see, that even $$$n$$$ is impossible. In other words, Polycarp does not like the string $$$s$$$ if there is an integer $$$j$$$ ($$$1 \le j \le n-2$$$), that $$$s_{j}s_{j+1}s_{j+2}=$$$"one" or $$$s_{j}s_{j+1}s_{j+2}=$$$"two". So, let's say that ss has |ss| elements, you can always split it into O(sqrt|ss|) fragments of O(sqrt|ss|) letters each. If you have several options for restoring the names, print any of them. Is it just intuition or it came from some kind of theorem, formula etc? The proofs of the below statements will not be mentioned here; It's advised to do such proofs on your own for exercise. Binary search. Could you please check my submission here: https://codeforces.com/contest/1788/submission/192980945. There are $$$\binom{k-h + 29}{29}$$$ possible ways (up to order, which does not matter). 2, A: 100: 2: Codeforces Div. held parallel to a Div 2 contest, they use some of the problems both for Div 1 and Div 2. Problem C. Using the formula for the sum of an arithmetic progression, we found that the sum of the first pair is (3n+3)/2. Least Number of Unique Integers after K Removals. In the other cases, $$$dp_{i,j} = ((dp_{i+1,j} + 1) + dp_{i,j+1})/2$$$. and does is have Cons (disadvantages)? The only programming contests Web 2.0 platform, Educational Codeforces Round 152 (Rated for Div. Can you find a better bound on the "best" number of moves? Though algorithms may seem complex, so tough, And bug-ridden code will make you frown, Stay true to your dreams, and find a way, And victory shall come to you one day. Zigzag Conversion. We will say something more about the choice of these values when we will describe how $$$a_1,a_2,\dots, a_h$$$ shall be chosen. The only programming contests Web 2.0 platform, Codeforces Round 606 (Div. Here's how alternate solution you said works. If there is at least one positive element $$$a_x$$$. palindrome[i][j] judges weather the substring ij is a palindrome string. F. A Random Code Problem. To me, it's truly depressing that a lot of these people genuinely believe that they can magically gain +1500 rating some day by just working really really hard. Codeforces Problem Statement : You're given the sequence a1,a2,.,an. We want to reach $$$(a, b)$$$. After that, you just add its length to the time it starts. Editorial of Codeforces Round 889 (Div. maximize the length of freetime until the current moment). Each swap will make one pair increase by $$$x$$$, and the other decrease by $$$x$$$. However, this can be rephrased as follows: for each pair of dots $$$i$$$ and $$$j$$$, count the total number of subsets where dot $$$i$$$ and dot $$$j$$$ form an RL substring, and output the sum for all pairs. Be careful not to use locked cards. Consider an $$$(n+1) \times (m+1)$$$ grid, where the $$$i$$$-th row initially contains a block in column $$$S_i$$$, and row $$$n+1$$$ contains a block in column $$$m+1$$$. Morning Sandwich | Educational Codeforces Round 152 (Rated for Div. This is my code. Therefore, if $$$n$$$ is an even number, the problem has no solution $$$=>$$$ $$$n$$$ is odd number. The most critical part is when we go from last odd number to first even number: we need to increase right by (last_odd 2) + 1 which is equals to n-1 because last_odd=n. Pattern finding is the case for me. By linearity of expectation, the answer is the sum of $$$dp_{S_i, S_{i+1}}$$$. These values are flexible (the solution would still work with $$$h\le 45$$$ and $$$k-h=45$$$ for example). I'm also trying to find my fault, because I have no idea what is the test 3, The only programming contests Web 2.0 platform, https://translate.google.ru/translate?sl=ru&tl=en&js=y&prev=_t&hl=ru&ie=UTF-8&u=http%3A%2F%2Fe-maxx.ru%2Falgo%2Fpalindromes_count&edit-text=&act=url, Editorial of Codeforces Round 889 (Div. Programming Insider 115 subscribers Subscribe 828 views 1 month ago codeforces contests Codeforces Round #851 (Div. Isn't it cruel to enforce this false belief? You are truly a genius . If there exist multiple $$$k$$$ that satisfy the given condition, print the smallest. Dual (Easy Version) . If you know of any tips, please write about it. Educational Codeforces Round 152 Editorial. I heard about it, though, but thought it was complicated. 2) | Codeforces solution CodeSolve 844 subscribers Subscribe 0 120 views 10 minutes ago Like and subscribe. 1481. while there are at least $$$2$$$ sad students, you can swap them and both of them will be happy; if there is exactly $$$1$$$ sad student left, you can swap it with any other student. The 4d version is not too different from the 2d one. Yes, Div 1 A/B are harder than Div 2 A, B. i don't use prefix multiply, i only use same hint like that. The official implementations of all the problems are here. Without loss of generality we may assume $$$0\le a\le b$$$. We can observe $$$L = L'$$$ or $$$L' >= 2*L$$$. The entire complexity of the algorithm is O(n*log^2(k*|s|)). In one move, you can make at most $$$2$$$ sad students happy (because you can change the position of at most two students), so you need at least $$$\lceil \frac{s}{2} \rceil$$$ moves. If you use at least once a card that is not on the top on the deck, you can prove that using the cards in order (from the top) would give the same number of victory points. For in this game, no player truly loses, For every problem solved, a victory chooses, And so keep pushing, keep trying, don't stop, For the greatest programmer, is the one who never stops. First, find $$$k$$$ nodes in the cycle, doing a binary search on the successor of the last node found. 0.2n2+1 Thank you. $$$dp_{i,j} =$$$ expected number of moves of block $$$x$$$ before it disappears, if the block $$$x$$$ is in position $$$i$$$ and the block $$$x+1$$$ is in position $$$j$$$. In my little experience with dp I haven't found a problem like this. Proof: The strategy is the same adopted to show Lemma 1. You can optimize the solution using a bitset. Ques C) Can anyone tell me what is the error? I was trying a tree solution for C, but was getting wrong answers and eventually decided to make a solution that supposedly took advantage of the fact that the string is cyclic. Since $$$x_1 + x_2 + y_1 + y_2 \leq 25$$$, $$$\min(x_1 + x_2, y_1 + y_2) \leq \lfloor \frac{25}{2} \rfloor = 12$$$, as we wanted. To count how many of these there, we can first count the number of dots, excluding $$$i$$$ and $$$j$$$, such that the presence of these dots will not prevent $$$i$$$ and $$$j$$$ from moving towards each other. Solution Link: https://ideone.com/ggzx42 One can prove it by induction on $$$k$$$ as follows. How should we choose $$$a_1, a_2,\dots, a_h$$$? I am trying understand the 158E Phone calls solution but it seems a little complicated for me. in C can anyone explain i got k = 3*(n+1)/2 now how to generate pairs like k, k+1, k+2----k+n-1 as sum. How can i find editorial of previous contest? I've seen submissions like this one 192978163 and can't figure out why it passes. (in this way one can solve the problem also with larger constraints on $$$A, B, C, D$$$; but it is tedious), Expand held parallel to a Div 2 contest, they use some of the problems both for Div 1 and Div 2. I suppose the edi glossed over it because there are a lot of valid constructions.
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