Because 64 * 10 ^ 9 is too many operations for 0.5 seconds :)

same thing. give some tests, please

I don't have tests, but my bug was the following: when I sorted sequences, ones with negative balance should be sorted in the symmetrical way with the sequences with positive balance, and I've compared them wrongly.

Edited by author 04.11.2012 17:36

You can solve it using DP[N].

You shouldn't use table 1000*1000. It could be solved using table 1000*2

int x[100],y[100],max[100];

I used: bool table[1000][1000];  which is 0,95367431640625 MB.
table[i][j] = 1, if the respective quarter can be crossed diagonally, and '0' otherwise.
I also used an array a[1001], in which to store the distance, and several auxiliary variables.
The total memory used is 1128 KB.

374

Add '1' to the input number and see if it's lucky, i.e. if the sum of the first 3 digits is the same as the sum of the second 3 digits of the newly formed number.
Subtract '1' from the input number and check the same.
Output 'Yes' if the newly formed number satisfies the condition, and 'No' otherwise.

Adjancency matrix will require 10000^2 elements. It's too many, u won't fit in memory limit.
Use adjancency list instead. For example, if you use C++, you can represent it as an array of vectors, where the vector at index i contains numbers of vertices that adjanced to the vertex with number i.
for (int i=0; i<N; i++) {
int a, b;
cin >> a >> b;
vect[a].push_back(b);
vect[b].push_back(a);
}

a@b
b@a
a@a
b@b
4

I use suffix array and get AC

The same error!
I don't understand why and what does mean acess violation?
Help, please!

program p1136;
  type arbore=^nod;
       nod=record
             info:word;
             left,right:arbore;
           end;
  var t,r,q:arbore;
      n,i:integer;
      a:array [1..3000] of word;
procedure postordine(t:arbore);
  begin
    if t<>nil then
      begin
        postordine(t^.right);
        postordine(t^.left);
        write(t^.info,' ');
      end;
  end;
begin
  readln(n);
  for i:=n downto 1 do read(a[i]);
  new(t);new(q);new(r);t^.info:=a[1];
  for i:=2 to n do
    begin
      r:=t;
      while q<>nil do
        begin
          if a[i]>r^.info then begin new(q);q:=r^.right;end
                          else begin new(q);q:=r^.left;end;
          if q<>nil then r:=q;
        end;
      new(q);q^.info:=a[i];
      if r^.info<a[i] then r^.right:=q else r^.left:=q;
    end;
  postordine(t);
end.

yes, if Holden is no Hater.

both

This test is incorrect. If a child says that he is innocent then mother write down number 0 for his answer. So test could look as follows:
1
2
0 0
Certainly, answer is
NO