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WA 3 | ~'Yamca`~ | 1940. Not So Simple Years | 13 Mar 2025 18:10 | 1 |
WA 3 ~'Yamca`~ 13 Mar 2025 18:10 Don't forget to add primes <= k |
cool problem, but overrated | ~'Yamca`~ | 2195. Binary Trees | 12 Mar 2025 20:16 | 1 |
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PLEASE_Python Time limit exceeded | Alice | 1196. History Exam | 11 Mar 2025 17:31 | 1 |
how can it be improved???? n = int(input()) sp = [] a = [] c = 0 for i in range(n): sp.append(int(input())) m = int(input()) for i in range(m): x = int(input()) if x in sp: c += 1 print(c) |
if a = 0 and b = 0?? | gooooooogol | 1420. Integer-Valued Complex Division | 11 Mar 2025 10:55 | 2 |
if a = 0 and b = 0 => r = 0 and q every complex number => count solation equation = infinity |
easy and overrated problem | ~'Yamca`~ | 1311. Stable Construction | 11 Mar 2025 00:59 | 1 |
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Is it possible to reset progress? | sm_programmer | | 10 Mar 2025 02:05 | 2 |
I would like to know if I can reset my progress here. I want to start afresh, but as I don't see a way to delete accounts, I thought that, at least, resetting my progress would be enough, but I can't do that either! Thanks in advance for your help! Edited by author 14.02.2025 08:40 You can request deleting your account by writing an email to support |
Strange statement | 🦄imosk72🦄[GTGU] | 2181. Student, or There and Back Again | 9 Mar 2025 16:50 | 1 |
> The student’s home is located at the coordinates (1, 1), and the university is at the coordinates (N, M). It means, for example, if N == 1 && M == 1 then university have same coordinates as home. Of course we can not use any roads in such case. But according to the sentence: > There are a total of N streets running from north to south and M streets running from west to east. there are some roads that can not be used. I think there should be N - 1 streets running from north to south and M - 1 streets running from west to east. |
correct code | Vlad | 1639. Chocolate 2 | 8 Mar 2025 12:26 | 3 |
int main(){ int x, y; cin >> x >> y; if ((x*y-1)%2==1) cout << "[:=[first]"; else cout << "[second]=:]"; return 0; } #include <iostream> using namespace std; int main () { int m,n; cin>>m>>n; if((m*n)%2==0)cout<<"[:=[first]"; else cout<<"[second]=:]"; return 0; } |
easy bfs | 👑TIMOFEY👑`~ | 1976. Game Optimization | 7 Mar 2025 12:34 | 1 |
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WA19 | 👑TIMOFEY👑`~ | 1703. Robotic Arm | 5 Mar 2025 16:43 | 1 |
WA19 👑TIMOFEY👑`~ 5 Mar 2025 16:43 |
WA 14 | ~'Yamca`~ | 1916. Titan Ruins: Waiting for Stability | 4 Mar 2025 16:52 | 1 |
WA 14 ~'Yamca`~ 4 Mar 2025 16:52 |
hint? | sailingoat | 2179. Need More Roads | 2 Mar 2025 19:06 | 1 |
hint? sailingoat 2 Mar 2025 19:06 can I have hint, at addflash@dmc.chat Please |
Please ! Need help..my algoritm is wright, but ! WA9 | gippotalamus | 1296. Hyperjump | 2 Mar 2025 01:13 | 5 |
> > i found my mistake Edited by author 20.11.2005 00:44 I have wa#9 but I can't found mistake :( what is test 9? I tried this and I get the correct answer, but test 9 code fails. Here is the code that fails test 9: from typing import List potentials: List[int] = [] N: int = int(input()) for p in range(N): potentials.append(int(input())) def list_max_right_sum(lst: List[int]) -> List[int]: amount: int = 0 left_min: int = 0 left_max_index: int = 0 k: int = 0 for i, p in enumerate(lst): amount += p k = i + 1 if amount < left_min: left_min = amount left_max_index = i + 1 return lst[left_max_index: k] result: List[int] = list_max_right_sum(potentials) result.reverse() result = list_max_right_sum(result) print(sum(result)) |
Я не понимаю что надо сделать? | eremeev.me.2012@gmail.com | 1336. Problem of Ben Betsalel | 26 Feb 2025 16:51 | 2 |
Что означает отношения квадрата и куба и что за число n дано число n, найти a и b где n = a^2/b^3 |
If you've got WA#13,here is why | PredaBoss | 1183. Brackets Sequence | 25 Feb 2025 08:48 | 3 |
Test 13 is an empty sequence :))) Thanks! Edited by author 25.02.2025 08:48 |
WA8 | 👑TIMOFEY👑`~ | 1996. Cipher Message 3 | 24 Feb 2025 19:59 | 1 |
WA8 👑TIMOFEY👑`~ 24 Feb 2025 19:59 |
hint | xurshid_n | 1819. Professional Approach | 21 Feb 2025 23:12 | 6 |
hint xurshid_n 19 Jul 2012 14:57 3-edge connected components of graph. A Simple 3-Edge Connected Component Algorithm by Yung H. Tsin. But I self do not found this article :) I have a stupid idea to solve this problem: first ,dfs a tree, then for each node compute the hash_value of edge set which cover this node.. for each query a,b if for every node on the path of (a,b) the set of edge cover by this node is not equal to any node outside the path of (a,b) and times of edge segments covered the path >=2 then result is Yes,otherwise it is no.. I'll try to use president tree to implement this idea... hope there is nothing wrong.. Edited by author 05.12.2016 14:35 Edited by author 05.12.2016 14:35 WA on test # 18,any one help?? AC 0.421s O(n*log(n)^2) its log(n)^2 because I use heavy_light decomposion+segment_tree ccz181078 has give a random solution to the sub problem of this problem:: give a tree with n node,n<=50000 for every edege there is a color 1<=c<=50000, give q<=50000 queries ,for each query give two node a,b judge for every color on path a-->b this color appear only on the path a--->b ,not outside path a--->b. sol :for every edge gives a random 32 bit number, so that for every col xor of edge with this color is zero.. for each query just judge if xor of random number on this path is zero. if it is ,answer is Yes, otherwise it is No You're wrong. The answer may be "Yes" but nodes not in one 3-edge connected component |
accepted python | stariy_bog | 1119. Metro | 20 Feb 2025 21:04 | 1 |
# works on pypy import sys n, m = sys.stdin.readline().split() n, m = int(n), int(m) k = int(sys.stdin.readline()) can_diag = {tuple(map(int, sys.stdin.readline().split())):0 for _ in range(k)} city_map = [[0]*(n+1) for _ in range(m+1)] for i in range(n+1): city_map[0][i] = i*100 for i in range(m+1): city_map[i][0] = i*100 for i in range(1, m+1): for j in range(1, n+1): if (j, i) in can_diag: city_map[i][j] = 100*(2**0.5)+city_map[i-1][j-1] else: city_map[i][j] = 100+min(city_map[i][j-1], city_map[i-1][j]) print(round(city_map[-1][-1])) |
AC long double | 👑TIMOFEY👑`~ | 2008. Swifty | 17 Feb 2025 20:00 | 1 |
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WA4 | 👑TIMOFEY👑`~ | 1936. Roshambo | 17 Feb 2025 12:18 | 1 |
WA4 👑TIMOFEY👑`~ 17 Feb 2025 12:18 i think i get this wa by precision, i try to use long double but this dont enough, maybe i have specific algorithm for this problem |