Сергей used only small letters or There can be Title LETTERS

1 - 2 3 1 4 5
2 - 3 1 2 4 5
3 - 2 3 1 4 5


1 and 3 are same

var a:array[0..10000]of integer;
    max,i,j,n,k,l,group1,group2:longint;
begin
 read(n);
 for i:=1 to n do
  read(a[i]);
 for i:=1 to n-1 do
  begin
   k:=i;
   max:=a[i];
   for j:=i+1 to n do
    if a[j]>max then
     begin
      max:=a[j];
      k:=j;
     end;
    a[k]:=a[i];
    a[i]:=max;
  end;
  for i:=3 to n do
  if group1<group2 then
   group1:=group1+a[i]
  else
   group2:=group2+a[i];
 writeln(abs(group1-group2));
end.

... Three different *postitve* integers (a, b and c) ...

Есть еще вот какое предложение:
1) в топе по лучшим отправкам создать возможность выбрать язык лучших отправок... чтобы видеть лучшие программы на С++, Паскале, С#, JAVA отдельно
2) Чтобы в лучших отправках сортировалось не только по времени, но и по кол-ву используемой памяти. (сейчас не так, видел отправки с лучшей памятью на худших местах)

Edited by author 16.08.2010 16:30

Hi, I tried to use java to solve this problem but failed, yet do not know where is the bug,
I have pasted the source code here, wish someone can help me out. I am very very grateful.

public class ReversRoot1001 {
    public static void main(String args[])throws IOException{
        BufferedReader br = new BufferedReader(new FileReader("ReversRoot1001.txt"));

        Vector<Integer> record = new Vector<Integer>();

        String line;
        while((line = br.readLine())!=null){
            Scanner sc = new Scanner(line);
            int a = sc.nextInt();
            record.add(a);
//            if(sc.hasNext()){
//                int b = sc.nextInt();
//                //int c = sc.nextInt();
//                record.add(b);
//            }
            while(sc.hasNext()){
                int b = sc.nextInt();
                record.add(b);
            }
        }

        for(int i = record.size();i<record.size();i--){
            System.out.println(record.get(i));
        }
        br.close();
    }

}

what would be the distance between point (0,0) and square (1,0) (-1,0)?? Is it equal 0 or 0.7??

Does anyone knows what is tested in the test 6??? Or anyone have examples????

Gettin' wrong answer. Can anybody post it, please!?

If you encounter any problems please describe it in this thread.

I think that test 16 have at least two codirectional vectors.

My AC solution crashes on this test
2 2
**
**

#include <stdio.h>
#include <math.h>
#define M_PI_4     0.785398163397448309616

double d[51] = {0};
char printed[51] = {0};

struct point {
double x;
double y;
};

struct square {
point a;
point b;
} squares[51];

point rotate(point a, double d_al){
point _a;
_a.x = a.x * cos(d_al) - a.y * sin(d_al);
_a.y = a.x * sin(d_al) + a.y * cos(d_al);
return _a;
}

double get_d(point a, point p) {
return sqrt((double)((a.x - p.x)*(a.x-p.x) + (a.y -p.y)*(a.y - p.y)));
}

int main () {
    int n;
    scanf("%d",&n);
    point p;
    int i;
    for (i=1;i<=n;i++) {
        scanf("%lf%lf%lf%lf", &squares[i].a.x,&squares[i].a.y,&squares[i].b.x,&squares[i].b.y);
    }
    scanf("%lf%lf", &p.x,&p.y);
    double tn;
    double alpha,d_al;
    point _a,_b,_p,a,b,temp;
    for(i=1;i<=n;i++) {
        b = squares[i].b;
        a = squares[i].a;
        if(a.x == b.x && a.y == b.y) {
            d[i] = get_d(p,a);
            continue;
        }
        tn = (double)(b.y - a.y)/(double)(b.x - a.x);
        alpha = atan(tn);
        d_al = M_PI_4 - alpha;
        _a = rotate(a,d_al);
        _b = rotate(b,d_al);
        _p = rotate(p,d_al);
        if( _a.x > _b.x) {
            temp = _b;
            _b = _a;
            _a = temp;
        }

        if(_p.y > _b.y) {
            if(_p.x < _a.x) {
                temp.x = _a.x;
                temp.y = _b.y;
                d[i] = get_d(_p,temp);
                continue;
            }
            if(_p.x > _b.x) {
                d[i] = get_d(_p,_b);
                continue;
            }
            d[i] = _p.y - _b.y;
            continue;
        }
        if(_p.y < _a.y) {
            if(_p.x > _b.x) {
                temp.x = _b.x;
                temp.y = _a.y;
                d[i] = get_d(_p,temp);
                continue;
            }
            if(_p.x < _a.x) {
                d[i] = get_d(_p,_a);
                continue;
            }
            d[i] = _a.y - _p.y;
            continue;
        }
        if(_p.x < _a.x) {
            d[i] = _a.x - _p.x;
            continue;
        }
        if(_p.x > _b.x) {
            d[i] = _p.x - _b.x;
            continue;
        }
        d[i] = 0;
    }

    double min;
    int j,k;
    for(i=1;i<=n;i++) {
        min = 99999999999999999;
        k = 0;
        for(j=1;j<=n;j++)
            if(!printed[j] && min > d[j]) {
                min = d[j];
                k = j;
            }
        printed[k] = 1;
        printf("%d ", k);
    }

return 0;}

Don`t forget about d = 0 and p = 1, =)

Edited by author 16.08.2010 03:58

What change if shift beginning time of the exam to more early
time ? Please explain to me. Thank!

who can explain the problem?

Let S4(n),S5(n),S6(n)- number of different
4-poligons,5-poligons and 6-poligons in n poligon.
I have AC using formula F(n)=4*S4(n)+5*S5(n)+S6(n).
But this formula can not be right because
in different 6-poligons 3 hords may intersect in
the same point and this triangle will be counted
more the once.

To do calculations while [kx,ky] and [(k+1)x,(k+1)y] not intersect is very slow solution