Common Board| Show all threads Hide all threads Show all messages Hide all messages | | To admins: Why 2 seconds limit? I think it's too much. (-) | Yuri K | 1066. Garland | 12 May 2026 10:26 | 1 | | | best tests | Dmi3Molodov | 1167. Bicolored Horses | 11 May 2026 18:33 | 1 | 131071 65536 0 1 0 1 0 1 ... 0 1 0 1 0 answer: 65535 (long time) 131071 65535 0 1 0 1 0 1 ... 0 1 0 1 0 answer: 65536 (long time) 432 234 0 1 0 1 0 1 ... 0 1 0 1 answer: 198 | | F(21)=138879579704209680000 ? | Dmi3Molodov | 1402. Cocktails | 9 May 2026 04:01 | 1 | | | To Admins: Weak tests | Dmi3Molodov | 1123. Salary | 8 May 2026 14:24 | 1 | I've got AC with program that returns 01 for input 01. | | Problem 1394 "Ships. Version 2" rejudged, number of tests limited | Vladimir Yakovlev (USU) | 1394. Ships. Version 2 | 8 May 2026 08:21 | 8 | The test data for this problem has been reworked. New tricky tests have been added. At the same time the majority of the old tests have been removed. From now on, the total number of tests in this problem is limited to 20 tests. It means if new good tests are discovered, they couldn’t be simply added but should replace some of the old ones. If you’re stuck, don’t spend time on figuring out the test data and hyperoptimizing, because the test could be replaced with a similar one any time. Instead, detect that your heuristics fail and launch another one in this case. There is no universal test in the test data that makes all the heuristics fail at the same time. All solutions have been rejudged. Only a few authors still have AC after the rejudge. Because of test removal some of the previously rejected solutions got accepted. Edited by author 03.05.2026 22:50 The test data for this problem has been reworked. New tricky tests have been added. Кто теперь создавал новые тесты? Есть публикация? Старые в рамках этой работы https://is.ifmo.ru/disser/buzdalov-dissertation.pdf были сделаны, насколько я понимаю. Edited by author 04.05.2026 18:18Частично мои тесты. Публикации нет. Доступа к чужим решениям у меня не было, то есть тесты достаточно общные. LLM при создании тестов не применялось. Под новыми тестами (и новым вашим быстрым решением) лежат какие-то новые, ранее неизвестные теоретические основания или просто переосмысление на базе нового опыта/свежий взгляд? Честно говоря сложно ответить, идея этих тестов появилась несколько месяцев назад, когда я решил улучшить свое старое решение, то есть в целом, это свежий взгляд. Но и теоретическое обоснование у них скорее есть чем нет. У меня из области поиска вырезаются огромные лакуны - в решении очень большое число магических констант, которые, впрочем, являлись оптимизацией для старых тестов, без них тоже проходило по времени. Ожидал, что если будут подбирать новые тесты, то будет TLE почти наверняка. Но раз без доступа к решениям тесты генерировались, то понятно, почему нет TLE с новыми тестами. Edited by author 07.05.2026 19:16 Edited by author 08.05.2026 08:21 Edited by author 08.05.2026 08:21 Могли бы Вы вложить тесты в открытый доступ или отправить мне на мою почту (107th at mail ru) ? Частично мои тесты. Публикации нет. Доступа к чужим решениям у меня не было, то есть тесты достаточно общные. LLM при создании тестов не применялось. | | this test help me | ~'Yamca`~ | 2168. Runes in the Field | 6 May 2026 05:10 | 1 | WA8: 6 aabcaa ans: 6 RE9: 7 aaaaaaa ans: 7 | | To admins | andreyDagger`~ | 2222. Cavalry Battle Advanced | 5 May 2026 20:47 | 2 | The letter С in the name of problem is russian, seems like this is not supposed to be russian) Re: To admins 🎧 Vadim Barinov \Frez_Fstilus/'``' :) 5 May 2026 20:47 Yeah, thanks) I fixed it. Chess Battle Advanced | | Subfactorial is the word :) | Yuri K | 1366. Presents | 4 May 2026 12:36 | 1 | | | Problem statement clarification | 👨🏻💻 Spatarel Dan Constantin | 2165. Cake Cutter | 3 May 2026 07:52 | 1 | For each possible triangulation, compute the smallest area triangle in it. Print the largest smallest area triangle. If that does not make much sense, let's go over the sample input: The given quadrilateral can be triangulated in only two different ways. The areas of the resulting triangles are: Cut #1: 11.5 + 16.0 (min. area triangle = 11.5) Cut #2: 10.0 + 17.5 (min. area triangle = 10.0) answer = max(11.5, 10.0) = 11.5 Good luck! | | What is test 5 ? | pizdec | 1808. Chapaev at the Planet Ocean | 3 May 2026 05:50 | 2 | What answer on test 10 2.0 5 10 0 -1 5 -10 0 1 ? I think answer "Yes" Give me some tests Edited by author 20.07.2018 19:14 My AC solution says the answer is: Yes | | To admins | Yuri K | 1600. Airport | 3 May 2026 00:36 | 1 | Your C compiler doesn't understand correctly formats like "%Lf" for "long double" data type. I have to declare "double" variables for print only and use "%lf" specifications. | | To admins | 👨🏻💻 Spatarel Dan Constantin | 1983. Nectar Gathering | 2 May 2026 09:13 | 2 | To admins 👨🏻💻 Spatarel Dan Constantin 5 Apr 2026 15:01 I believe there might be some precision issues with the tests. Here is why: My AC solution (ID = 11182582) has a parameter STEPS set to 1000. As I set it higher, the accuracy increases. On this input: 54 46 -161 -82 -150 -33 100 127 -175 My AC program gets: 48.217529 But when I set STEPS to 10000, the improved AC program gets: 38.806852 In addition, I have another program (ID 11182578 - WA #22) outputting: 37.855475851615 I believe test #22 is wrong (as well as my AC solution) while my WA #22 solution is in fact correct. To strenghten that conviction, when I set STEPS to 30000, the improved AC program gets: 38.191302 Thus, the two answers are within a relative error of less than 10^-2. Please look into this issue. Thank you! After further investigation, I was able to get to the bottom of things: (1) My (ID 11182578 - WA #22) solution was indeed wrong and I was able to fix it (ID 11197306 - AC). (2) My (ID 11182582 - AC) solution continues to give a completly different result on the above mentioned test. Conclusions: (1) To pass WA #22, I had to handle cases where the origin (0, 0, 0), when projected on the plane containing the triangle, would fall on one of the vertices or on one of the edges of the triangle. (2) Since both of my AC solutions give widely different results on the above mentioned test it means the tests are weak and at least the above test should be added to the testset. The correct answer is the one returned by my (ID 11182578 - WA #22) and (ID 11197306 - AC) submissions. For those who look for tests: Input #1: 10 0 0 10 100 0 10 0 100 Output #1: 4999.748318598260 Input #2: 54 46 -161 -82 -150 -33 100 127 -175 Output #2: 37.855475851615 This is a very nice problem! I learned a lot from solving it. Good luck in your attempt! | | WA #21 | 👨🏻💻 Spatarel Dan Constantin | 1412. Autumn Tide | 2 May 2026 08:45 | 1 | WA #21 👨🏻💻 Spatarel Dan Constantin 2 May 2026 08:45 Input: 10 10 1 5 5 Output: 100 | | overrated | ~'Yamca`~ | 1771. A Deputy's Morning | 29 Apr 2026 02:23 | 1 | Simple problem without any algo and math | | easy bfs | 👑TIMOFEY👑`~ | 1976. Game Optimization | 28 Apr 2026 19:32 | 2 | | | Checker failed #19 | 👨🏻💻 Spatarel Dan Constantin | 2213. A Small Fix | 19 Apr 2026 21:58 | 3 | Hi! I got this verdict: Checker failed #19 Can someone please look into this and fix it? Thank you! | | Which algo to use? | T_be_GP | 1002. Phone Numbers | 19 Apr 2026 00:05 | 4 | Look for Word Break Problems | | Test 11 | mouse_wireless2 | 1621. Definite Integral | 18 Apr 2026 21:29 | 2 | Test 11 mouse_wireless2 15 Jan 2018 20:09 If you're struggling with test 11, the test is: 1 0 1000000 -2000 1 Exact (up to 1e-14) answer is 0.00628318530714 If your solution passes this, it will most likely pass all tests :) Thank you! This test helped me a lot. And I've got AC. (I used 'decimal' with 100 digits in Python and Newton-Cotes method with n=8.) | | RE #10 | Yuri K | 1291. Gear-wheels | 14 Apr 2026 21:12 | 1 | RE #10 Yuri K 14 Apr 2026 21:12 I have RE for test # 10. Don't understand why. | | Instruction how to solve. | Mahilewets | 2099. Space Invader | 14 Apr 2026 01:02 | 6 | Check (AB,CD)==0 (orthogonality). Check (AB, CD, DA) ==0 (planarity). Check AD>AC>AB, AC>BC, BD>BC (order). Check whether the projections to XY, YZ, XZ craddle each other continuations. It is sufficient to check only the projections to XY plane to get Accepted verdict. Sounds too complex. If we replace C, D with their orthogonal projection on AB, then all steps except first collapse to only one step (check D = C*a, a > 1). Still too complex, though. Over 20 lines in Python. There should be more simple solution... Edited by author 26.12.2017 04:30 No lengths or projections XY, etc. Using scalar product (sp) and triple product (tp) it is at most five conditions: sp(ab,cd) == 0 and tp(ab,bc,cd) == 0 and # ⟂ and planar sp(ab,bc) >= 0 and # C after B sp(cd,bc) >= 0 and # BC goes in direction of CD sp(cd,bd) >= sp(cd,bc) # D after C And you probably made a mistake. Your 3rd step would make a mistake if intercection is incide AB, incide CD. I am now stuck with 19test and I used some real bulletproof solution, without using float at all. ---------- I do 3rd step buy solving 3x3 matrix and finding t and z, just instead of dividing i only check +or- for i in range(3): if mtrx[i][1]!=0: if ( mtrx[i][0]>=0 and mtrx[i][1]>0 )or( mtrx[i][0]<=0 and mtrx[i][1]<0 ): t=True else: t=False for i in range(3): if mtrx[i][2]!=0: if ( mtrx[i][0]>=0 and mtrx[i][2]>0 )or( mtrx[i][0]<=0 and mtrx[i][2]<0 ): z=True else: z=False return t and z ------------ I start to believe, that it is incorrect test result not my soultion Edited by author 13.04.2026 01:15 Edited by author 13.04.2026 01:15 Edited by author 13.04.2026 01:15 Edited by author 13.04.2026 01:18 Edited by author 13.04.2026 01:50 counterexample 4 0 0 -1 0 0 0 4 0 0 -6 0 It is invalid, but your your order is OK #Finally did it. Now i see why indie games are so bugged def Vect3d(start,end): #int to int return (end[0]-start[0],end[1]-start[1],end[2]-start[2]) def multSc(vecta,vectb):#int to int rslt=0 for i in (0,1,2): rslt+=vecta[i]*vectb[i] return rslt def multVct(va,vb): #int to int x=va[1]*vb[2]-va[2]*vb[1] y=va[2]*vb[0]-va[0]*vb[2] z=va[0]*vb[1]-va[1]*vb[0] return (x,y,z) #vc=t*va+z*vb def TZbool(va,vb,vc): #always int and bool, no division mtrx=[[0,0,0],[0,0,0],[0,0,0]] for i in (0,1,2): mtrx[i][0]=va[i] mtrx[i][1]=vb[i] mtrx[i][2]=vc[i] c1=0 r1=0 while c1<2: if mtrx[r1][c1]!=0: for r2 in (0,1,2): if r2!=r1: for c2 in (0,1,2): if c2!=c1: mtrx[r2][c2]=mtrx[r2][c2]*mtrx[r1][c1]-mtrx[r1][c2]*mtrx[r2][c1] mtrx[r2][c1]=0 c1+=1 r1=(r1+1)%3 ######################### t=False #t>=0 for i in (0,1,2): #only one mtrx[i][0]!=0 in solved 3x2 mtrx if (mtrx[i][0]>0 and mtrx[i][2]>=0) or (mtrx[i][0]<0 and mtrx[i][2]<=0): t=True z=False #z>=0 for i in (0,1,2): if (mtrx[i][1]>0 and mtrx[i][2]>=0) or (mtrx[i][1]<0 and mtrx[i][2]<=0): z=True return t and z A=tuple(map(int,input().split())) B=tuple(map(int,input().split())) C=tuple(map(int,input().split())) D=tuple(map(int,input().split())) goNext=True if goNext: #check if turning 90 AB=Vect3d(A,B) DC=Vect3d(D,C) if multSc(AB,DC)!=0: goNext=False print('Invalid') if goNext: #check if AB and CD intersect AD=Vect3d(A,D) if multSc(AD,multVct(AB,DC))!=0: goNext=False print('Invalid') if goNext: BC=Vect3d(B,C) CD=Vect3d(C,D) # -DC if TZbool(AB,CD,BC): print('Valid') else: print('Invalid') Edited by author 14.04.2026 01:12 |
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