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

I got WA test 5!!!
why???

input :
5
1 6
2 5
3 4
10 100
20 30
wrong output :
2
5 4
right output :
3
3 2 1

Edited by author 13.08.2010 14:02

4 4
1 2 0
2 3 0
3 4 0
4 1 0
Impossible

My solution resulted in Crash {FLT_INVALID_OPERATION} several times, before I understood, that the square of an integer number can be negative!

Oh! What a stupid mistake!-)

The crash mentioned above was caused by the following function in my source code:

function Dist( const p1, p2: Point ): extended;
begin
    Result := sqrt( sqr(p2.x-p1.x) + sqr(p2.y-p1.y) );
end;

I couldn't even imagine, that sqr() function can produce negative results!

So, If you get Crash on test #6, just use extended (^: