Common Boardimport java.util.Scanner; public class javasucc { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int h = 2, s = 10; for(int i=0; i<n; i++) { int a = in.nextInt(); String b = in.next(); if(b=="hungry"){ if(a>h) {h = a;} } else if(b=="satisfied"){ if(a<s) {s = a;} } } if(h >= s) System.out.println("Inconsistent"); else System.out.println(s); } } and the problem is that program doesn't want to do if and i can't change s to a 1) WA1, task example isn't passed  it's shame to cry "help" on WA1. You should better debug your code. 2) https://stackoverflow.com/questions/513832/howdoicomparestringsinjava My algorythm has maximum O(n^2) complexity. I have TLE 18 test. But, when i artificially set limits for n (amount of layers) to 200 and for k (amount of bricks in a single layer) to 290, my program still gets TLE 18 test, i really have no idea how is it possible. It should get WA\AC if inputs are more than limits since i do not see any possibility of an endless cycle in my algo. Edited by author 27.02.2020 21:19 Edited by author 27.02.2020 21:36 Could anybody give me some tricky tests? This problem seems to be trivial but I got WA. I guess it's all about format of result. try this: 3 6 1 2 1 2 1 1 Thank you! We should pay attention to the case when there is any candidates that no one votes for him or her Thanks! Most helpful answer here! Edited by author 13.07.2017 16:25 Not possible to say anything if you won't post your code. Check your syntax.
a = int(input()) b = int(input()) print(a+b) What is the problem i try with with python 3 & C but every time it says compilation error. What is the problem here??? del Edited by author 06.01.2015 21:37 My 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 After so many calculation and math I have solved the problem..... Here we can have two worst cases... Case 1: having all the right shoes first.so here needed time is 2*b and we have now all the left shoes remaining...so total time is 2*b+40... Case 2: we may have 39 right shoes so time needed here is (39*2=78)...then we have only one right foot left but we may encounter all the left shoes and here needed time is 40+2*(a40)....> 40 for the first 40 shoes and 2*(a40) is for the remaining shoes as they needed to be thrown away...then we have the only one right foot left and it need 1 second... so total time = 78+40+2*(a40)+1 = 119+2*a80 = 2*a39 ans=max(Case 1,Case 2) Edited by author 22.02.2020 03:07 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 heavylight 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 dfschild the right child if turnout is good and do nothing otherwise. All parents in dfstree 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/vD6SAm Well, you have to know how to determine the distance between two points first. It can be calculated by, dis = sqrt((x1x2)^2+(y1y2)^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 01:31 List<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 00:41 I 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=nn%10; if(n>39&&((x/10)%2)) t=t+(x30)/20; if(n>49&&((nn%10)%4==0)) t=t+(x40)/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 
