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Общий форумarun ac in one go // Задача 1016. Кубик на прогулке 4 июн 2016 10:38 alot of time in debugging though ! :) Ivan Metelev UFU SESC Give me some tests please [4] // Задача 2010. Юный гроссмейстер Саша 28 окт 2014 07:14 I have WA#16, give me some tests please Edited by author 09.11.2014 18:38 Ivan Metelev No subject [3] // Задача 2010. Юный гроссмейстер Саша 9 ноя 2014 17:56 Edited by author 09.11.2014 18:35 Edited by author 09.11.2014 18:35 Ivan Metelev Re: No subject [2] // Задача 2010. Юный гроссмейстер Саша 1 июн 2016 11:12 I can poste my bad code, can someone help me? var a, b, n, x, y: int64; begin readln(n, x, y); write('King: '); if n = 1 then writeln(0) else if ((x = 1) and (y = 1)) or ((x = n) and (y = 1)) or ((x = 1) and (y = n)) or ((x = n) and (y = n)) then writeln(3) else if (x = 1) or (y = 1) or (x = n) or (y = n) then writeln(5) else writeln(8); write('Knight: '); if (n = 1) or (n = 2) or ((n = 3) and (x = 2) and (y = 2)) then writeln(0) else if (((y = 1) and (x = 1)) or ((y = n) and (x = n)) or ((y = 1) and (x = n)) or ((y = n) and (x = 1))) or (n = 3) and ((x = 2) or (y = 2)) then writeln(2) else if ((x = 2) and ((y = 1) or (y = n))) or ((x = n - 1) and ((y = 1) or (y = n)) or ((y = 2) and ((x = 1) or (x = n))) or ((Y = N - 1) and ((x = 1) or (x = n)))) then writeln(3) else if (((x >= 3) and (x <= n - 2)) and ((y = 1) or (y = n))) or (((y >= 3) and (y <= n - 2)) and ((x = 1) or (x = n))) or ((x = 2) and ((y = 2) or (y = n - 1)) or ((x = n - 1) and ((y = 2) or (y = n - 1)))) then writeln(4) else if (((x >= 3) and (x <= n - 2)) and ((y = 2) or (y = n - 1))) or (((y >= 3) and (y <= n - 2)) and ((x = 1) or (x = n))) then writeln(6) else writeln(8); if n = 1 then a := 0 else if (n - x >= x) and (n - x >= n - y) and (n - x >= y) then a := (n - 1) + 2 * (x - 1) else if (x >= n - x) and (x >= n - y) and (x >= y) then a := (n - 1) + 2 * (n - x) else if (y >= x) and (y >= n - y) and (y >= n - x) then a := (n - 1) + 2 * (n - y) else if (n - y >= x) and (n - y >= n - x) and (n - y >= y) then a := (n - 1) + 2 * (y - 1); if n = 1 then b := 0 else b := (n - 1) * 2; writeln('Bishop: ', a); writeln('Rook: ', b); writeln('Queen: ', a + b); end. Edited by author 01.06.2016 11:32 d_m Re: No subject [1] // Задача 2010. Юный гроссмейстер Саша 2 июн 2016 14:44 I don't know where is problem in your code. But you can insert in the beginning something like that: if x > (n + 1) div 2 then x := n + 1 - x; if y > (n + 1) div 2 then y := n + 1 - y; and remove all comparisons with n. It will make your code much simpler. Jane Soboleva (SumNU) Re: No subject // Задача 2010. Юный гроссмейстер Саша 2 июн 2016 17:15 A much more simple approach would be creating something like function InBounds(x0, y0, xshift, yshift: longint): boolean; begin inc(x0, xshift); inc(y0, yshift); InBounds:=((x0 >= 1) and (x0 <= N) and (y0 >= 1) and (y0 <= N)); end; This will allow to check king and knight in a much more simple and elegant way. King: res:=-1; for i:=-1 to 1 do for j:=-1 to 1 do if InBounds(x0, y0, i, j) then inc(res); writeln('King: ', res); Knight: res:=0; for i:=1 to 2 do //length before turn for j:=0 to 1 do //sign 1 for k:=0 to 1 do begin //sign 2 if InBounds(x0, y0, (j + j - 1) * i, (k + k - 1) * (3 - i)) then inc(res); end; writeln('Knight: ', res); For rook, it's always N + N - 2, even for N = 1, so there was no need to specifically bring up that case... Queen is rook + bishop, and bishop i'd say is the "hardest" part here, but well, you just take the minimum of squares he can move to in each 4 directions, just be attentive... Good luck! SabraO Why do I get Runtime Error for solution in Python 3.4 [1] // Задача 1001. Обратный корень 1 июн 2016 19:15 import sys; import math; class Stack: def __init__(self): self.items = [] def isEmpty(self): return self.items == [] def push(self, item): self.items.append(item) def pop(self): return self.items.pop() def peek(self): return self.items[len(self.items) - 1] def size(self): return len(self.items) size = 0 intSize = sys.getsizeof(int()); s = Stack(); while (size<100): line = input(); if (not line.isspace()): values = [int(s) for s in line.split() if s.isdigit()]; size += intSize; for value in values: s.push(math.sqrt(value)); while (s.size()>0): print("%.4f" % (s.pop())); ToadMonster Re: Why do I get Runtime Error for solution in Python 3.4 // Задача 1001. Обратный корень 2 июн 2016 12:46 > while (size<100): How did you receive magic number "100"? Shouldn't you read input until EOF? al13n How to implement this with Voronoi diagram [1] // Задача 1369. Тараканьи бега 28 фев 2012 00:18 Ok, I understand Fortune's algorithm for building Voronoi diagrams. But how to implement answering the nearest point queries? Can it be done without explicitly building full-featured Voronoi diagram and then using another sweeping line on it? I guess, the queries can be answered during building the diagram, without actually building the diagram. But how to generate events "a query point meets parabolic front"? Is it even possible in O(log n)? Orient Re: How to implement this with Voronoi diagram // Задача 1369. Тараканьи бега 31 май 2016 20:45 Edited by author 10.11.2016 18:14 Юрий Что не так?! C# [1] // Задача 1068. Сумма 31 май 2016 18:54 using System; namespace Consoleapp { class Program { static void Main(string[] args) { int sum = 0; int len = Convert.ToInt32(Console.ReadLine()); for (int i = 2; i <= len; i++) { sum = sum + i; } Console.WriteLine(sum); } } } Oleg Baskakov Re: Что не так?! C# // Задача 1068. Сумма 31 май 2016 19:00 Absolute value not exceeding 10000 means the number can be in range -10000..10000. lxn Precision // Задача 1369. Тараканьи бега 31 май 2016 18:16 I have figured out that the 6th test contains numbers with 8 digits after the decimal point. Is it the maximum alowed or are there tests with more than 8 digits after decimal point. What tollerance shouled be used to compare distances between points? It it neccessary to do it with no error or is there an accepted tollerane? I have generated several test cases with only 6 digits after decimal point, and doubles doesn't allow to resolve this tests correctly. If there are 8 digits after decimal point and 4 before, it is necessary to store about 24 digits. 1) is is necessary to compare distances between points with no error? 2) What is maximum alowed number of digits after decimal point in input? test cases: 2 999.969732 999.984915 1414.181493 0 1 0 0 >>>>> (999.969732 * 999.969732 + 999.984915 * 999.984915 == 1414.181493 * 1414.181493 == 1999909.295143709049) 2 999.856428 999.897768 1414.039753 0 1 0 0 >>>>> (999.856428 * 999.856428 + 999.897768 * 999.897768 == 1999508.423064301008 < (1414.039753 * 1414.039753 == 1999508.423064301009)) 2 999.725264 999.971561 1413.999196 0 1 0 0 >>>>> (999.725264 * 999.725264 + 999.971561 * 999.971561 == 1999393.726288646417 > (1413.999196 * 1413.999196 == 1999393.726288646416)) Edited by author 31.05.2016 18:21 GastonFontenla TEST CASE // Задача 1208. Соревнование легендарных команд 30 май 2016 06:01 Here's the big test case: 18 a b c d e f g h i j k l m n o p q r s t u v w x y z 1 2 3 4 5 6 7 8 9 0 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 If you are using bitmask be careful with: *Use long long to store the bitmask. There can be 58 contestant (18*3) as showed above *Be careful with casting in C/C++. I was using the following: #define ll long long void SB(ll &m, ll x) { m = ((m) | (1 <<(x))); } But the (1 << (x)) part auto-cast to int. Hence the overflow The correct should be: #define ll long long void SB(ll &m, ll x) { m = ((m) | (1LL <<(x))); } Hope it help! Edited by author 30.05.2016 06:02 DSamokhina Where is error? test 17 // Задача 1820. Уральские бифштексы 29 май 2016 23:59 import java.io.*; import java.util.*; public class t1820 { public static void main(String[] args){ Scanner in = new Scanner(System.in); PrintWriter out = new PrintWriter(System.out); int n = in.nextInt(); int k = in.nextInt(); int rez = 2; if (n>k){ if (n % k == 0) { rez = 2*n; } else{ if (n % k > k / 2) { rez = (n/k+1)*2; } else { rez = (n/k-1)*2+3; } } } out.println(rez); out.flush(); } } rkhapov OLE on test #4 [3] // Задача 1438. Time Limit Exceeded 29 май 2016 18:15 Hi I have OLE on #4 test Please, give me some test P.S. Sorry for my english :( rkhapov Re: OLE on test #4 [2] // Задача 1438. Time Limit Exceeded 29 май 2016 18:26 i also had tried forward and backward labels, i think my program work correct Jane Soboleva (SumNU) Re: OLE on test #4 [1] // Задача 1438. Time Limit Exceeded 29 май 2016 22:05 I had OLE #4, and it turned out to be a minor slip in my code. That is, for commands of type «if smth1 <operator> smth2 goto label, i was searching smth1 and smth2 indices in my variable indices array; label i should have searched in my goto indices array, but i accidentally was searching a label index in variable array too; as a result, it returned 0 (which it returned in case if the needed element was the first one, and if the element wasn't found, which shouldn't be happening). So it jumped to the first label and continued from there. Example: label1: A = 10 print 100 label2: goto label2 Due to a mistake, "label2" has been looked for in variables array, which only had "A" in there. It was not found, it returned 0, it jumped to label1, tons of print commands were executed. After mistake is fixed, "label2" is looked for inside labels array, and correctly returns 1. I can't say for sure you have the very same mistake, but still it is likely just a minor slip like this one. rkhapov Re: OLE on test #4 // Задача 1438. Time Limit Exceeded 29 май 2016 23:01 I use std::map<std::string, int> for variables and labels. On your example my program writes: 100 Program terminated. Variables state: A: 10 I think, its correct. I had tested many similar program and on each of them program's output was correct. My code: https://gist.github.com/rkhapov/a8dd783fe154813fc2f76649a6f7d9ec Litvinov problem in Test6 // Задача 1014. Произведение цифр 28 май 2016 19:49 if is_test_script: print True if begin_script(10) == 25 else False print True if begin_script(9) == 9 else False print True if begin_script(150) == 556 else False print True if begin_script(11) == -1 else False print True if begin_script(10000000000) == -1 else False print True if begin_script(1000000000) == 555555555888 else False print True if begin_script(167) == -1 else False print True if begin_script(24) == 38 else False print True if begin_script(180) == 566 else False print True if begin_script(360) == 589 else False print True if begin_script(0) == 10 else False print True if begin_script('test') == -1 else False print True if begin_script(1013) == -1 else False print True if begin_script(27) == 39 else False print True if begin_script(1) == 1 else False print True if begin_script(0001) == 1 else False print True if begin_script(45) == 59 else False print True if begin_script(40) == 58 else False print True if begin_script(22) == -1 else False that tests are true Edited by author 29.05.2016 05:08 Edited by author 29.05.2016 05:09 shkodlaildar i think you need to add anti-hash tests [2] // Задача 1989. Подпалиндромы 16 май 2016 12:38 Hello admins, please add anti-hash tests for this task. Alexey Dergunov [RIP Samara SAU] Re: i think you need to add anti-hash tests // Задача 1989. Подпалиндромы 17 май 2016 12:20 How to solve it without hashes? transistor Re: i think you need to add anti-hash tests // Задача 1989. Подпалиндромы 27 май 2016 17:04 Actually, u can write double-hashes and easy pass these anti-hash tests. umid5566 Test 1 is wrong. [1] // Задача 1563. Баяны 28 мар 2016 22:55 Takanashi Rikka Re: Test 1 is wrong. // Задача 1563. Баяны 27 май 2016 15:11 why? Soporboy Botirov [TUIT Urgench] This is some Tests to WA#8 and WA#9 [5] // Задача 1892. Утро в Кольцово 16 ноя 2011 17:33 WA#:8 4 7 1 10 4 00:00:00 00:00:01 00:00:03 00:00:19 4 00:00:02 00:00:04 00:00:19 00:00:20 ans: 00:00:00 00:00:18 00:00:36 00:00:40 00:00:07 00:00:17 00:00:25 00:00:35 WA#9 1 2 3 1 5 0:0:1 0:0:5 0:0:7 0:0:8 0:0:5 5 0:0:2 0:0:2 0:0:5 0:0:7 0:0:10 ans: 00:00:01 00:00:13 00:00:14 00:00:15 00:00:16 00:00:03 00:00:04 00:00:05 00:00:07 00:00:10 WA#9 1 2 3 1 5 0:0:5 0:0:7 0:0:8 0:0:14 0:0:17 5 0:0:1 0:0:3 0:0:5 0:0:10 0:0:15 ans: 00:00:08 00:00:09 00:00:14 00:00:19 00:00:20 00:00:01 00:00:03 00:00:05 00:00:11 00:00:16 This answers from My ACC program LuckyI Re: This is some Tests to WA#8 and WA#9 [1] // Задача 1892. Утро в Кольцово 22 дек 2011 14:57 It's seems your answer on first test is wrong. Point is that my AC program answer is 00:00:00 00:00:18 00:00:38 00:00:42 00:00:07 00:00:17 00:00:27 00:00:37 Otrebus Re: This is some Tests to WA#8 and WA#9 // Задача 1892. Утро в Кольцово 27 май 2016 03:26 Contrary to what Luckyl said, I get the same result as the original poster on the first test in my accepted program. alp Re: This is some Tests to WA#8 and WA#9 [2] // Задача 1892. Утро в Кольцово 24 янв 2012 22:35 Your test 2 inconsistentб because "The times in each list are pairwise different and are given in ascending order." 1 2 3 1 5 00:00:01 00:00:05 00:00:07 00:00:08 00:00:09 5 00:00:02 00:00:03 00:00:05 00:00:07 00:00:10 Answer 00:00:01 00:00:13 00:00:14 00:00:15 00:00:16 00:00:03 00:00:04 00:00:05 00:00:07 00:00:10 Edited by author 24.01.2012 22:39 falicos Re: This is some Tests to WA#8 and WA#9 // Задача 1892. Утро в Кольцово 8 июл 2013 13:48 this test help me to pass WA8 input 500 500 500 500 50 09:59:11 09:59:12 09:59:13 09:59:14 09:59:15 09:59:16 09:59:17 09:59:18 09:59:19 09:59:20 09:59:21 09:59:22 09:59:23 09:59:24 09:59:25 09:59:26 09:59:27 09:59:28 09:59:29 09:59:30 09:59:31 09:59:32 09:59:33 09:59:34 09:59:35 09:59:36 09:59:37 09:59:38 09:59:39 09:59:40 09:59:41 09:59:42 09:59:43 09:59:44 09:59:45 09:59:46 09:59:47 09:59:48 09:59:49 09:59:50 09:59:51 09:59:52 09:59:53 09:59:54 09:59:55 09:59:56 09:59:57 09:59:58 09:59:59 10:00:00 50 09:59:11 09:59:12 09:59:13 09:59:14 09:59:15 09:59:16 09:59:17 09:59:18 09:59:19 09:59:20 09:59:21 09:59:22 09:59:23 09:59:24 09:59:25 09:59:26 09:59:27 09:59:28 09:59:29 09:59:30 09:59:31 09:59:32 09:59:33 09:59:34 09:59:35 09:59:36 09:59:37 09:59:38 09:59:39 09:59:40 09:59:41 09:59:42 09:59:43 09:59:44 09:59:45 09:59:46 09:59:47 09:59:48 09:59:49 09:59:50 09:59:51 09:59:52 09:59:53 09:59:54 09:59:55 09:59:56 09:59:57 09:59:58 09:59:59 10:00:00 answer(from Ac program) 16:55:51 17:04:11 17:12:31 17:20:51 17:29:11 17:37:31 17:45:51 17:54:11 18:02:31 18:10:51 18:19:11 18:27:31 18:35:51 18:44:11 18:52:31 19:00:51 19:09:11 19:17:31 19:25:51 19:34:11 19:42:31 19:50:51 19:59:11 20:07:31 20:15:51 20:24:11 20:32:31 20:40:51 20:49:11 20:57:31 21:05:51 21:14:11 21:22:31 21:30:51 21:39:11 21:47:31 21:55:51 22:04:11 22:12:31 22:20:51 22:29:11 22:37:31 22:45:51 22:54:11 23:02:31 23:10:51 23:19:11 23:27:31 23:35:51 23:44:11 09:59:11 10:07:31 10:15:51 10:24:11 10:32:31 10:40:51 10:49:11 10:57:31 11:05:51 11:14:11 11:22:31 11:30:51 11:39:11 11:47:31 11:55:51 12:04:11 12:12:31 12:20:51 12:29:11 12:37:31 12:45:51 12:54:11 13:02:31 13:10:51 13:19:11 13:27:31 13:35:51 13:44:11 13:52:31 14:00:51 14:09:11 14:17:31 14:25:51 14:34:11 14:42:31 14:50:51 14:59:11 15:07:31 15:15:51 15:24:11 15:32:31 15:40:51 15:49:11 15:57:31 16:05:51 16:14:11 16:22:31 16:30:51 16:39:11 16:47:31 † SiriuS † [TWYT Union] Re: This is some Tests to WA#8 and WA#9 // Задача 1892. Утро в Кольцово 6 апр 2014 17:07 My program gives another answers for all these tests. But I got WA#15. Who can help me? Vladimir Yakovlev (USU) Problem 1548 "Sakura and Statistics" has been rejudged [7] // Задача 1548. Сакура и статистика 28 янв 2016 15:17 A bunch of new tests have been added. 33 of 38 authors lost their AC. A little hint: this problem do has a polynomial solution, and it's not so tricky as you may think. Edited by author 28.01.2016 15:20 bsu.mmf.team Re: Problem 1548 "Sakura and Statistics" has been rejudged [6] // Задача 1548. Сакура и статистика 18 фев 2016 22:36 OUCH Jane Soboleva (SumNU) Re: Problem 1548 "Sakura and Statistics" has been rejudged [5] // Задача 1548. Сакура и статистика 18 май 2016 05:36 spam random @ pray Didn't really expect it to work, especially after such a dramatic rejudge. And i didn't even get to use sorts of «clever» random with GA and mutating solutions. Apparently, most of my WAs were because of inappropriate process of building candidate rectangles, or so i believe... Also, here's a test that helped me, though quite likely it won't be too helpful for anyone else: 7 5 10001 11011 11111 11111 11110 11010 10010 --- 5 1 1 7 1 2 1 6 2 3 1 5 4 2 4 7 4 1 5 4 5 Orient Re: Problem 1548 "Sakura and Statistics" has been rejudged [4] // Задача 1548. Сакура и статистика 26 май 2016 23:21 Jane Soboleva (SumNU) писал(a) 18 мая 2016 05:36 spam random @ pray The hint is clear: there is deterministic polynomial algorithm (~O(s^2)). Some useful tests (for that algorithm): 4 9 010000010 111010111 101111101 000101000 9 --------- 4 4 1100 1110 1111 0111 3 ---- 4 4 1100 1110 0111 0010 3 ---- 5 5 11100 11110 11111 01111 00111 3 ----- 4 3 001 111 011 010 3 --- 4 3 010 111 011 001 3 --- 4 4 0010 1110 0111 0101 4 ---- 4 4 0101 1111 1110 0010 4 ---- 4 3 011 010 110 100 3 --- Jane Soboleva (SumNU) Re: Problem 1548 "Sakura and Statistics" has been rejudged [3] // Задача 1548. Сакура и статистика 26 май 2016 23:44 Oh, good job, and thanks for the tests~ I do have a certain algo in mind actually, but i'm afraid i'll be too lazy to try and code it yet — not until some test that breaks my program is added, haha. Well, and here's another test case then, simple but could be useful. I had a bunch of WAs on 3 and 4 initially; i was building candidate rectangles by expanding them from corners. But then i realized there can be H-like cases, and that horizontal row isn't supported by any corners. 3 5 10001 11111 10001 Orient Re: Problem 1548 "Sakura and Statistics" has been rejudged [2] // Задача 1548. Сакура и статистика 26 май 2016 23:51 Jane Soboleva (SumNU) писал(a) 26 мая 2016 23:44 I do have a certain algo in mind actually, but i'm afraid i'll be too lazy to try and code it yet — not until some test that breaks my program is added, haha. Your smart randomized brute force approach is incredible at my mind. I sure eventually there will be Sakura v2 (something like N=10000 (nonsensible) and M=500) and optimized O(s^2) algorithm became unavoidable. Edited by author 27.05.2016 06:42 Jane Soboleva (SumNU) Re: Problem 1548 "Sakura and Statistics" has been rejudged [1] // Задача 1548. Сакура и статистика 27 май 2016 00:02 Edit: affirmative, remove it~ Edited by author 27.05.2016 00:09 Orient Re: Problem 1548 "Sakura and Statistics" has been rejudged // Задача 1548. Сакура и статистика 27 май 2016 00:03 Roger that. Removed. Edited by author 27.05.2016 00:11 IlushaMax Just for interest [2] // Задача 2066. Простое выражение 14 мар 2016 00:37 [code deleted] This code Accepted so A can be 0 or 1; B can be 0 or 1; C can be 0 or 1 in tests. But if we will see on combinatons then there are only 4 combinations (if a<=b<=c) and there are 8 tests as min. The paradox Edited by author 14.03.2016 01:18 Edited by moderator 24.11.2019 13:45 Ivan Metelev Re: Just for interest // Задача 2066. Простое выражение 25 май 2016 17:34 Please do not writeln your code there. ToadMonster Re: Just for interest // Задача 2066. Простое выражение 25 май 2016 18:45 > This code Accepted so So no run-time checks if a, b, c are in [0-1] interval. Example: [code deleted] Edited by author 25.05.2016 18:45 Edited by moderator 24.11.2019 13:45 ruwelcome i got Ac use 0.02s [4] // Задача 1118. Нетривиальные числа 20 май 2003 12:51 [code deleted] Edited by moderator 19.11.2019 23:41 Madi Re: i got Ac use 0.02s [2] // Задача 1118. Нетривиальные числа 21 дек 2008 19:20 nuuu.. che skazat nu krasavchik. Nguyễn Cảnh Toàn Re: i got Ac use 0.02s // Задача 1118. Нетривиальные числа 30 дек 2008 21:00 thanks you , my pro is TLE :) Ivan Metelev Re: i got Ac use 0.02s // Задача 1118. Нетривиальные числа 25 май 2016 14:39 Your english is so good) Сергей Re: i got Ac use 0.02s // Задача 1118. Нетривиальные числа 4 май 2014 22:47 умный, молодец Adrian Some tips (got AC using 0.031s) [3] // Задача 1118. Нетривиальные числа 3 окт 2010 19:35 First, triviality of 1 is 0. Second, we can judge number i from 2 to 1000000, if i is a prime number, then prime[i] is true. Third, we traversal i from b to a, because if j>i and prime[j]==true, prime[i]==true, we can get 1/j<1/i. if we get prime[i]==true,just output it and break. if we can not get such a number, calculate triviality of i and output the minimum. Thus, we can get AC with a little time spended,but larger memory. My English is poor, if you cannot understand me, I would say sorry for it.^__^ Egor Stepanov Re: Some tips (got AC using 0.031s) [1] // Задача 1118. Нетривиальные числа 30 дек 2012 03:02 your english are good) i got it. may be i got it because my solution is similar yours=) Egor Stepanov Re: Some tips (got AC using 0.031s) // Задача 1118. Нетривиальные числа 30 дек 2012 03:02 but i got AC using 0.1s Ivan Metelev Re: Some tips (got AC using 0.031s) // Задача 1118. Нетривиальные числа 25 май 2016 14:37 Your english good enouth to speak with Russians) Ivan Metelev Why compilation ERROR?!?!?!?!? WTF?!?!?!? [2] // Задача 1964. Китайский язык 29 ноя 2014 19:14 var a,i,b,n,k: int64; begin readln(n,k); for i:=1 to k do begin read(a); b:=b+a; end; writeln(b-n); end. Mescheryakov_Kirill Re: Why compilation ERROR?!?!?!?!? WTF?!?!?!? [1] // Задача 1964. Китайский язык 3 дек 2014 19:14 In the "For" can not be used "int64" Ivan Metelev Re: Why compilation ERROR?!?!?!?!? WTF?!?!?!? // Задача 1964. Китайский язык 25 май 2016 11:30 Thank you Nivranshu Hint // Задача 2000. Grand Theft Array V 24 май 2016 14:45 To play optimally, both players would first move towards each other and then from the point of their intersection they will take all the numbers from that point to their nearest boundaries. Let Player 1 start from X and Player 2 start from Y. Then, if (X < Y), player 1 will take all the numbers from A[1] to A[X] and player 2 will take all the numbers from A[Y] to A[N] and also player 1 will take the numbers from A[X] to A[(X+Y)/2] and player 2 will take the numbers from A[(X+Y)/2+1] to A[Y]. if (X > Y) same as above but now player 1 will work from X to N and player 2 will work from 1 to Y. if (X == Y), calculate the sum from A[1] to A[X-1] and another sum from A[X+1] to A[N], add A[X] to the greater sum and give it to player 1. Arderun Java. Runtime error on #4 test. Why [1] // Задача 1100. Таблица результатов 24 май 2016 13:55 here is my code import java.io.PrintWriter; import java.util.*; public class Problem1100 { public static class Busket extends LinkedList<Long>{} public static void main(String[] args) { Scanner in = new Scanner(System.in); PrintWriter out = new PrintWriter(System.out); int times = in.nextInt(); Busket[] list = new Busket[100]; long id =0; int score =0; for(int i =0;i<times;i++){ id = in.nextLong(); score = in.nextInt(); boolean insert=false; if(list[score]==null){ Busket busket = new Busket(); busket.add(id); list[score]=busket; } else{ list[score].add(id); } } StringBuffer result = new StringBuffer(); for(int i = list.length-1;i>=0; i--){ Busket busket =list[i]; if(busket!=null) for(Long mId: busket) { result.append(mId); result.append(" "); result.append(i); result.append("\n"); } } out.println(result.toString()); out.flush(); } } Edited by author 24.05.2016 13:57 Edited by author 24.05.2016 14:01 Edited by author 24.05.2016 14:09 Edited by author 24.05.2016 14:11 Oleg Baskakov Re: Java. Runtime error on #4 test. Why // Задача 1100. Таблица результатов 24 май 2016 14:31 new Busket[100] means indices 0..99. Use [101] for 0..100 indices. Edited by author 24.05.2016 14:33
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