50 38 17 9 9

Here it is my code:

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

#define  eps 1e-16

struct plate
{
  int  id;
  double  r, x, y;
} p[3], pp[3], tp;

double  h, w, temp;

double  sq(double a) {
    return  a*a;}
int  possible();
void  output(int k);

int  main()
{
  int  i, j, k;

  scanf("%lf%lf%lf%lf%lf", &w, &h, &pp[0].r, &pp[1].r, &pp[2].r);
  pp[0].id = 0;
  pp[1].id = 1;
  pp[2].id = 2;

  for  (i = 0;  i < 3;  ++i)
    for  (j = 0;  j < 3;  ++j)
      for  (k = 0;  k < 3;  ++k)  if  (i != j  &&  i != k  &&  j != k) {
        p[0] = pp[i];
        p[1] = pp[j];
        p[2] = pp[k];
        if  (possible())  {
          output(0);
          return 0;
        }
      }

  printf("0");
  return 0;
}

////////////////////////////////////////////////////////////////

int  possible()
{
  p[0].x = p[0].r;
  p[0].y = p[0].r;

  p[1].x = w - p[1].r;
  p[1].y = p[1].r;

  if  (sq(p[0].x - p[1].x) + sq(p[0].y - p[1].y)  - sq(p[0].r + p[1].r)  <  -eps) {
    p[1].y = p[0].y + sqrt(sq(p[0].r + p[1].r) - sq(p[1].x - p[0].x));
    if  (p[1].y + p[1].r - h > eps)
      return 0;
  }

  p[2].x = p[2].r;
  p[2].y = h - p[2].r;

  if  (sq(p[0].x - p[2].x) + sq(p[0].y - p[2].y)  - sq(p[0].r + p[2].r)  <  -eps) {
    p[2].x = p[0].x + sqrt(sq(p[0].r + p[2].r) - sq(p[2].y - p[0].y));
    if  (p[2].x + p[2].r - w > eps)
      return 0;
  }

  if  (sq(p[1].x - p[2].x) + sq(p[1].y - p[2].y)  - sq(p[1].r + p[2].r)  <  -eps)
    return 0;

  return 1;
}

void  output(int k)
{
  if  (k == 1) {
    double  t;
    for  (k = 0;  k < 3;  ++k) {
      t = p[k].x;
      p[k].x = p[k].y;
      p[k].y = t;
    }
  }

  int  i, j;

  for  (i = 0;  i < 3;  ++i)
    for  (j = i+1;  j < 3;  ++j)
      if  (p[i].id > p[j].id) {
        tp = p[i];
        p[i] = p[j];
        p[j] = tp;
      }

  for  (k = 0;  k < 3;  ++k)
    printf("%.4lf %.4lf ", p[k].x, p[k].y);
}

Is it forbidden to post WRONG programs here?
Anyways my idea is :
If there exists a solution then there exists a solution where two circles are in the corners of the rectangle. It is obvious because there is always the rightmost and the leftmost circle and we can drag them all the way to the corner.
So what I do is: I take every pair of circles, try to put each of them in every corner (16 variants for a pair of circles). Then I "erase" all the area, where the center of the 3rd circle can not be located. Firstly, the last circle (I mean it's center) has to be located inside a rectangle (W-2R)x(H-2R) whre R - is the radius. We also must exclude two circles with radii R1+R and R2+R where R1 and R2 are the radii of the already existing circles.
I store all the intersections of the 2 circles with the 4 lines (sides of the rectangle) and check all these points if they are possible positions for the last circle's center.
Please tell me where am I wrong, or give me some critical test data.
Thanks in advance.
alex[LSD], a dumb russian programmer.

My solution is a full search! We should find all places for every plate: between two plates, plate & wall, 2 walls!!!

Edited by author 04.09.2005 15:17

Edited by author 04.09.2005 15:17

Edited by author 05.09.2005 21:11

Set one circle in the up right corner, other in the down left corner, and third set where I can (after setting first two).
Thanks.

I think my pure-geometry based solution is correct, and I can not imagine test which it fails on, but it received WA(18). So I used some brute force optimizations and got AC, but I still don't know why was my first solution wrong...