ограничение по времени для Питона не выполняется...

I get WA 13 again and again and I can not find a mistake. My main logic looks like this:
for(j=0;j<n;j++)
{l=(x[j]-X)*(x[j]-X)+(y[j]-Y)*(y[j]-Y)+(z[j]-Z)*(z[j]-Z);
if (l-D<1e-10) D=l;}
printf("%0.6f\n",acos((2-D)/2.0)*r);

I believe it is correct..So can anybody help me with finding a bug, please? Does anyone have a same problem?

Thanks.
I find my mistake.

Edited by author 10.06.2005 01:33

61 submits lost AC verdict (many because of tricky 12 test).
Try to find a bug in your source code. Good luck!

P.S. Thanks to Alias (Alexander Prudaev) for idea of new tests.

Notice, that scalar product of two normalized vectors can be more than 1 on the module.

Заметьте, что скалярное произведение двух нормализованных векторов по модулю может быть больше 1.

Good luck!

Give me limitations for R, wi, li, Wi and Li. I do not know how small they are. So if R = 0.00000000000000001, this test is relly "tricky", I understand. Mischief managed :)

P.S. To get AC after rejudge you should avoid floating point division in your program. Just simplify your formula.

if program writes -0, your [censored] checker says WA#3
inspite of accuracy  in 2 digits. abs(-0-0) < 0.001 !

Edited by moderator 06.04.2007 22:54

I use it in my prog, but it gets TLE4. Help please. Here is my sol

[code deleted]

Edited by moderator 02.07.2006 13:25

I've implemented an octtree (I used http://www.nist.gov/dads/HTML/octree.html as a reference on this data structure, it's easy to understand. Just try it!)

I've got AC, but I cann't make it run quicker, than 0.4 sec. If you need my source code to help me, or have anything else to tell me, feel free to e-mail to urrel AT rambler DOT ru.

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

const double PI = 3.141592653;
int n,m,i,i2;
double radius;
double stp[5000],dtp[5000];
double sa,da,mindist,tmp;

double dist(double s1, double d1, double s2, double d2)
{
    double x1,y1,x2,y2,z1,z2,s,d;
    s = s1*PI/180.0;
    d = d1*PI/180.0;

    x1 = cos(s)*cos(d);
    y1 = cos(s)*sin(d);
    z1 = sin(s);

    s = s2*PI/180.0;
    d = d2*PI/180.0;

    x2 = cos(s)*cos(d);
    y2 = cos(s)*sin(d);
    z2 = sin(s);

    double a,b,c,alp;

    a = 1;
    b = 1;
    c = sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) + (z1-z2)*(z1-z2));

    alp = acos((a*a + b*b - c*c)/(2*a*b));

    return alp;
}

int main()
{
//    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
    scanf("%d %d",&n,&m);
    scanf("%lf",&radius);

    for (i = 0; i<m; i++)
    {
        scanf("%lf %lf",&stp[i], &dtp[i]);
    }

    for (i = 0; i<n; i++)
    {
        scanf("%lf %lf",&sa, &da);

        mindist=9999999;

        for (i2 = 0; i2<m; i2++)
        {
            tmp = dist(sa,da,stp[i2],dtp[i2]);
            if (tmp<mindist) mindist = tmp;
        }
        printf("%.2lf\n",mindist*radius);
    }

    return 0;
}

some of my code:
{
  now:=sqr(x1[i]-x)+sqr(y1[i]-y)+sqr(z1[i]-z);
  if now-ans<zero then ans:=now;
}

If I make zero=-1e-6,WA.But if make zero=0 or zero=-1e-10 it got AC...

If it is about precision, why zero=0 OK?

Edited by author 04.03.2005 22:20