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Common BoardIf you have WA on test #3 or on test #1 or on test #2 you must use DP! Else if you have WA on test #4 you must use unsigned int or int 64 or long. If you have WA on test #8 you must change start and finish points ( 6-2 and 2-6 is equal ) !^_^! Thanks a lot for WA 8.I got AC.^^ Thank you. I got WA at #8.And I have ACed now; Thank you from reminding! Thank you for your suggestion Thank you very very much! thx a lot!! If you have WA8 and use Binary Search - check that it work properly, i.e. return far station instead of nearest Edited by author 03.12.2010 00:05 thank u very much for the 2nd suggestion. first i got wa4 and then wa8..your message help a lot! thank you. Thanks a lot for WA #8! It's really tricky. first i got wa14 and forgot to use long long.your message help a lot! thank you. l1=[] l2=[] l3=[] for i in range(0,1000): l3.append(False) def funct(a): b=a.split() if b[0]=='register': if b[1] not in l1: l1.append(b[1]) l2.append(b[2]) print("success: new user added") else: print("fail: user already exists") elif b[0]=="login": if b[1] not in l1: print("fail: no such user") else: k=l1.index(b[1]) if l3[k]: print("fail: already logged in") else: if l2[k]==b[2]: l3[k]=True print("success: user logged in") else: print("fail: incorrect password") elif b[0]=="logout": k=l1.index(b[1]) if l3[k]: l3[k]=False print("success: user logged out") else: print("fail: already logged out") f=int(input()) for m in range(0,f): n=input() funct(n) f**king time limit!!! List is O(n) lookup, while dict is O(log n), so use that using DP you can achieve O(n) как сделать ввод на питоне в этой задаче? how to make python input in this task? Use scanline and sort events by right border abd after that just count segments that their l > now so after this make now = r, ans++; firstly ans = 0 and now = 0; Sorry for my English i'm from Russia My AC program in Java has 396 lines =) And you?))) C++, 220 :) 143, C++ :) C++, 144 lines, 3026 characters (including comments, 15 empty lines, and the tracing utilities). pascal 196 o(∩_∩)o... It's not the shortest, but I am really proud of it! Edited by author 22.09.2009 21:13 Java - 330 :) Good architecture is the key for the shortest program and saving nerves.IMHO Edited by author 04.02.2010 21:26 That's funny. This is the first question that comes to mind after the AC. (300 PASCAL lines) Java 186 Edited by author 29.11.2020 17:38 Edited by author 29.11.2020 17:43 Python (74) is really good at these kinds of problems. p1 = Polynomial.parse(input()) p2 = Polynomial.parse(input()) print(p1*p2) please tell me the 16 test I have Wa #9 please give some tests Try to find minimum amount of peoples with complexity o(1) like that: if ((m * d1) % d2 == 0) { b = (m * d1) / d2; } else { b = (m * d1) / d2 + 1; } Edited by author 26.04.2010 19:58 i've changed for(int j=1; j<=i; ++j) cout<<floor(fr)<<' '; for(int j=i+1; j<=d2; ++j) cout<<ceil(fr)<<' '; to for(int j=1; j<=i; ++j) cout<<fixed<<setprecision(0)<<floor(fr)<<' '; for(int j=i+1; j<=d2; ++j) cout<<fixed<<setprecision(0)<<ceil(fr)<<' '; Edited by author 03.07.2021 11:15 in python: -5 % 7 == 2 in c++: -5 % 7 == -5 For God's sake, use ans/100 instead of ans*0.01f But i don't know why Maybe you forgot that the order of the numbers in the answer is important. you can implement your simple fraction class and use it instead of floating point numbers I don't know why, but in Python, code points of frame characters are: upper_left = 1066 upper_right = 1111 lower_left = 1040 lower_right = 1065 horizontal = 1044 vertical = 1110 That's really strange, because there are no common codecs encode those characters as such. Could Admin fix this or at lease give some explanations please? Also, don't put non-ascii characters in your source code, but use: c == chr(1066) or ord(c) == 1066 ord('┌') == 9484 list('┌'.encode('cp437')) == [218] list('┌'.encode('utf_16_le')) == [12, 37] # 12+37*256 == 9484 and Unicode code point is: U+250C == 9484 Any submission on Kotlin or Java return 'Restricted function' error. Even solutions that previously could pass at least some tests. Здравствуйте, решение из примеров для задачи "1000. A+B" object Main extends App { println(readLine().split(" ").map(_.toInt).sum) } выдает ошибку "Restricted function" Edited by author 23.06.2021 07:43 Edited by author 23.06.2021 07:43 Edited by author 23.06.2021 07:43 Remember: maximum accuracy can not be more than 10 in minus the secong degree. Запомните: максимальная точность числа не больше 10 в -2 степени. I have O(len(s) * k^2), and it is definitely too slow (TL#26). BTW, and here is some tests (don't want to create separate topic): 3 CC 00 1 01 ***** ans=-1 3 BAB 10 111 0 ***** ans=7 3 BA 00 01 1 ***** ans=3 3 CA 001 000 01 ***** ans=5 By applying suffix automaton, O(len(s) + k^2) complexity is received. Input: 3 '╣' like '[╣-I]' '╣' like '╣' '╣' like '_' Output: YES YES YES My AC solution gives NO to the first one; there should be no tests where the start character in a range is greater than the end character. |
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