| 
 | 
Common Boardtry 6 6 0 3 0 5 0 5 6 0 0 0   and   7 2 0 5 0 6 4 0 5 7 0 0 0 0 Well, I've passed that tests, but still getting WA 13 Any advices? When build the graph, make it undirected rather than directed one.     import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.*;   public class Lesson {       public static void main(String[] args) throws Exception {           BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));         int countPrepod = Integer.parseInt(bufferedReader.readLine());         TreeSet<Long> prepod = new TreeSet();         for (int i = 0; i < countPrepod; ++i) {             prepod.add(Long.parseLong(bufferedReader.readLine()));         }           int countStudent = Integer.parseInt(bufferedReader.readLine());         ArrayList<Long> student = new ArrayList();         for (int i = 0; i < countStudent; ++i) {             student.add(Long.parseLong(bufferedReader.readLine()));         }           student.sort(((o1, o2) -> Long.compare(o1,o2)));         int result=0;         for (long find:prepod)         {                 while (Collections.binarySearch(student,find)>=0)                 {                     ++result;                     student.remove(find);                 }         }         System.out.println(result);     }   }   1. This problem can be reduced to the following:       Given X, find two vertices such that path weight is equal to X. 2. It probably cannot be solved with naiive DP. Use centroid decomposition. 3. std::unordered_map gets TL while std::map gets AC. i have some problems with (), please give answers for some other cases 1   Edited by author 06.11.2015 00:47 Input: 4   Output: (((sin(1)+4)sin(1-sin(2))+3)sin(1-sin(2+sin(3)))+2)sin(1-sin(2+sin(3-sin(4))))+1 What it is it? My program overcome all my tests,but system(TIMUS) written me about WA#9!  WHY? me too! any ideas, please send to:   williamm2006@126.com   thanks!   program p1406; var   a:array[1..2000]of longint;   ch:char;   i,j,k,n,m:longint; begin   while not eof(input) do     begin       read(ch);       if ch in['0'..'9']         then begin           inc(n);           a[n]:=ord(ch)-48;         end;     end;   for i:=n downto 1 do     if a[i]>0       then break;   if((i=1)and(a[i]=0))or(n=1)     then begin       writeln(-1);       halt;     end;   j:=i-1;m:=a[i]-1;a[i]:=0;   for j:=i-1 downto 1 do     if a[j]<>9       then break;   if(j=1)and(a[j]=9)     then begin       writeln(-1);       halt;     end;   inc(a[j]);   for i:=j+1 to n do     begin       inc(m,a[i]);       a[i]:=0;     end;   i:=n;   while m>0 do     begin       if m>=9         then a[i]:=9         else a[i]:=m;       dec(i);       dec(m,9);     end;   for i:=1 to n do     write(a[i]);   writeln; end. Don't be afraid of geometry, this problem is really easy Please, fix English in the statement:   is maked -> is made You standing -> You're standing you have made -> you made Thank you! Is it supposed to be solved without floating-point numbers at all? This problem is definitely can be solved using integers only Don't be afraid of geometry, this problem is really easy  I confirm :) #include<iostream>   using namespace std; int sum(int temp); int main(){     int a[6];     scanf("%1d%1d%1d%1d%1d%1d",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5]);     int first=a[0]*100+a[1]*10+a[2];     int second=a[3]*100+a[4]*10+a[5];     if(a[3]==9&&a[4]==9&&a[5]==9){         cout<<"No";         return 0;     }     int pre=second-1;     int next=second+1;     if((sum(pre)==(a[0]+a[1]+a[2]))||(sum(next)==(a[0]+a[1]+a[2]))){         cout<<"Yes";     }else{         cout<<"No";     }
  return 0; }   int sum(int temp){     int count=0;     do{         count+=(temp%10);         temp/=10;     }while(temp/10!=0);     count+=temp;     return count; } #include<iostream> #include<vector> #include<string> #include<algorithm> using namespace std;   class team{     public:         string id;         int n; };   bool comp(const team &t1,const team &t2);   int main(){     int k;     cin>>k;     vector<team> T;     while(k--){         string temp;         int num;         cin>>temp>>num;         team t;         t.id=temp;         t.n=num;         T.push_back(t);     }     stable_sort(T.begin(),T.end(),comp);     for(int i=0;i<T.size();++i){         cout<<T[i].id<<" "<<T[i].n<<endl;     }
  return 0; }   bool comp(const team &t1,const team &t2){     return t1.n>t2.n; } Why Wring answer? import java.util.Scanner;   public class T1654 {     public static char[] arr;     public static String out = "";     public static void main(String[] args) {         // TODO Auto-generated method stub         Scanner sc = new Scanner(System.in);         String inp = sc.next();         sc.close();         arr = new char[inp.length()];         arr = inp.toCharArray();         format();         if(out != "") {             System.out.println(out);         } else {             System.out.println("");         }
      }
      public static void format() {         boolean isFormat = false;         int l = arr.length;         for(int i = 0; i< l-1; i++) {             if(arr[i] == arr[i+1]) {                 arr[i] = 0;                 arr[i+1] = 0;                 isFormat = true;                 } else if(arr[i] != arr[i+1] & arr[i] != 0){                 out += arr[i];             }         }         out+= arr[arr.length-1];         if(isFormat == true) {             arr = new char[out.length()];             arr = out.toCharArray();             out = "";             format();         }     }   } получил АС, долго искал где долбанный баг, т.к. ТЕСТ 22 выдавал ошибку по времени, если у вас такая же проблема и вы используете преобразование double в int то в 22 тесте точность P и Q 3 знака, поэтому ТЛ и выдает, вероятно на вход подаются какие нибудь числа 22.333 и 22.334 удачи You are wrong! P and Q are given with up to 2 digits after decimal point in all tests. похожая фигня, точности 2 знаков после запятой не хватало, впилил третий знак, всё заработало code like:  size_t out = 0; printf ("%zu\n", out);  result - > wa#1 (((  if i use  printf ("%d\n", out);  all  test accepted (( Why? https://wandbox.org/ & gcc 7.1  - The first version works correctly ((( i have ac in 0.187 s, but on this test my program runs at least 0.9:   35 123 2 3 4 5 6 0 8 9 10 11 0 13 14 15 0 17 18 0 20 21 0 23 0 25 0 27 0 0 0 0 0 0 0 0 0 I've got accepted after adding the condition of end of file in creating new line. May be, it will be useful.   Edited by author 31.01.2016 15:50   Edited by author 31.01.2016 15:50 2 5 11 as fg asdf asdfr a qwer as s s qwer e   Correct answer is 5 #include<iostream> #include<vector> #include<algorithm>   using namespace std;   int main(){     int k;     cin>>k;     vector<int> line1;     vector<int> line2;     vector<int> sum;     while(k--){         int m,n;         cin>>m>>n;         line1.push_back(m);         line2.push_back(n);     }     bool plus_1=false;     for(int i=line1.size()-1;i>=0;i--){         int temp=line1[i]+line2[i];         if(plus_1){             temp++;         }         if(temp>=10){             plus_1=true;             temp-=10;         }else{             plus_1=false;         }         sum.push_back(temp);
      }     for(vector<int>::reverse_iterator iter=sum.rbegin();iter!=sum.rend();++iter){         cout<<*iter;     } return 0; } I've been wa on this case for a long time: abaabaaba The right answer is abaabaabaaba Can anyone please suggest the test case which might give me WA35...   I have used string , also tried with character array...  and KMP algorithm   ..... please do reply     Edited by author 27.08.2018 13:31 Aren't 'abaabaaba' a palindrome? Can anybody please give me cases for Test2 ? If you only go through the matrix and saved the minimum value, it is not the solution for that problem. Really fast.   for (int k = 0; k < n; ++k) {   for (int i = graph[k]._Find_first(); i < n; i = graph[k]._Find_next(i)) {     graph[k] |= graph[i];   } }  |  
  | 
|