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 (lD<1e10) D=l;} printf("%0.6f\n",acos((2D)/2.0)*r); I believe it is correct..So can anybody help me with finding a bug, please? Does anyone have a same problem? I had exactly this kind of problem. WA13. then I wrote: double tmp= (2  minL)/2.0; tmp=0.00000000001; if (tmp<1) tmp=1; return r*Math.acos(tmp); and AC. It is problem with precision, of course. Thanks. I find my mistake. Edited by author 10.06.2005 01:33 Could anybody give my some difficult sample? I wanna check not trevial source data 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. I have get AC again, but problem with test#3 is keep open 0.00 != 0.00 Edited by author 08.04.2007 11:42 Problem with "0.00" has been resolved yesterday. But your WA3program outputs "1.#R". Edited by author 08.04.2007 13:01 I have send my program on your email what is mean ? 1.#J 1.#R ??? Can be result of sqrt(1) or acos(5), etc... 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! omg! i've read my first message and have terrified! some new hint: it's the problem with precision =) 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. No, I use more than one floating point division and square roots in my program and I managed to get AC. My only bug was with the test, where points lie in the opposite to each other. There are no "bad" and "unpity" tests in this problem. And simple solution with floats can be AC'ed just after a bit of thinking ;) Edited by author 09.04.2007 04:47 if program writes 0, your [censored] checker says WA#3 inspite of accuracy in 2 digits. abs(00) < 0.001 ! Edited by moderator 06.04.2007 22:54 I am tired of you who offends Timus Online Judge again and again. If you continue writing your posts to this board such a way, I will just remove them completely. Edited by author 06.04.2007 22:55 After your edit, observers could have an impression that I have used an abusive word. It's not true, it was word "stupid" you are wrong there is fragment of my program : double Norm(double x) { // if (x>0.001) return x; // else // return 0; } ... printf("%.2lf\n",Norm(abs(...))); ... with these commets program gets WA#3 without  AC if you want, i can send you my program entirely Edited by author 07.04.2007 22:15 Your WA and AC solution have are wrong both but tests are weak. You have WA after rejudge. program ural1331; const maxm=5000; zero=1e6; var w,l,x,y,z:array[1..maxm]of real; n,m,i,j:word; r,ww,ll,xx,yy,zz,ans:real; procedure calcoord(var w,l,x,y,z:real); begin w:=w*pi/180; l:=l*pi/180; x:=cos(w)*cos(l); y:=cos(w)*sin(l); z:=sin(w); end; function min(a,b:real):real; begin if a<b then min:=a else min:=b; end; function arcsin(s:real):real; begin if s>1zero then arcsin:=pi/2 else arcsin:=arctan(s/sqrt(1s*s)); end; begin read(n,m,r); for i:=1 to m do begin read(w[i],l[i]); calcoord(w[i],l[i],x[i],y[i],z[i]); end; for i:=1 to n do begin read(ww,ll); calcoord(ww,ll,xx,yy,zz); ans:=1e38; for j:=1 to m do ans:=min(ans,sqr(xxx[j])+sqr(yyy[j])+sqr(zzz[j])); writeln(arcsin(sqrt(ans)/2)*2*r:0:2); end; end. Edited by author 26.10.2004 00:27 Middle time of speed can be taken by using simple qsort of points by key=max(y[i],i=1,..3) and then logtime search of pozition of point x[k] under consideration in sorted array with cutting off parts of array y which far enough from the pozition. 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 2moderators: Why some people can print their solutioncod on forums, but some can't? What it depend on? Number of solved problems? Nickname(for example if I'll write my email in brackets after my name it will be more respectful...)? Noone is allowed to post any code here in accord with the fourth rule of Timus Online Judge Webboard ("Messages should NOT contain source code..."). Despite the rank and respect rate. Even I cannot :) But there are some old topics which contain source code  the moderators just have no time to delete them all. THX for explanation. But I misunderstood sth also. Some people put at the end of their's report's headline '(+)' or '()'. What it's meen? '()' means the body of the message is empty, because everything is said directly in the subject. If my message is short enough to fit the subject, I just warn the reader that there is nothing more in the body... '(+)' means I want to say something else in the body. 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 email 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((x1x2)*(x1x2) + (y1y2)*(y1y2) + (z1z2)*(z1z2)); 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 nowans<zero then ans:=now; } If I make zero=1e6,WA.But if make zero=0 or zero=1e10 it got AC... If it is about precision, why zero=0 OK? Edited by author 04.03.2005 22:20 
