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Общий форумWA #5: 2 -2000 0 -1000 0 0 -1000 0 -2000 > 367.0796326795 WA #7: 2 935 973 -16 438 867 30 196 463 > 1040.9330224527 WA #13: 2 0 7 0 8 6 10 6 1 > 12.7006778631 What can cause a "wrong answer" in 7-th test? I used sqrt_decompostion and always get WA7. what is tricky test 7? I found my bug, I renum all vertexes, but used old indexes :) (renum all child of vertex to as id....id+k, sequencial indexes). Edited by author 15.02.2012 18:35 Since description says p[i]<i it was tempting to assume that initial sequence is already in DFS order, but it's not necessarily so. Got RE7 when explicitly checked for this. The total length of all strings (including the first one with the number N) exceeds the specified size of 4000000 bytes. Hi, Kindly add support for ocaml to execute timus problems. Thanks. 1) write in the assignment that family ties and victims cannot be repeated in the input, or add such tests. this is a real problem when searching for maniacs in databases. 2) write in the assignment that a werewolf cannot kill himself. #include <cstdio> int main() { freopen("output.txt", "w", stdout); printf("99999\n"); printf("0 0 10 100\n"); for(int i =0 ; i < 99998; ++i) { printf("%d %d %d %d\n", (i+1)*10,0,(i+2)*10 , 100); } return 0; } My early solution get AC, but must get TL. Edited by author 19.01.2011 16:05 Also this test 4 0 0 2 6 2 2 4 4 4 0 6 6 6 2 8 4 My AC solution gives 12 when correct answer is 8 What is the answer for 2 0 0 3 3 3 2 5 7 Answer is -1? - What?! Why? ^ | __ | | | | | | | | | |___| | Such input gives that picture, so why there is no path?! | | | | |__| |___| ----------- > Your picture has 3,1 for lower-left corner of second rectangle It's just brute-force task :) There 36 lines in my AC code versus yours 165 lines :) Edited by author 08.03.2013 19:59 thanks, now AC with set :/ It can actually be solved with trie. My solution uses 0.4s (cin) and 20MB. Just store a map in every node. Solved with C++ with tree of static arrays of links, without any STL structures Same here. 0.062 sec, 1.8Mb 5 3 1 2 2 2 2 2 3 4 6 ans: 11 5 3 3 1 2 3 4 5 1 4 7 ans: 12 9 9 123 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ans: 9 Why does the last test have an answer 9? on 9th second he can't pick up the last stone, instead he drops all of them, so isn't the answer 27? I have solved this problem by Fenvic tree, at first I got WA on 5 test, then I changed int-s into long long-s and now I have WA on 6 test. If you had this kind of experience please tell me what types should I use to get AC? Thanks. The test 6 aint about data types' ranges. იდეაში, ეგ ხო ფენვიკის ხით უნდა გავიდეს, მე ვთვლი რაოდენობას, რამდენი გაყიდვა შესრულდება და შემდეგ ვამრავლებ 0.01–ზე და ვაბეჭდინებ. და ეგრე უნდა მუშაობდეს წესით ხო? Try this test BID 0.01 SALE 0.02 10 Answer:0.00 How I can do it on 0.001? Edited by author 31.10.2008 02:29 Edited by author 16.06.2009 00:42 Mine is O(N^2) How do you do ? I track w0[i][j] as length of longest horizontal stride of a[i][j]==0 starting at i,j (0 if a[i][j]==1), symmetric for w1[i][j]. Now w1[i][j] defines the length to be tested (as 2*w[i][j]+1), this is done by descending downwards. At first it seems to be O(N^3) but due to nature of the pattern you won't descend all that frequently, so I believe it's O(N^2), and it gave 0.001 AC in C++ You should find maximum matrix which is square, black and also contains white square inside rotated by 45 degrees. For instance: 1) 1 1 1 0 1 1 1 1 1 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 1 1 1 2) 1 1 1 1 0 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 0 1 1 1 1 are desired matrices maximum width of which you must find 1st matrix is invalid (pre-last row) My solution runs in O(N logN) and it uses heap data structure. The discussion forums hint that there may be a O(N) time solution. Is there a linear time algorithm? heap?:D just use lower_bound And, BTW, Yes there is. At first I used Binary Search on each station but on the next station you can start with the previous index. code: int canReach1 = A[i] + L1; while (ind1 <= end && A[ind1] <= canReach1) ind1 ++; i've used dp (which is easy to notice) with some optimization. suppose we have three stations s1 s2 s3 and there exists path s1-s2 covered with l1, and path s1-s3 also covered with l1. then we can skip the analysis from s2. we can easily reduce used time if we create list of "allowed" stations to analyze (like s3) You can keep deque (FIFO) of pairs (max_x,total_price) if using C1 on last trip, same for C2, same for C3. All three deques will be non-descending on price (and naturally ascending on coordinate). At each step pick minimum from non-empty deques (their front element) and push new advances to their end. Edited by author 11.06.2026 06:16 What means if gear-wheel have only one cog? Edited by author 26.09.2004 20:25 In this test have some gear not connect from other Try this test 2 1 0 1 0 1 6 answer 6/1 0/1 Edited by author 18.04.2005 12:50 There is simple O(N+M) algo (core part is like 5 lines), and it's "online" (no preprocessing of queries required) Edited by author 10.06.2026 04:20 Notice, that you can move duons alongside diagonal (A->F, A->H, G->E, ...) Set up your own camera numbering. Change the C-D and H-G numbers. Who know O(H*W) or, O(H*W*5) algo ??? You can dynamically update amount of branches per-type per-distance for each horizontal step leading to O(W*H*9) Edited by author 09.06.2026 11:10 It is much more interesting to look for good tests for this task than just to solve it. Here is the AC program (with a small error in order not to publish solutions). I will look for where it will fail. #include<iostream> #include<cmath>//too lazy to write sqrt() yourself int main() { unsigned l, h, o; std::cin>>l>>h>>o; unsigned phase = 0; if(l<h*2) phase = o*sqrt((2*h-l)/981)/15+1; char const*answer[]{"Butter","Bread\n"}; std::cout<<answer[(phase>>1)&1]<<std::endl; } Tree for minimum is enough. It's O(N) problem with O(1) memory (apart from string itself) Input 0 50 10 0 0 100 200 -1 Output -5.00 I thought at first, that an absolute value has to be printed, which gets WA 3. |
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