Edited by author 18.07.2008 17:47

is operations of int64(pascal) slower than long long(c++)?

http://timus.h17.ru/

Edited by author 06.02.2007 03:15

Can any one tell me how to solve this problem using greedy approach pls..?

I can't find my mistake.
Can you sent me test#2 to <<mihran91@mail.ru>>?
this is my code


# include <iostream.h>
# include <math.h>
main ()
{
    double dx[50001],dy[50001];
    int i,n;
    double x0,y0,xk,yk,x1,y1;
    cin>>n;
    for(i=0;i<n;i++)
        cin>>dx[i]>>dy[i];
    dx[n]=dx[0];dy[n]=dy[0];
    xk=0.5*(dx[0]+dx[1]);
    yk=0.5*(dy[0]+dy[1]);
    x1=dx[0]-dx[1];
    y1=dy[0]-dy[1];
    y0=yk+0.00000001;
    x0=(xk*x1-y0*y1+yk*y1)/x1;
    double ma,mb,a[2],b[2],det,otv=0;
A:    for(i=0;i<n;i++)
    {
        a[0]=dx[i]-x0;
        a[1]=dx[i+1]-x0;
        b[0]=dy[i]-y0;
        b[1]=dy[i+1]-y0;
        ma=sqrt(a[0]*a[0]+b[0]*b[0]);
        mb=sqrt(a[1]*a[1]+b[1]*b[1]);
        det=a[0]*b[1]-b[0]*a[1];
        if(det>0)
            otv+=acos((a[0]*a[1]+b[0]*b[1])/(ma*mb));
        else
            otv-=acos((a[0]*a[1]+b[0]*b[1])/(ma*mb));
    }
    if(fabs(fabs(otv)-6.28318530717)>0.00001)
    {
        otv=0;
        y0=yk-0.00000001;
        x0=(xk*x1-y0*y1+yk*y1)/x1;
        goto A;
    }
    if(otv<0)
        cout<<"cw";
    else
        cout<<"ccw";
    return 0;
}

May number C has more digest, then numbers A and B?

I think that answer is amount of digits |A+B-C|

Sorry, it's incorrect

Edited by author 03.01.2007 23:51

please, tell me, what type of array i must take?
i took MLE on this : array[1..125000] of int64;

What's the answer for
------------------------------
30 8
15 16 19 17 18 20 21  27 30 29 28  26 25 24 23 22  14 13 12 11 10 5 4 3 2 1 6 7 8 9
------------------------------
30 10
15 16 19 17 18 20 21  27 30 29 28  26 25 24 23 22  14 13 12 11 10   5 4 3 2 1 6 7 8 9

------------------------------

The last line of the input file in the example reflects that the right limit of the black interval is 47. But the white interval in the output file begins with 47. How can it be??

If one supposes that limits of these intervals are not included then they are shouldn't be included in the 3rd and the 4th lines either. Then the white interval ends with 299 and another one begins with 301. But the answer would be "301 633" this way.

What is the right way to understand the terms of this problem?

Can you give me the test which answer is =0.
And I got WA#4 please give me the tests.

Edited by author 25.06.2008 12:15

Edited by author 25.06.2008 12:15

New tests were added. 112 authors lost AC verdict.

() 1

??
() ->0 ???

Are you sure that 23th test is correct ?

You should remove Black figures and White too.

Can you please recheck the problem statement because of disparity  of English and Russion versions. Coordinates of cockroaches are integer without any limitation in russian version, and english version says that all cordinates are real with some restriction on it.

Edited by author 16.06.2007 15:51

Help me!
Program Contest_B;
Var
  I, R : LongInt;
  Ans : Int64;
  Tmp : Extended;
Begin
  ReadLn(R);
  If R = 1 Then
    Ans := 4
  Else
    Begin
      Ans := 0;
      For I := R - 1 Downto 0 Do Begin
        Tmp := Sqrt(Sqr(R) - Sqr(I));
        If Tmp - Trunc(Tmp) <= 1E-9 Then
          Inc(Ans, Trunc(Tmp))
        Else
          Inc(Ans, Trunc(Tmp) + 1);
      End;
      Ans := Ans * 4;
    End;
  WriteLn(Ans);
End.

For people who will solve this problem in the future: it is possible to use floating point arithmetics and get AC with n^2*logn solution.

who can give me some test data?

My program is wa #19?

puzzling ?