Could someone give me a hint (or test) to help me to fix WA on test #11?

Just use secondary diagonals of (1,1), (2,2),...,(n,n) matrixs (this wasn't regular plural of noun matrix , excuse me for my grammar).

p.s that was my way, maybe there is another

Edited by author 05.05.2005 07:06

Heh, i take some day to solve this problem. Solution, by Master Herr Nikita Glashenko (Никита Глащенко), because my solution got (i move sqare from 0 to pi/2) TLE (slow code). Do next: take one point and mirror it in front of and sheer from O(0,0). Example: 3 4, -3-4, 4-3,-4 3. And take the equalization of the line catched on  two points(x1,y1) (x2,y2). if size of chord(lying on the line catched on  two points(x1,y1) (x2,y2))=r*sqrt(2)[use approximation 1e-7 for example] in one of the four variants (for four point, what you get), then 'yes' else 'no'.
PS: Division by zero mit uns(with us), you should control it.

Edited by author 18.01.2009 21:44

How can some of the programs solve this problem within 0.001 s?

If Hobbit moves in order 1->4->2->3 (like in basic test answer), he will pay 2816. But if he moves in order 1->3->4->2, he will pay only 2050.
So why minimal price way is "1 4 2 3"?

Why there is not a section, how to write C# solutions???

Edited by author 18.01.2009 18:00

could anyone here tell me if there are any problem about "maximal matching problem"

I've read in one of the posts below that the problem is to be solved by recursive deleting of comments/arithmetics (at least i've got so).
I suppose the problem is solvable with char-by-char reading. It's necessary to remember only one previous character and to update some conditions like "we are in comment", "we are in arithmetic expression", ... at the proper time. I've got AC with such an algo.

Test:
2
aa
Answer: 6

Why not just Inf?

What's wrong with my program?Could you give me some test?
Thank you!
Sorry for my poor English!:)
Here is my program:
#include<iostream.h>
int main()
{
 freopen("financial.in","r",stdin);
 freopen("finiancial.out","w",stdout);
 int testcase;
 cin>>testcase;
 int i,m,n;
 bool flag;
 for (int t=1;t<=testcase;t++)
   {
    cin>>n;
    m=1;
    while (n%2==0)
      {
       n=n>>1;
       m+=1;
      }
    if ((m%2==1) && (n>1))
      cout<<"YES"<<endl;
    else {
          flag=false;
          for (i=2;i<=n-1;i++)
            if (n%i==0)
              {
               flag=true;
               break;
              }
          if (flag) cout<<"YES"<<endl; else cout<<"NO"<<endl;
         }
   }
 return 0;
}

Thank you for your help!

Hello! Sorry for my language. I have not found in C++ (Microsoft Visual Studio 2008) function for digit Pi, and input it self - "float Pi=3.1415926" and i recived WA in test 3, but when i use Pi=3.141_!6!_926 (It's wrong!) I got Accepted! What you think about this?

What is 5th test? could anybody help me?? Some ideas, some tests...

Give me plz some tests. Please only good permutation :)

I keep getting WA on testcase 4, but I'm sure my program is correct. Can someone give me this test or send me AC program or just tell me what is the trick in this testcase?

Can somebody help me with this test?

I think there is some small bug in my program or in the statement understanding, but I can't find it...
An appropriate test would help me a lot.

моя прога
Uses SysUtils,Math;
  {$APPTYPE CONSOLE}
var i,f,j,n,c,x,y,x1,y1:integer;
    a:array [-1..10,-1..10] of byte;
    b:array[1..8] of char=('a','b','c','d','e','f','g','h');
procedure proc(k,l:integer);
var min,i1,j1:integer;
begin
  for i1:=-2 to 2 do
    for j1:=-2 to 2 do
      if (abs(i1)+abs(j1)=3) and (a[k+j1,l+i1]=0) then
        inc(min);
  if min<c then
  begin
  c:=min;
  x1:=k;
  y1:=l
  end;

end;
begin
  {$ifndef online_judge}
  reset(input,'input.txt');
  rewrite(output,'output.txt');
  {$endif}
  fillchar(a,sizeof(a),1);
  readln(n);
  for i:=1 to n do
    for j:=1 to n do
      a[i,j]:=0;

  case n of
  1:writeln('a1');
  2:writeln('IMPOSSIBLE');
  3:writeln('IMPOSSIBLE');
  4:writeln('IMPOSSIBLE') else
  begin
    x:=1;
    y:=1;
    f:=0;
    repeat
    c:=10;
    a[x,y]:=1;
    writeln(b[x],y);
    for i:=-2 to 2 do
      for j:=-2 to 2 do
        if (abs(i)+abs(j)=3) and (a[x+j,y+i]=0) then
        proc(x+j,y+i);
    inc(f);
    x:=x1;
    y:=y1;
    until f>=sqr(n);
    end;
  end;

  end.
Фишка в том, что прога оптимально(!) обходит псведошахматную доску с заданными размерами ребер n<=8 и находит решение по обходу доски конем, чтобы он только один раз наступал на одно поле(я проверял), но правда, поскольку решение отличается от того что в тестах, приишлось сдать массивом конастант... обидно. Если есть возражения, пишите!