DM for solution telegram link:- https://t.me/codechef7343 % Working Codechef & Codeforces & Leetcode Solution Provider of any contest of codechef Update this pointer as you walk through values of $$$S_r$$$. For all $$$l, r$$$ that the interval $$$[l, r]$$$ is good, we have $$${r-l+2}\choose 2$$$ sub-intervals of $$$[l, r]$$$ that is good. Reply. Make A Equal to B | Codeforces Round 825 | Division 2 | Div 2 | HindiA. And it is easier to code talking first node as the source of dfs. If the first friend with a car is able to send $$$p_1$$$ home (but not $$$p_1$$$ and $$$p_2$$$ at the same time), then we will start checking $$$p_2$$$ for the second friend with a car. Also I have created a validate function which can check if it is possible to form an increasing sequence. Hello Codeforces! It's pathetic. Make A Equal to B time limit per test The resulting sequence will be [9, 6, 3] [ 9, 6, 3]. In the second test case integers $$$a$$$ and $$$b$$$ are already equal, so you don't need to perform any operations. Now, we just need to check $$$2$$$ cases, one in which the re-arrangement operation is used, and one in which it is not. Calculate the minimum number of operations required to make a and b equal. The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) the number of numbers. It can be updated by considering the next index but it will take $$$O(n^4)$$$. I code part of c1 why declaring dp with size n + 5 when n + 1 can also do the work. Make A Equal to B | Codeforces Round 825 | Division 2 | Div 2 | HindiA. $$$p_t > p_{t-1}$$$ Let us make some observations. I would like to share my DFS solution approach for problem E. There can be at most O(2*n) valid subarrays. We can basically find out for each index i, if i is the start of a subarray, what are the first two bad points (bad points being points at which the given condition gets violated). Host and manage packages. The transitions can be made in the following manner: Iterate over $$$i$$$ from $$$n$$$ to $$$1$$$, and then iterate over $$$L_i$$$ from $$$i$$$ to $$$1$$$, notice that all that really matters about the state which you transition from is its second parameter, so basically maintain an array which stores best possible transition for each value of second parameter, and keep trying to improve it as $$$L_i$$$ decreases. - New control tag: "!RuntimeMode:S7A". CF-userinfo - GitHub Pages Enough to compare the first element with middle value. Another Solution For D. We don't need to swap or check sorted. $$$a[l,r]$$$ represents the subarray $$$a_l,a_{l+1}, \dots a_r$$$. This means that each segment will now have a sum of total_sum / number_of_segments. Make A Equal to B B. 2) Problem Solution Click Here C2. Rudolf and Snowflakes (hard version) codeforces1846E2, D1. Contribute to kantuni/Codeforces development by creating an account on GitHub. Make A Equal to B || Codeforces Round 825 (Div 2) || Codeforces CodeForces | Make A Equal to B - StopStalk Can u pls elaborate your second point. The same item-ID address format has to Make A Equal to B Codeforces solution |Codeforces Round #825 (Div. 2 information see the document "S7A_as_Service.PDF" on the installation CD . As done in the editorial, traverse the segments after sorting by left coordinate, and again after transforming $$$(l,r)$$$ to $$$(-r,-l)$$$. A. Make A Equal to B codeforces solution div#825 - YouTube Now we can see that the indices we wil be comparing on the next step will be a subset of the current step. In case you found C2 tedious to implement or found many cases to deal with, I would recommend you to have a look at the intended solution. Surprisingly we had predicted that this round would be easier than usual Div 2 rounds. The rack and slot number of the S7-CPU must be stated in the corresponding cfc++20(64), timidcatt: I'm not sure what the editorial is trying to say, and I'm finding it difficult to decipher the code in the given solution. Hello Codeforces! So $$$track[i]=\sum_{j=i}^{q-1} dp[j] + track[q]$$$. With that claim in mind, let's make a two pointers approach. Now $$$dp[9]=min(dp[8]+1,a[9])=8$$$ and $$$dp[10]=min(dp[9]+1,a[10])=8$$$. Take a look at Ticket 16316 from CF Stress for a counter example. - The string write from the OPC Server sometimes failed . Here in this video we have discussed the approach to solve" A. March 16, 2020 03 . The above algorithm nearly works, except for when a segment is only adjacent to differently colored segments inside itself. Time complexity is $$$O(n)$$$. they are spelled differently. If you've seen these problems, a virtual contest is not for you - solve these problems in the archive. Start with $$$l = 1$$$ and $$$r = 1$$$, and $$$antr = 0$$$ ($$$antr$$$ is the last $$$r$$$ that we compute for an answer). The first line contains two integers, n and m (1 n 3000, 1 m 3000) the number of words in the professor's lecture and the number of words in each of these languages. I wish editorialists weren't this lazy. The same address format has to be used for the tag-I/O The only line of each test case contains two integers a and b ( 1 a, b 10 9 ). Problem C1 ( Good Subarrays ) Easy O(N) Solution: While traversing in the array, I sum the minimum of the index and the element i.e. Now we can iterate over all indices of array a, and binary search the farthest length of good subarray [i,R] for all 1<=i<=n. You choose the number of these operations yourself. Zero-One (Easy Version) codeforces 1733D1. Sex scene with Florence Pugh in 'Oppenheimer' becomes latest target of So $$$q$$$ is the smallest index greater than $$$p$$$ such that $$$a_q \leq z + q - p$$$. Problem - 499B - Codeforces I have a slightly different idea of C 1)I sum up all values of array 2)build a prefix sum of array 3)went through all divisor of Sum 4)now let Div be the current divisor then this div should exists In prefix sum,it multiple should exists, 5)I took the longest i In prefix sum where Div exist and updated the answer! This solution is $$$O(m)$$$, you can check my submission (Python:175604510, C++:175815111) for further details. Good Subarrays (Hard Version) D. Equal Binary Subsequences E. Swap and Take 1736; A. For problem D, we can first check all leaves. Divide both sides by nine, and we are left with x is equal to negative five. GitHub: Let's build from here GitHub Problem - 1624B - Codeforces Take an integer ccc ( c>1c>1c>1 and a should be divisible by ccc ) and replace a with ac\frac acca; DI) or analog (i.e. We make $$$n$$$ pairs of the form $$$(s[2i-1],s[2i])$$$ for $$$(1 \leq i \leq n)$$$. What is the minimum number of moves that we need in order to obtain two equal strings? Implementation: https://codeforces.com/contest/1741/submission/175824953, Alternative implementation, with an additional factor of $$$O(\log c)$$$ in space and time complexity, but uses no data structures outside std::set: https://codeforces.com/contest/1741/submission/175826713, (EDIT: the use of a segtree can be circumvented in my original solution with the bitwise maximum technique in this implementation, which should also improve execution time. - The driver can now be registered as a service with the local system account as the log-on account. {"payload":{"allShortcutsEnabled":false,"fileTree":{"":{"items":[{"name":".cph","path":".cph","contentType":"directory"},{"name":".vscode","path":".vscode . My expected ratings(after testing) were : A 800, B 1200, C1 1400, C2 2100, D 1900 and E 2200, Am I the only one who receives RTE on test 4 for no reason? Make A Equal to B | Codeforces Round 825 | Division 2 | Div 2 | HindiA. I've solved problem B with LCM approach, however I saw some solutions checked if((a[i] % gcd(a[i 1], a[i + 1])) == 0) for 1 <= i <= n 1 how this solution also worked ? It is supported only ICPC mode for virtual contests. see gcd(k1*lcm(a1,a2) , k2*lcm(a2,a3)) will always have the gcd(lcm(a1,a2),lcm(a2,a3)) rest will be contribute by gcd(k1,k2) the contribution from both the terms must take us to a2. 0. So we have $$$3$$$ ranges to deal with $$$(1,p-1)$$$, $$$(p,q-1)$$$ and $$$(q,n)$$$. ? 1 + Div. be completely cleaned from S7A registry entries after uninstall. global memory was exhausted, the driver crashed. It is supported only ICPC mode for virtual contests. Imagine a segment $$$S$$$ with left and right points being $$$S_l$$$ and $$$S_r$$$. Now $$$a_q-q \leq z-p$$$. 5, Zach_Tang: min(i+1, a[i]) {i+1, because of 1-indexing) But in this the problem is that if we encounter a number which is smaller than its index, then it gives wrong answer. Of course, during the lecture you write down each word in the language in which the word is shorter. Notice that the array $$$l$$$ is non-decreasing. Make A Equal to B | Codeforces Round 825 | Division 2 | Div 2 | HindiA. Output exactly n words: how you will record the lecture in your notebook. But another thing is, how to precalculate the track[i]. How do I set my home country on Codeforces? Update divide-by-three-multiply-by-two.cpp. Thank you for participation! There are some problems where you discover later that you might need the extra indices, e.g., due to switching between 0-indexing and 1-indexing, or because the nature of your solution to the problem would benefit from stepping further ahead, etc, and it's more convenient to mindlessly declare it as a larger size every time than to rely on consciously adjusting the declaration when you need to, which you might forget (like if you were really deep into writing some part of the solution at the time) or miss (like if there are multiple such arrays affected and you overlook one). If a[r]rl+1 and a[r+1]" Lastly, the answer is the maximum among all the permutations. Let's start with an example We have an array like 4 3 2 , According to your assumption. int pos = res.size()-1; Now we only left with range $$$(p,q-1)$$$. We can use a number which is a product of many different primes for an example of this. You are given the text of the lecture the professor is going to read. class Solution { If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive. Now we claim that $$$dp[i]=dp[i-1]+1$$$. A - Make A Equal to B: 1.2. A B , 0,1 Ans1= ( +1) A B , 0,1 I tried this problem and found that making every element equal to the median of array is optimal is there any proof for this. Educational Codeforces Round 151 [Rated for Div. 2] dp[i] = min(dp[i-1] + 1, a[i]) adp[i] = min(adp[i-1] + 1, b[i]), dp: dp[1] = 1 dp[2] = 2 dp[3] = 3 dp[4] = 4 dp[5] = 5 dp[6] = 6 dp[7] = 1 dp[8] = 2 dp[9] = 3 dp[10] = 4, adp: adp[1] = 1 adp[2] = 2 adp[3] = 3 adp[4] = 4 adp[5] = 5 adp[6] = 6 adp[7] = adp[p] = 7 adp[8] = 8 = a[8] adp[9] = 8 = a[9] adp[10] = 8 = a[10]. Suppose $$$dp[i]=a[i]$$$. So you can easily compute the sum of this arithmetic progression. - When the signal conditioning function "S&M" was used for a byte value, the driver wrote wrong values to Can anyone explain me what this code is doing ? Build 222 : New Functions: The most straightforward way to improve it to $$$O(n^3)$$$ is to use prefix maximums. Yes, it was my bad. DI) or analog (i.e. Is there an error in my understanding of the editorial? Any suggestions on etiquette and improvements are welcome. Why will this assumption fail? The intuition is to check for sizes of multiple of 2. In the third test case you have to apply the first, the second, the third and the fourth operation to $$$b$$$ ($$$b$$$ turns into $$$20 + 1 + 2 + 3 + 4 = 30$$$). GitHub - kantuni/Codeforces: Solutions to Codeforces Problems Initially, all elements of the array are, Another Problem About Dividing Numbers, You are given two integers aaa and bbb. I tried something similar, but it was giving TLE, so I had to use sparse table for RMQ to make it $$$nlogn$$$. $$$d(N)\le N^{\frac{1.5379 \log(2)}{\log(\log(N))}}$$$ is a known upper bound for the divisor function for $$$N \le 3$$$, though. For example, let the permutation be $$$p$$$. This approach is similar to the one in the tutorial, in that in each iteration we fix the sum of each segment and check to see if we can get a split satisfying the given conditions. It simply needs a Thus our selected characters = $$${0,1,0,1, \dots ,0,1}$$$. So number of $$$1$$$ in $$$s$$$ is $$$x+2 \cdot y$$$. 1 <= C1 <= 10 1 <= C2 <= 10^4 1 <= A [i] <= 10^8 - Communication error messages will be now displayed in the alarm history of iFIX. It could be done by finding c = lca(a, b) c = l c a ( a, b). So, if $$$\gcd(b_i, b_{i + 1}) > a_i$$$ for any $$$i$$$, we should also have $$$\gcd(c_i, c_{i+1}) > a_i$$$. This function is also 2) will start.. 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Make A Equal to B | Codeforces Round 825 | Division 2 | Div 2 | HindiA. And $$$1\cdot (-1) \cdot (-1) = 1$$$. Good Subarrays (Easy Version) Codeforces Round #825 (Div. 2 - Blogger Make A Equal to B | Codeforces Round 825 | Division 2 | Div 2 | HindiA. Time complexity is $$$O(n \cdot \log(n))$$$. Each test case's second line contains n space-separated integers d1,d2,,dq (ai is 0 or 1) representing the array x. Evey test case's 3rd line contains n space-separated integer value e1,e2,,eq (bi is 0 or 1) going to represent the array y. increase any element by 1 decrease any element by 1 What is the minimum number of operation we can do to make all elements in array equal. Also my binary search + rmq solution runs in less than 600ms. So we won't be able to arrange the whole tree in increasing order. Codeforces Round #825 (Div. 2) - - dp[i]=min(dp[i1]+1,a[i]). There's another solution to B, which I believe to be more intuitive. Codeforces 1154B. Make Them Equal Codeforces GitHub A digital (i.e. After the Finding minimum moves required for making 2 strings equal You can see that there is no way to move elements from the left subtree to the right subtree (or vice versa), other than to swap the two subtrees entirely. The numbers of factors do not have a log2N upper bound. Time complexity is $$$O(n)$$$. For each test case print one integer the minimum numbers of operations required to make $$$a$$$ and $$$b$$$ equal. array a = 4 3 2 array b = 4 12 6 2 Here you can see a1 is gcd of b1&b2 but a2 & a3 case fail cause gcd of b2 & b3 is not equal to a2 and similar to a3. Does anyone have a solution to question C2 except that the queries are dependent I thought a lot but did not find a solution. html - How to make two buttons the same size? - Stack Overflow So $$$track[8]=dp[8]+dp[9]+dp[10]=24$$$ if $$$dp[8]=a[8]$$$. i get it it's about finding k1 and k2 .but i'm confused can you explain ? In one turn, you can do one of the following operations: Tea with Tangerines C. Phase Shift D. Meta-set E. House Planning F. Pebbles and Beads 1735 b2 should be multiple of a1 {gcd(b1,b2) = a1}, b3 should be multiple of a3 {gcd(b3,b4) = a3}. Output the words of the lecture in the same order as in the input. For this range, the worst element is the element a[j] for which a[j]-j is too low (minimum on the range [L,R]). Solutions. However, as we want to know all segments with left points smaller than or equals to our current right point $$$S_r$$$, the idea is iterate over all segments $$$S$$$ in increasing order of right points. Sorry For bad English. The first line contains two integers, n and m (1n3000, 1m3000) the number of words in the professor's lecture and the number of words in each of these languages. The time complexity is $$$O(nm + k!\times f)$$$. detect, trace and solve communication issues. This round will be rated for the participants with rating lower than 2100. You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.. the context-sensitive help works properly. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. Make A Equal to B | Codeforces Round 825 | Division 2 | Div 2 | HindiA. It is guaranteed that each of the strings ci belongs to the set of strings {a1,a2, am}. Suppose $$$track[i]$$$ denotes $$$\sum_{j=i}^{n} dp[j]$$$ if $$$dp[i]=a[i]$$$. And as we will refer to a subset of the current step on the next step, just swapping the two elements suffices for counting the swaps. Links: "slot no. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) the numbers. That's why I commented this on the editorial, not contest announcement. address "!RuntimeMode:S7A" delivers the value 0 (in the F_CV property), when the driver is running Of course this solution gives you a TLE. Finally all the points with indices between $$$L_i$$$ and $$$R_i$$$ will have $$$a_i$$$ added as their contribution to the score. Did the same. As the constraints suggest, we should use dp to solve this problem. The first line includes a single numeric p (1<= p <=400), which represents the number of tests. vector func(int num){ D can also be done in O(n*log(n)) , variation of mergesort, here is the code. Let's consider each index as a node. Product. In C1, how can we solve it using two pointers, why is this approach failing : Is the algorithm mentioned in the editorial for C2 always correct? you need to set a width and height for your button and to centre text horizontally and vertically use the following: width:120px; height:50px; text-align:center; line-height:1.1em; font-size:1.1em; width and height can be set to match your desired measurements. Educational Codeforces Round 152 [Rated for Div. 2] - Codeforces Take a two blocks [5, 7] [8, 9] here min element in block 1 is 5. iterate over first block and check if absolute difference is less than block size. The problem statement has recently been changed. GitHub - SRD2705/Codeforces-Python: Many solutions of Codeforces Trying this with some of examples above and block of even larger size might clear things out. {"payload":{"allShortcutsEnabled":false,"fileTree":{"":{"items":[{"name":"1 A. Theatre Square.cpp","path":"1 A. Theatre Square.cpp","contentType":"file"},{"name . Make A Equal to B | Codeforces Round 825 | Division 2 | Div 2 | HindiA.