Remember, that (0,1) belong to the FIRST Card.
If your get RE - check, that after Sorting every number < nCards =)

Well I really don't know where my mistake is....
My solution becomes right solution for a test case like this:
IN:
40 39
AB
BA AB BB AA AA BB BB BB AA AA AB BA AB BA AB BA AB BA AB BA
BA AB BB AA AA BB BB BB AA AA AB BA AB BA AB BA AB BA AB BA
OUT:
YES
EDIT:(5712638724) - my mistake
( i hope this test case is right :)

But what ever I do... I become WA on test 4 :S

Edited by author 13.04.2009 23:49

give me some test 8 please!!!

in test 2 checkers go out of desk
in test 6 I don't know but something terrible:)
solution without common sense got AC but you add some test with common sense and now I don't know how to get AC on this problem
we give you some tests...
OK you added them? but you don't delete wrong existing tests

It's not clear if black player can continue movement or not after eating white bishop.

var i,j,n,b2,n0,mid:integer;
x,y,al:array[1..10000] of real; num:array[1..10000] of integer;

begin
readln(n);
for i:=1 to n do begin
readln(x[i],y[i]);
num[i]:=i;
end;

n0:=1;
for i:=1 to n do begin
if x[i]<x[n0] then n0:=i;
if (x[i]=x[n0]) and (y[i]<y[n0]) then n0:=i;
end;

for i:=1 to n do
if i<>n0 then begin
if x[i]=x[n0] then al[i]:=pi/2
else al[i]:=arctan((y[i]-y[n0])/(x[i]-x[n0]))
end;

for i:=1 to n-1 do         {Sorting by angle}
for j:=i+1 to n do
if (al[j]<al[i]) and (i<>n0) and (j<>n0) then begin
b2:=num[i]; num[i]:=num[j]; num[j]:=b2; end;

for i:=1 to n div 2 do if i<>n0 then inc(mid) else inc(mid,2);
writeln(num[n0],' ',num[mid]);
end.

Edited by author 05.01.2009 13:11

Edited by author 05.01.2009 13:12

Can anyone provide me atleast initial test case fro problem# 1068??

Edited by author 22.04.2009 14:45

I used only 0.001 sec... it`s just O((log_B^N)^2).

u can check it here.
http://acm.timus.ru/status.aspx?space=1&pos=841174

http://cm2prod.baylor.edu/ICPCWiki/Wiki.jsp?page=Results%20World%20Finals%202009

Problems: http://icpcdmt.csc.kth.se/2009Problems.pdf

Edited by author 21.04.2009 23:04

Do we have to find two smallest mst that differ from each other by one edge?

#include <iostream>
using namespace std;
int main()
{
bool t;
int n,nn;
cin>>n;
short a[n];
for(int i=0;i<n;i++)
{
t=true;
cin>>nn;
if((i-1>=0&&a[i-1]<=nn)||(i==0))
a[i]=nn;
else
for(int j=0;j<i;j++)
{
if(a[j]>nn)
{
for(int jj=i;jj>j;jj--)
a[jj]=a[jj-1];
a[j]=nn;
t=false;
}
if(!t)break;
}
}
int c=0;
for(int i=0;i<(n/2)+1;i++)
c+=a[i]/2+1;
cout<<c;
}

compiling error??

Edited by author 07.10.2007 21:56

Hi,

For input 4 17, how come the output is 2 15 ? We have to do this right ?

sqrt(4 % 0) = undefined
sqrt(4 % 1) = 0
sqrt(4 % 2) = 0
sqrt(4 % 3) = 1
sqrt(4 % 4) = 0
sqrt(4 % 5) = 2
sqrt(4 % 6) = 2
sqrt(4 % 7) = 2
sqrt(4 % 8) = 2
sqrt(4 % 9) = 2
sqrt(4 % 10) = 2
..
..
sqrt(4 % 16) = 2

So where is 2 15 coming from ? Is there something wrong with the problem ?

Edited by author 20.04.2009 03:46

Edited by author 20.04.2009 03:46

I tried with Scanner and BufferedReader, but have no luck :(
Help????

Could anyone give me the algorithm? I have been trying to solve this problem for a week, but I couldn't. My problem is TLE#3. My idea is to brut-force from sqrt(N), but it doesn't work.
I would be very glad for the solution!

I wanted to put my solution , but I had some problem
   Sorry Kula meybe you didn't take my solution
                 write here your e-mail adress
                                   I'll send you

Edited by author 08.02.2006 14:05

Could you give me test #10?
I made an algorithm which runs in O(n*logn*logn)which gives the same results as the trivial one (sort for suffix array and then comparing every two adjacent O(n^3)), but it doesn't pass this test, while the previously mentioned trivial one does (it gets stuck at #29). I supose I have some kind of trivial error, so it would be rather helpfull.