n aibii0~aijbj, ,0ii~i-a[i]O(nlogn), Codeforces Round #751 (Div. So even if we insert the elements in other fist best place, the answer won't change. I guess you are same to me : observe that $$$b$$$ will be sorted, and we will find a way to put $$$b[i]$$$ in $$$a$$$. ans How do I set my home country on Codeforces? The independent diamond originated in France, is a world-famous puzzle game and the Chinese-invented "Hua Rong Dao", the Hungarian invented the &qu [Codeforces Round #627]1324C-Frog Jumps[Thinking], cf--contest1324: C. Frog Jumps + Greedy Thought, Frog jumps stairs and metamorphosis jumps, Sword refers to the offer, the frog jumps a step abnormally (C++ implementation), C++ implementation leetcode#10- II frog jumps the stairs problem, Sword refers to offer 8, frog jumps steps python and c++, Codeforces Round #627 (Div. Codeforces Round #751 (Div. 2) Frog Traveler Solution (Java/C++) is there any proof for the observation that the array will remains constant after n operations? 1nL,Rd All we have to consider is between $$$p_j$$$ and $$$p_i$$$. Unfortunately, after a poor jump, he fell into a well of n n meters depth. Your divisors array is only of size 31. i 688 $$ - $$$ American, Bar, Pub. When we are adding the nodes into the queue we would simply query the desired interval of possible jumps and check if there is any that are not INF. Codeforces Round #751 (Div. 2)D. Frog Traveler - CSDN Now Gorf is on the bottom of the well and has a long way up. standard output. When you process the next place, it is impossible to create a new disjoint continuous segment (you start within the segment, which was covered by the initial jumps and you make another continuous segment of jumps). Codeforces-1601B: Frog Traveler codeforces1601B 3 a3=2a_3=2a3 =21b1=1b_1=1b1 =12a2=2a_2=2a2 =2 (You can check the code out in my submissions.) (C2020jzm told me.). It jumps by the following algorithm: the first jump is aa units to the right, the second jump is bb units to the left, the third jump is aa units to the right, the fourth jump is bb units to the left, and so on. Initially the segment tree will have all elements set to something that is not INF. Seeking the minimum number of times to climb to the ground, and the output scheme, the height of each climbing (before rest). Then it gets removed from all sets with j [1, k 1]. Hi all, I need help to figure out why my submission for Codeforces #751 Div 2 Problem D Frog Traveler, is giving TLE. just iterete over k and check whether cnt[i] % k == 0 ,where cnt[i] mean count of numbers in which i-th bit is on. m #define ub upper_bound Now , see where all can you slip back from these x nodes (only one node for each ). d In this way, each dual group is visited at most, deleted \(1\) Second, so the complexity is \(O(nlogn)\) Essence For the output scheme, directly in \(bfs\) Just record the front drive, pay attention to the height before output, and \(bfs\) It is the height after rest, just make some modifications. I guess it relate to dp optimize technique (you can see it here : https://codeforces.com/blog/entry/8219 ). The first line contains a single integer $$$n$$$ ($$$1 \le n \le 300\,000$$$) the depth of the well. How many kinds of jumps are there? tags:4.0 data structure8.0 search8.2 bfs. So you record the lowest height you jumped so far and make all the jumps that go past that height. So if it is written logN, it may be logN+1, 5*logN, etc. Ohhhh my, it's absolutely perfect! It is guaranteed that it is always possible to reach n + 1 n + 1 from 0 0. 0 For now, we will use a heap/priority_queue to maintain, in increasing order of the dp value, all the states that haven't been propagated yet. This is the video in which I describe how I went from nowhere to India Rank 1 on Codeforces. What conclusion do you prove? It is OK to give an implementation? A frog can jump up to 1 step or 2 steps at a time. I have a solution using DP with divide and conquer and I have TLE. 2. Codeforces Round #751 (Div. This is, it's the main idea. The $$$i$$$-th test case is described as a string $$$s$$$ consisting of at least $$$1$$$ and at most $$$2 \cdot 10^5$$$ characters 'L' and 'R'. Did anyone use the method in the editorial to solve Div.1 C? Thanks! My Competitive Programming Journey (Codeforces Grandmaster) Find the total number of jumping methods the frog jumps on an n-level step. What if I pick any increasing $$$p_i$$$? Give a string containing only two characters Oh, indeed, I'm dumb. Isn't 1601A - Array Elimination complexity should be $$$O(n \log C + \sqrt{C})?$$$. big thanks for the round! If there are more than two possible f D. Sorting the Coins time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Recently, Dima met with Sasha in a philatelic store, and since then th D. Winter is here Winter is here at the North and the White Walkers are close. Though the DP idea definitely helped me find the solution. $$$ O (n \ log \ n) $$$. Impressum. In the fourth test case of the example, the frog can choose $$$d=1$$$ and jump $$$5$$$ times to the right. i , C. Array Elimination Until now=n. Prik Thai Maine. We can use a similar way of pairing parentheses to discretize the jumpable range of each point on the number line. Since $$$b_i=s_j$$$ holds too. | FrogTheTraveler You have to answer $$$t$$$ independent test cases. The Horned Frogs will travel 3,588 miles in the 2023 with its farthest trip being 781 miles to Iowa State in Ames, Iowa. After picking x, we are only picking elements from (last, x) but not from behind last. I use Java. checks for positions from j 1 to j a[j]. a Problem setters: LHiC NiceClock voidmax meshanya fedoseev.timofey AlesyaIvanova napstablook ch_egor vintage_Vlad_Makeev GlebsHP isaf27 Endagorion Anti-Light 4eT_llpuyHblJl KiKoS cdkrot Zlobober Tikhon228 grphil. can u tell ur solution for E pls,coz i cant understand one described in editorial? . We will use $$$dp_i$$$ maximal number of moves needed to travel from $$$i$$$ to $$$0$$$. 1601B - Frog Traveler was authored and prepared by KiKoS 1601C - Optimal Insertion was authored and prepared by isaf27 1601D - Difficult Mountain was authored and prepared by Tikhon228 1601E - Phys Ed Online was authored by jury and prepared by fedoseev.timofey 1601F - Two Sorts was authored by cdkrot and prepared by cdkrot, L HiC I've developed my expertise in strategic planning working in advertising agencies, and I now use it as a consultant to create new products and services in line with . May 28, 2014 6 min read. , linkk b O Editorials of previous contests are all very fast! s Now lets suppose after a few iterations of the while loop, you are at depth 'j', where i > j >= i a[i]. Otherwise, firstly print integer $$$k$$$ the minimum possible number of jumps. This is one of the key parts of the O(n) solution: you have to visit each dp state only once. The picture describing the first test case of the example and one of the possible answers: In the second test case of the example, the frog can only jump directly from $$$0$$$ to $$$n+1$$$. L n If a_i changes, the number of integers which equal to a_i will multiply. R Korney Korneevich and XOR (easy version) (1800) When we are at the ith position, segmenttree[j] will store dp[j] + n(elements in (j, x) having skill >= a[i]). I use Java. minn It is guaranteed that the sum of lengths of strings over all test cases does not exceed $$$2 \cdot 10^5$$$ ($$$\sum |s| \le 2 \cdot 10^5$$$). The frog wants to reach the n + 1 n+1 -th cell. If our query function returns INF then we know that we cannot make a jump as every value in the desired interval would have been visited. It means that if the frog is staying at the $$$i$$$-th cell and the $$$i$$$-th character is 'L', the frog can jump only to the left. And so forth. I have found $$$O(N)$$$ solution for 1602D Frog Traveler. The answer needs to be modulo 1e9+7 (1000000007). You can Have a look at my approach here : Fast Editorial! L For some reason I've decided to not use the Fenwick trees like I did before and switched to Order Statistics Tree from GNU C++ PBDs. In the second example, Gorf can jump to one meter below ground level, but will slip down back to the bottom of the well. If we can maintain such a segment tree, then each time we just have to query in [1, maxj] and dp[i] = queryAnswer + 1. Topic transfer door First of all, see this question is like DP (alley is not used). After mulling on this for a while I think I understand. Hi! It is not difficult to find that if we can jump to x, we must also can jump to y. Let us now look at a simple form of the optimal sequence again: Inc1 NonInc1 Inc2 NonInc2 Inc3, Now we know for a fact that: a[NonInc1] <= a[Inc1] and a[NonInc2] <= a[Inc2](otherwise noninc points would have been inc points). = But $$$j\lt u$$$ would mean $$$u$$$ can't be reached from $$$j$$$ in one step. So there is a contradiction. Having element $$$j$$$ in this stack means that $$$min(A_l,\dots,A_i) \space \forall \space stk_j \leq i \le stk_{j+1} = A_{stk_j}$$$. If you only prove that $$$p_i$$$ increasing, this won't be enough for my solution. I.e. So we can solve the problem by two pointers. Find the total number of jumping methods the frog jumps on an n-level step. e Codeforces Round #751 Editorial - Codeforces The frog doesn't want to jump far, so your task is to find the minimum possible value of d d such that the frog can reach the cell n + 1 n + 1 from the cell 0 0 if it can jump by no more than d d cells at once. A. i Thanks. My gut feeling was that, sort of like in a BFS, the first time you propagate to a certain dp state creates the optimal cost for that state. while (T--) Each item will be removed no more once per $$$ log \ n $$$ and for each index we can make 1 extra lower_bound because bfs can visit each index only 1 time. Create variable last and set it to 0 and a[0] = -infinity initially. So we can maintain a variable lim, which shows the leftmost visited position in the propagation. is video tutorial present for div2D if yes plz share the link if not ak2006 can u please make a video tutorial for div2D , i am not able to understand anything in this , help plz. int T;cin>>T; 2), Codeforces Practice Tracker Browser Extension, Educational Codeforces Round 152 Editorial, Educational Codeforces Round 152 [Rated for Div. I don't understand the editorial. Consider any fixed j and increasing i (j < initial i). The idea is to keep 2 variables Reach and NextReach and a counter for the jumps taken so far. It will store prefix sum up to next $$$a$$$ taking into consideration all inserted elements. Copyright 2020-2023 - All Rights Reserved -, Codeforces Round #751 (Div. , The frog chooses some positive integer value $$$d$$$ before the first jump (and cannot change it later) and jumps by no more than $$$d$$$ cells at once. Maybe this was your first time seeing this trick so you should remember it now. Next trick, we shift both, and make $$$f(True) = 1$$$ and $$$f(False) = -1$$$. 1602E - Optimal Insertion is super similar to a recent contest's 1579E2 - Array Optimization by Deque. It is guaranteed that it is always possible to reach $$$n+1$$$ from $$$0$$$. The answer needs to be modulo 1e9+7 (1000000007). you can always take non-visited $$$j$$$ with minimum $$$j - a_j$$$. L'Equilibre. My solution, which differs from the editorial's, is $$$\mathcal{O} ((N + M) \log (N + M))$$$. And you are trying to access the element out of bounds of 31. To get the traversal order we can keep track of each node's parents. , Now in (x, last) pick all the elements with skill >= a[x]. Let cnt(i) = number of a[i] has i-th bit. The surface of the well's walls vary in quality: somewhere they are slippery, but somewhere have convenient ledges. Or, can you find all divisors faster than $$$O(\sqrt{C})$$$ without too much complicated number-theory algorithms? Will it have any climbers after which diff does not increase and remain the same? 1492A - Three swimmers. With this new workaround, we can use segment-tree with operation: range-add, get index of maximum. Oh, I realized something just a few minutes ago. bottom of page . Unfortunately, Gorf has to take a break after each jump (including jump on $$$0$$$ meters). I think O(n) code is easiler to write than O(nlogn) code. My O(n) solution to Div2D/Div1B: let dp[i] be the minimal number of steps from n to i. Finally \(f_0\) That is the answer. 3) C. Frog Jumps, Codeforces Round #627 (Div. Input Then, to find the index where you can get from i, you need to take j = lower_bound(i - a[i]) and check that j exists and *j <= i. Here is my code if you want to take a closer look: seggy is my segment tree. The new algorithm has time complexity of $$$O(nlog n)$$$. The only programming contests Web 2.0 platform, Editorial of Codeforces Round 889 (Div. Site: CodeForces: Links: Problem. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. s Here, we will perform a BFS from the bottom of the well to all nodes it can reach above it and store the maximum height we reach as integer $$$start$$$ . L, Go one step to the left, if Manager I Strategy & Innovation @frog. Assume that there are two positions that can reach 0 with one jump. Minimum will take same place again. n<=3e5n<=3e5n<=3e5 And it's best to reach this height for the first time. First, the described solution uses a Segment Tree which keeps for each index $$$j$$$ value $$$j - a_j$$$. Java #include We can find them using our set. a_i,b_i Thus, we can store list of values we are planning to insert between two numbers. 742A - Arpa's hard exam and Mehrdad's naive cheat. 1256A - Payment Without Change. . Keep going until we reach 0. i In the third example, Gorf can reach ground level only from the height $$$5$$$ meters below the ground level. These ro Division + Dij I didn't expect it to write, it is a light micro. An important conclusion is, there exists an answer with $$$a_i+b_i$$$ increasing. In fact, you don't have to construct a solution meeting $$$p_i$$$ increasing. D. Frog Traveler (1900) E. Optimal Insertion (2300) F. Difficult Mountain (2700) 94. I have another solution for 1602D - Frog Traveler which works in $$$O(N\cdot logN)$$$. Formally: Also remember to update segmenttree[i] = dp[i] before going to i + 1. Problem 1601B - Frog Traveler is totally adorable. Description There was a frog named Freddy sitting on a rock in the middle of the lake, and suddenly he spotted another frog (her name is Fiona) sitting on another rock. 236 $$ - $$$ French, European, Wine Bar. We use x and y to denote these two positions, where x is closer to 0 than y.So we can clearly find from the figure below that y can not only jump to 0 meters, but also jump to the position of x. 979A - Pizza Pizza Pizza. Maybe a little similer to the Tutorial. For div1B/div2D, I understand that we want to find all j's that can land on u, but how is this achieved? Consider an operation on k elements, if all of these elements have i-th bit equal to 1, their i-th bit'll be set to 0. For 1C/2Ein the editoralcould you tell me what's the This sum differs by a constant " in the "solve"I can find the constantplease. Can't KiKoS provide a sample implementation to the logic given the editorial of Frog Traveler? It can be seen that it work in O(nlogn),because there is at most n "edge",and log is from set. i Frog i i has a hop length of ai a i. Consider that this problem can be transferred into a shortest path problem which all edges' weights are 1. I created a graph in the form of a segment tree and ran a 0/1 BFS on it. bi () \(bfs\) . It jumps by the following algorithm: the first jump is aa units to the right, the second jump . Travel Frog (, Tabikaeru) is a free-to-play mobile game developed by Hit-Point Co.,Ltd that operates in Nagoya, Kyoto, released on 24 November 2017 for iOS and Android devices.. Notice that we can't just go with a for-loop from n to 0 and call it a day, because some dp[x] states can only be calculated from dp[y] (y < x) states. A problem in our school's private contest is similar with 1D/2F, even stronger, the difference is every climber has a weight and you need to maximize the sum of weight.There is a quite simple solution(but I wrote another in CF contest). a_i,b_i, a Length is N (1 \leq N N N \leq 2 &lowa Codeforces Round #751 D Frog Travelerdp, Codeforces Round #751 (Div. 2) 2022-06-13. n In other words, if Gorf is on $$$x$$$ meters below ground level, then in one jump he can go up on any integer distance from $$$0$$$ to $$$a_x$$$ meters inclusive. Just exactly like my style. a_i n Started serious coding for my intern project today. For the first part,we can sort it by the order of $$$s$$$. Get to the heart of le-de-France with one of our in-depth, award-winning guidebooks, covering maps, itineraries, and expert guidance. The difference is that suppose you are at depth 'i' now, you would visit i to i a[i]. When I calculates it by myself, it works right. I had no time to debug it during round. n i Frog Traveler - CodeForces 1601B - Virtual Judge Time limit 2000 ms Mem limit 524288 kB Source Codeforces Round 751 (Div. You can reverse the strings in it, so that you can put them together at the end but make sure that there are no same ones at th A. Tritonic Iridescence Topic link:Click to open the link The meaning of the question: Use CMY to fill the string, and the two adjacent ones must not be the same. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. The next $$$t$$$ lines describe test cases. so that's why we can't. No, it involves noticing that if you made a jump to some height, you should not jump to that height again. Most tricky part, is that it would be awesome if we could also insert, but it's a lot of work, because simplest workaround which I know is to implement dynamic balanced tree (treap or red-black tree or B-tree or Splay tree) with segment-tree like information. The only thing players have control over in the . Each character of $$$s$$$ is either 'L' or 'R'. Consider placing b_mid somewhere in the array A, particularly, the position that minimizes inversions between the elements of A and b_mid. Problem - 1850F - Codeforces However, it may turn out that it would be better if they used p_mid instead. Let's first switch the jump step. from a[i] to 1 and add a break statement whenever you find that a depth is visited. It's optimal for $$$b_i$$$, so $$$r_i>=s_i$$$, otherwise $$$p_j$$$ would be a better position for $$$b_i$$$. Editorials. https://codeforces.com/contest/1602/submission/133185501. Can someone pls suggest where I went wrong. It's easy to guess the result: just look at the parity of the total time. Now consider what happens when B is in place of b_mid. Then print the sequence $$$d_1,\,d_2,\,\ldots,\,d_k$$$ where $$$d_j$$$ is the depth Gorf'll reach after the $$$j$$$-th jump, but before he'll slip down during the break. add the climber(i 1) into set if his skill is >= a[i]. 3. #include b Problem B (Div. minn. Virtual contest is a way to take part in past contest, as close as possible to participation on time. Here is my solution to 1601C - Optimal Insertion. It's still hard to solve, but I gradually figured out the style of . 2), Codeforces Practice Tracker Browser Extension, Educational Codeforces Round 152 Editorial, Educational Codeforces Round 152 [Rated for Div. A. EhAb AnD gCd 1n-1 #include #include #include #include #include #include #include #include #include #include #define pb push_back #define lb lower_bound con. [0,a_i], b . For future readers, here's how I convinced myself. 2) Prove that at most $$$\log n$$$ steps after transformation, $$$a$$$ becomes repetitive. CodeForces | Frog Traveler - StopStalk 1602D - Frog Traveler CodeForces Solutions - pro-sequence Let's first switch the jump step. minn, a When a_i stops changing and becomes the minimum integer(least occurrences) at the same time, it's locked and never change. This allows us to sort original array $$$b$$$ and then maintain array $$$p$$$ in segment tree with operations: add on segment, and get position of maximum. She discovered that the layout of the city is grid-like, that is, there are n roads from east to west and m roads from south to north. Ok, so ideal solution picking can be visualised like this: Let us firstly just remove all climbers with skill < Initial Difficulty. If the optimal choice is p[mid] = p_mid and instead you would choose p_mid-1, some of the values in the left call might get worse because the max value for them would be p_mid-1. dp[n][m]011122223k (ans+k) int main() { std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); int ans=(1& vector22 #include #include #include #include #include #include #include #include #include #include # nn2n2n2aaa222bbbkkk bbb12nnnn1n-1n1 E. Cyclic Components -E. Cyclic Components nnnmmm 222cnt[] 0~n-2 B. i Now, the only part left to optimize is adding nodes into the queue. And Gorf can reach this height using a series of jumps $$$10 \Rightarrow 9 \dashrightarrow 9 \Rightarrow 4 \dashrightarrow 5$$$ where $$$\Rightarrow$$$ is the jump and $$$\dashrightarrow$$$ is slipping during breaks. In problem B div 1, you can do without a segment tree by storing a std::set of indices that bfs has not yet visited. i 908B - New Year and Buggy Bot. Hello, it was a great contest and nice editorial. It turns out that if we insert values in first best place as $$$p_i$$$, the answer will be optimal. Obviously the string a will only have one letter. ai,bi Why ? cin>>s; If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.