does anybody knows what is this fucking test 5?
if you know write the answer please:)

type
    point=record x,y:longint; end;
var
    vert:array[1..10000] of point;
    visited:array[1..10000] of byte;
    min:point;
    n,i,k,imin,minvert:integer;
Procedure Search_Min_Polar_Angle;
var
    i:integer;
    min_angle,angle,tg,a,b:real;
begin
    min_angle:=pi;
    for i:=1 to n do
        if (visited[i]=0) then
            begin
                a:=(vert[i].y-vert[imin].y);
                b:=(vert[i].x-vert[imin].x);
                if (b=0) then angle:=pi/2
                         else
                begin
                    tg:=a/b;
                    angle:=arctan(tg);
                end;
                if (angle<min_angle) then
                    begin min_angle:=angle; minvert:=i; end;
            end;
    visited[minvert]:=1;
end;

Begin
    readln(n);
    for i:=1 to n do visited[i]:=0;

    // naxojdenie samoy nijney levoy to4ki
    min.y:=maxlongint;
    for i:=1 to n do
        begin
            readln(vert[i].x,vert[i].y);
            if (vert[i].y<min.y) then
                begin min:=vert[i]; imin:=i; end
            else
            if (vert[i].y=min.y) and (vert[i].x<min.x) then
                begin min:=vert[i]; imin:=i; end;
        end;
    visited[imin]:=1;
    //-----------------------------------

    for k:=1 to n div 2 do Search_Min_Polar_Angle;
    writeln(imin,' ',minvert);
End.

Edited by author 09.05.2010 19:47

It says:
17=2^4+2^0
18=2^4+2^1
20=2^4+2^2

But, since the limit is [15;20], 16 must also be included:
16=2^3+2^3

I do not know why I am wrong, Netbeans works OK, but can not be accepted, below is my code.

public class Main {

    public static void main(String[] args) {
        System.out.println("Addition of Two Numbers(a,b) the sum is c");
        int a = Integer.parseInt(args[0]);
        int b = Integer.parseInt(args[1]);
        int c = a+b;
        System.out.println( "c="+c );
    }

}

thanks.

my program variation:

#include <stdio.h>
#include <math.h>

int main()
{
    int nrstones, i, min, w1, w2;
    int *weights;

    scanf("%d", &nrstones);
    weights = new int[nrstones];
    for(i = 0; i < nrstones; i++)
    {
        scanf("%d", &weights[i]);
    }

    min = weights[0];

    int perm = 1 << nrstones;

    for(i = 0; i < perm; i++)
    {
        w1 = 0;
        w2 = 0;
        int j;
        for(j = 0; j < nrstones; j++)
        {
            if((i >> j) % 2 == 0)
                w1 += weights[j];
            else
                w2 += weights[j];
        }

        if(min > abs((double)w1 - w2))
        {
            min = abs((double)w1 - w2);
            if(min == 0)
            {
                printf("%d", min);
                return 0;
            }
        }
    }
    printf("%d", min);
    return 0;
}

Edited by author 09.05.2010 11:37

Nice problem!
To solve it your needto use the topological sort on the graph (DAG).
It can be done with DFS algorithm.

2
10

4
100


14
266095289560

20
187019610394369600


Edited by author 09.05.2010 03:36

{$N+}

var x, y, temp:extended;
    n, l, r, m, sum, res:extended;

function rnd(t:extended):extended;
begin
     if frac(t)>=0.5 then rnd:=trunc(t)+1
                     else rnd:=trunc(t);
end;

begin
     readln(x, y, n);
     l:=10; r:=10*n; x:=x*10; sum:=-1;
     while l<=r do begin
           m:=trunc((l+r)/2);
           temp:=rnd(m/n*10);
           if abs(temp-x)<10e-12 then begin
                                     sum:=m;
                                     l:=m+1;
                                     continue;
                                     end;
           if temp<x then l:=m+1
                     else r:=m-1;
           end;
     {writeln(sum:0:0);}
     if sum=-1 then begin
                    writeln('Impossible');
                    exit;
                    end;
     l:=0; r:=2000000000; res:=-1;
     while l<=r do begin
           m:=trunc((l+r)/2);
           temp:=(sum+m)/(n+m);
           if rnd(temp*10)<=y*10 then begin
                                      res:=m;
                                      r:=m-1;
                                      end
                                 else l:=m+1;
           end;
     if res=-1 then writeln('Impossible')
               else writeln(res:0:0);
end.

Eight times in a row i'm given "compilation error"! The following online C compiler: http://www.delorie.com/djgpp/compile/ says everything it's okay. No errors/warnings there...

Кто знает 6 тест напишите пожалуйста

Can it be solved using qsort+n*bs(on x coordinates of points)
greetings

var
   x:array[1..20] of integer;
   y:array[1..20] of longint;
   min,s:longint;
   n,i,v:longint;
procedure outit;
var i:integer;
    s,c:longint;
begin
   s:=0;
   for i:=1 to n do
    inc(s,x[i]*y[i]);
   c:=v-s;
   if c>s then  s:=c-s else s:=s-c;
   if s<min then min:=s;
end;
procedure search(k:longint);
var i:integer;
begin
   if k>n then begin outit;exit;end;
   for i:=0 to 1 do
         begin x[k]:=i;search(k+1);end;
end;
begin
   readln(n);
   min:=0;
   v:=0;
   for i:=1 to n do read(y[i]);
   for i:=1 to n do begin min:=min+y[i]; end;
   v:=min;
   search(1);
   writeln(min);
end.

I think I'm right.

I don't know how could i choose next point.
I have some ideas about it:
1. The solution isn't very complexity(maybe it uses sorting)
2. I think, Elephpotamus always can eat all pumpkins
3. And our path should be spin or smth like it.
Help plz

I WA at Test 4.
Who can tell me what the test 4 is?

Edited by author 04.04.2010 10:48

what if in one way we can reach , but on opposite way we can't?
for example,what is output for this test?
input:
0 3

program temp;
var
  a:array [1..1000000] of qword;
  i,m:longint;
begin
  while not eof do begin
    inc(i);
    read(a[i]);
  end;
  for m:=i downto 1 do writeln(sqrt(a[m]):0:4);
end.

What's the problem?
为何#1就不过？

Edited by author 05.05.2010 19:51

Must we perform the quests in the same order, that they were listed in input?

#include <stdio.h>
int main()
{
    int n1,c1,n2,t,c2,n3,k,i;
    scanf("%d %d",&n1,&c1);
    scanf("%d %d %d",&n2,&t,&c2);
    scanf("%d",&n3);
    scanf("%d",&k);
    int (*A)[2] = new int[k][2];
    for( i=0; i<k; ++i )
    {
         scanf("%d:%d",&A[i][0],&A[i][1]);
    }
    int basic = n1,combined = n2,unlimited = n3,bas=0,com=0;

    for( i=0; i<k; ++i )
    {
         if(A[i][0]!=0)
         {
                        bas+=A[i][0]; com+=A[i][0];
         }
         if( (A[i][0]==0 && A[i][1]>6) || (A[i][0]!=0 && A[i][1]!=0 ) ) { ++bas; ++com; }
    }
    if ( com>t )
         {
               com-=t; com*=c2;
         }
    bas *= c1;
    basic += bas; combined += com;

    printf("Basic:     %d\n",basic);
    printf("Combined:  %d\n",combined);
    printf("Unlimited: %d\n",unlimited);
    delete [] A;
    return 0;
}