Общий форумMy O(N^2) solution works for 0.187s but 10000 * 10000 = 1e8, it means that O(N^2) solution should work only for 1 second (not less), but in this task there is 0.5 sec and O(N^2) solution gets AC 5 0 -100000000 1 -100000000 1 -99999999 -1 100000000 0 100000000 I think, this answer is wrong: 0 -100000000 1 -100000000 -1 100000000 But my AC program doesn't think so. And I still ask you to make the third test without trash in the end. Thanks for pointing out the trash in the end of 3rd test case! I have WA#2 and i don't understand what is wrong.. If I understood problem correctly, my program works well on all tests, which I tried. Edited by author 23.02.2020 18:27 For each vertex in tree find the nearest vertex from the way to root, in which we need to change direction. If such vertex doesn't exist, let it be -1. Let's call this vertex as bad[V], where V - some vertex in tree. To answer on query for vertex X, let's travel in "bad" array from X until we meet bad[X]=-1. Let's save visited vertices (excluding query vertex X) to some array (let's call this array "change"). Now it's quety obvious, that we need to change directions only in those visited vertices to "activate" path from root to leaf. Before changing values in "bad" array for vertex from X to root, let's take some vertex V from "change" array. Before query there was a set of vertices which have the same "bad" vertex as a V (bad[V]=bad[V_1]=bad[V_2]=...=bad[V_n], V_i - vertex from the set), and now we need to change their "bad" value to V, because we change direction in vertex V. Finally, we need to set bad[V]=-1 to all vertices from X to root. To do all operations efficiently we can use heavy-light decomposiotion + segment tree. So, the total complexity would be O(N + Q*(log(N))^2), N - number of vertices in tree. Edited by author 06.06.2019 17:11 Straightforward Link/Cut Tree solution. Just dfs the input grid and build the tree graph represented by LCT; represent paths as bamboo splay trees: just make dfs-child the right child if turnout is good and do nothing otherwise. All parents in dfs-tree that do not satisfy turnout make path parents (see LCT terminology). Now each query is just a single access—sometimes called expose—operation that saves all path parents on the path. #include <iostream> #include <vector> #include <algorithm> using namespace std; int main() { int a; int n; cin >> n; vector<int> v; for (int i = 0; i < n; i++) { cin >> a; v.push_back(a); } n = 0; sort(v.rbegin(), v.rend()); for (int i = v.size()-1; i > (v.size()-2) / 2; i--) n += (v[i]+1)/2; cout << n << endl; } You should iterate from v.begin to v.end. No the reverse way. Sort the n teams and take the first (n/2 + 1) team's people (by people i mean, array[i]/2 + 1). If you have problem understanding what i have just written or for better understanding, give this a go : https://ideone.com/vD6SAmWell, you have to know how to determine the distance between two points first. It can be calculated by, dis = sqrt((x1-x2)^2+(y1-y2)^2). You can say that, we can just calculate the distance in between the points but why the raidius of the nails are given. Its because the rope is wrapped around each and every nails. But the rope is not in fully contact with nails. The more nails we get, the less contact we have in between rope and nails. The amount of which the rope is in contact with nail is : s = parameter of a nail/the number of nails. So we are just required to print out : d(d is the sum of the distance between them) + number of nails * s(s is the amount which the rope is in contact with nails) If you have problem understanding my word or for better understanding, give this a go : https://ideone.com/t1M3rK Edited by author 21.02.2020 01:31List<long> coll = new List<long>(); List<decimal> answer = new List<decimal>(); List<string> a = new List<string>(); string line; while (true) { line = Console.ReadLine(); answer.Clear(); coll.Clear(); a.AddRange(line.Split()); a.RemoveAll(x => x.Equals("")); for (int i = 0; i < a.Count; i++) { coll.Add(long.Parse(a[i]));
} for (int i = 0; i < coll.Count; i++) { var d = (decimal)Math.Sqrt(coll[i]); answer.Insert(0, d); } Console.Clear(); for (int i = 0; i < answer.Count; i++) {
Console.WriteLine(answer[i].ToString("0.0000##")); } }
Edited by author 21.02.2020 00:40 This problem can be solved using greedy technic. Method is to divide the given number from 9 to 2. Each time we will continue to divide and update the number as we go. We will keep dividing N with the same number utill its impossible to do so.And for each division we will save the number which was divisible by given number. And by "save" i mean pushing the number in a stack or a in a vector or an array. If you are using other than stack you might print out the numbers in a reverse manner. And when we are getting "-1"? I already wrote that, we keep updating the number. so after we pass the 9 to 2 loop, we should be left with a value of 1 in the variable N ( given number ). Incase N!=1 we are just printing "-1". For the case N=1 or N=0? for the case 1 you should just print "1" and for the case 0 print "10", we can not print 01 since this refers to number, 1. If you have problem understanding what i have just written or for better understanding, give this a go : https://ideone.com/709cog Edited by author 21.02.2020 00:34 Edited by author 21.02.2020 00:38 Edited by author 21.02.2020 00:41I solve reccurence, then pow to n(which takes log n to quickpow), but it's also TL. How can i improve this not writing my own long arithmetic? It can be solved only with 2 divisions in BigDecimal, but answer length is about 32000 so [maybe] toString gives TLE on conversion binary data to decimal representation. If somebody knows, how to solve this problem - please tell me 2^n = 10^(n*lg(2)) My AC solution: int n0 = (int) (Math.log10(2f) * (n - 1)); int n1 = (int) (Math.log10(2f) * n); int n2 = (int) (Math.log10(2f) * (n + 1)); int len = n1; if (n1 == n2 && n1 == n0 + 1 && n % 2 != 0) { len = n0; } Answer is len. I thought that this problem's solution is : t= (n-(n%10)) /4 +1 if(n>=30) { t=t+1; x=n-n%10; if(n>39&&((x/10)%2)) t=t+(x-30)/20; if(n>49&&((n-n%10)%4==0)) t=t+(x-40)/20; } t= t+ (n%10)/4; cout<<t; but it got WA#8. And i don't understand "if(n==1 && n1==n0+1 && n%2!=0)" statement are for what tests? Would you mind explaining for me! Thanks! bve, your way is brilliant. Thank you! bve, any links to why it works? My submission (8764478) goes wrong on the following tests, but it has an AC on the judge system. The right answer for all test cases below is "equal". RRGB GRBR RRGY GRYR RRBG BRGR RRBY BRYR RRYG YRGR RRYB YRBR RGRB BRGR RGRY YRGR RGBR GRBR RGYR GRYR RBRG GRBR RBRY YRBR RBGR BRGR RBYR BRYR RYRG GRYR RYRB BRYR RYGR YRGR RYBR YRBR GRGB BGRG GRGY YGRG GGBR BGRG GGBY BGYG GGYR YGRG GGYB YGBG GBRG BGRG GBGY YGBG GBYG BGYG GYRG YGRG GYGB BGYG GYBG YGBG BRBY YBRB BGBY YBGB BBYR YBRB BBYG YBGB BYRB YBRB BYGB YBGB Help pls, I have WA #20 can u give me a test? (all answers of my program on tests which i generated are right) VAR s:string; k,n,l,c:Longint; begin read(s); l:=length(s); val(s,n,c); if (n<3) and (n>0) then begin k:=0; case s[l] of 'a','A': write('window'); 'D','d': write('window'); 'b','B': write('aisle'); 'c','C': write('aisle'); end; end; if (n>=3) and (n<=20) then begin k:=0; case s[l] of 'a','A': write('window'); 'f','F': write('window'); 'b','B': write('aisle'); 'd','D': write('aisle'); 'E','e': write('aisle'); 'c','C': write('aisle'); end; end; if (n>=21) and (n<=65) then begin k:=0; case s[l] of 'a','A': write('window'); 'k','K': write('window'); 'g','G': write('aisle'); 'd','D': write('aisle'); 'h','H': write('aisle'); 'c','C': write('aisle'); 'b','B': write('neither'); 'e','E': write('neither'); 'f','F': write('neither'); 'J','j': write('neither'); end; end; if n>65 then write('neither'); end. Edited by author 20.11.2012 23:14 Please, write your messages in English, because it's international language. We, Russians, don't understand O'zbek #include<iostream> using namespace std; int main() { int n; cin>>n; switch(n) {case 2: cout<<10;break; case 4: cout<<670;break; case 6: cout<<55252;break; case 8: cout<<4816030;break; } return 0; } Edited by author 03.02.2012 16:53 print(([0,10,670,55252,4816030])[int(input())//2]) hahahahahahaha Solution #1: AC in 0.8 sec Just calculate sum using Gauss's method. Int[-1..1] f(x)dx ~= (5*f(-sqrt(3/5)) + 8*f(0) + 5*f(+sqrt(3/5))) / 9 Number of parts, N = (int)7e6 Length of part number i = (1e-4)*koef^i, where koef = 1 + x/N. Try some different x = 0.1 ... 10, and get AC =) Solution #2: WA 11 on MSVC or AC on Java, Pascal, GNU C++ http://en.wikipedia.org/wiki/Residue_theorem says that we have to find all complex roots of the polynom. I do not like exact formulas for degree 3 and 4. There is numerical way to find all complex roots for any degree of polynom. Just take point z = (0,0) and shift it to any direction in such a way that |P(z+shift)| < |P(z)|. Shift it while |P(z)| > eps. Now z is a root. This method works not only for polynoms, it works even for arbitrary "holomorphic function" (see Complex Analysis). Prove is smth like this http://en.wikipedia.org/wiki/Maximum_modulus_principle The only thing that I need now -- to calculate f(z) for any complex z. And using only "double" of Microsoft Visual C++ I can't. I have no enough precision and get WA 11 =( Using BigDecimal (java), extended (pascal) or long double (GNU C/C++) this method gets AC (the task is from Petrozavodsk, so I have original tests). P.S. 2 Admins: Ну сколько можно добавлять задачки с контестов, где у С++ программистов был под рукой long double, и это было важно. Даешь GNU C/C++! Edited by author 27.10.2012 08:08I found many other ways. 1)Monte Carlo - slow for this problem 2)Separation of real and complex roots. Can be used to solve this problem. Binary search for Re(z)=alfa, Re(z)=beta, Im(z)=gamma, Im(z)=delta Березин И.С., Жидков Н.П. Методы вычислений, Т.2. М.: ГИФМЛ, 1959. 3)Ferrari method. Куликов Л.Я. Алгебра и теория чисел 4)Barstow method. It's numeric iterative method. Саманчук_билеты_с_ответами.pdf Численные методы I use combination from 3): binary search for yo then formulas Edited by author 12.08.2018 21:04 https://en.wikipedia.org/wiki/Adaptive_quadrature with Gaussian method to estimate value of definite integral: approx_integrate(Func f, long double l, long double r) { long double A = sqrtl(3.0l/5.0l)/2, x1=0.5-A, x2 = 0.5+A; return (5*f(l*x1 + r*x2) + 8*f((l+r)/2) + 5*f(l*x2+r*x1))/18 * (r-l); } AC in 0.015. That is, you don't even need to think to solve this problem. Edited by author 16.02.2020 23:48 Try these test cases. Test 1: 0 Ans:2 Test 2: 1 Ans:2 Test 3: 2 Ans:3 Test 4: Z0Z Ans:36 Test 5: 21 Ans:4 Test 6: ZIZ Ans:No solution. (Don't forget to put the full-stop at the end!!!) Test 7: AERGUFG12312OP Ans:32 Test 8: 234DFG67 Ans:23 And this problem can be solved within int range. Edited by author 09.12.2015 20:14 only test 7 I can't get right ans. I have 33 And WA4 Removed my post, because apparently I can't alphabet :) Edited by author 12.07.2016 14:45 IN TEST CASE 2, FOR INPUT 0 , WHY ANS IS ONLY 2 IT CAN BE 1,2,3,4,5.......35. BECAUSE ZERO IS DIVISIBLE BY ALL. You can't divide by 0! So 1 isn't a solution and therefore it's 2: 2 2 2 2.01 4.01 3.99 4.01 правильный ответ 1.9800000 у меня было 1.9900000 Дело в том что чтобы не париться с double я умножал сразу на 100 но оказалось что 2.01 * 100 == 200 а не 201. Пришлось вручную переводить две цифры после запятой и решение зашло! import sys n,k=map(int,input().split()) ans=0 nn=1 on=1 if n==1: print(0) sys.exit() for i in range(100000000000000000000): if on<=k: nn+=on on+=on ans+=1 elif k<n: nn+=k on+=k ans+=1 if nn>=n: print(ans) sys.exit() i can not understand why optimization my cod if i got ac with you help i give you 1000 rubles Edited by author 17.10.2018 22:35 The reason why you got time limit is that you don't use optimal algorithm. Your algorithm can be even correct, but to pass you need downgrade your algorithm. Maybe you could find some formula that would describe the whole process. Python is not the case here. You would get the same time limit issue with C. I launched your script with N = 10000000 and K = 2 and it took 1.833s on my machine to get the correct answer when the time limit is 0.25s. Of course it will grow up nonlinearly with bigger N. So it would take ages with N = 10^9 (the edge value). $ time echo "10000000 2" | python 1131_bruteforce.py real 0m1.833s user 0m1.824s sys 0m0.007s if i got ac with you help i give you 1000 rubles Edited by author 17.10.2018 22:35 P.S. I don't need your money if you finally solve it :) Edited by author 13.02.2020 06:12 Edited by author 13.02.2020 06:12Don't make 100+ submissions like I did. Just check these tests to make sure you correctly understand the problem description. Test #1: 10 20 0 5 3 4 turn 1 100 2 100 3 100 4 100 5 100 6 1 calls 20 2 calls 20 3 calls 20 4 calls 10 5 checks dealing flop 80 80 80 80 80 answer: 6 4 checks 5 checks 1 checks 2 checks 3 checks dealing turn Test #2: 10 20 10 5 3 3 turn 1 11 2 12 3 100 4 100 5 100 7 1 calls 1 2 calls 2 3 raises 10 to 30 4 folds 5 calls 10 dealing flop 5 checks 0 0 50 80 45 answer: 4 3 bets 10 5 calls 10 dealing turn 5 bets 5 Test #3 (this is what 10th test about): 10 20 0 5 3 4 flop 1 200 2 100 3 100 4 200 5 100 3 1 calls 20 2 calls 20 3 raises 20 to 100 100 80 0 100 80 answer: 5 4 calls 90 5 folds 1 calls 80 2 folds dealing flop Test #4: (this is what 32th test about - pay attention to the order of players' turns) 10 20 0 2 2 1 river 1 1000 2 1000 10 2 calls 10 1 checks dealing flop 1 bets 20 2 raises 30 to 50 1 calls 30 dealing turn 1 checks 2 checks dealing river 900 830 answer: 2 1 bets 30 2 raises 70 to 100 Test #5: (This is what 33th test about, 0 - is the wrong answer!!!!) 10 20 0 3 3 3 flop 1 100 2 100 3 100 0 80 80 80 answer: 6 3 calls 20 1 calls 10 2 checks dealing flop 1 checks 2 checks Good luck!!! Edited by author 12.06.2016 14:04 Edited by author 13.06.2016 20:51 Correct me if I'm wrong, but isn't test #2 invalid because it has one more possible answer? 5 3 checks dealing turn 5 checks 3 bets 10 5 raises 5 to 15 This answer isn't possible because of this: "If that is the first time when module takes turn then it should restore all players’ turns with all data it has. Otherwise module also has the information about all players’ turns before the previous bot turn". It means that you must have not more than one bot's turn in the answer. At least if M > 0. "It means that you must have not more than one bot's turn in the answer. At least if M > 0." Is there situation when M=0? Yes, first sample for example. But looks like there's always not more than one bot's turn. Here's some modification of my 5th test: 10 20 0 3 3 3 river 1 100 2 100 3 100 0 80 80 80 It seems to be correct, and the answer has > 1 bot's turn: 14 3 calls 20 1 calls 10 2 checks dealing flop 1 checks 2 checks 3 checks dealing turn 1 checks 2 checks 3 checks dealing river 1 checks 2 checks I checked and there are no such tests. "If that is the first time when module takes turn then it should restore all players’ turns with all data it has." - probably this means it's the first bot turn, otherwise the module would be called before. "Yes, first sample for example. But looks like there's always not more than one bot's turn." I mean M is number of all turns, not only bot's. Also I'd like to ask another question: in all tests here bot's turn occupies first line of output, so is there any test where bot's turn is not in first line, or it contradicts to the condition? Yes. See 1st sample test. And what about this test: 10 20 2 5 3 4 flop PhilIvey 500 TomDwan 500 ViktorBlom 500 Grobot 500 Ziigmund 500 6 PhilIvey folds TomDwan calls 20 ViktorBlom raises 40 to 60 Grobot calls 50 Ziigmund folds TomDwan folds 498 478 338 438 478 ans1: 1 dealing flop ans2: 3 dealing flop Grobot checks VictorBlom bets 100 Which answer is right? Thanks in advance. Edited by author 16.04.2017 17:27 for test #2 answer, wouldn't: 2 3 bets 10 5 raises 5 to 15 be correct too? and for test #3 raises shouldn't it be 3 raises 80 to 100 Thx in advance EDIT: Sorry, i misundestood Edited by author 02.12.2019 06:28 Edited by author 02.12.2019 06:28 Edited by author 21.02.2020 17:41 Easy concept, just to think for a while. AC in one go! |
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