[code deleted]

Edited by moderator 10.08.2006 10:07

for(i=0;i<l-1;i++)
for(j=i+1;j<l;j++)
{
if(Max>l-j)break;
z=b[j].x-b[i].x;
M[0]=b[i].pos;
M[1]=b[j].pos;
l1=2;
xb=b[j].x+z;

for(k=j+1;k<l;k++)
{
if(Max>l-k+l1)break;
bs=BSearch(xb,k);
...

Maybe it is O(N^3*logN)?

PS: Even if it's O(N^2*logN) it won't pass TL, only O(N^2) can do it.

Any hints ?

Edited by author 09.08.2006 21:48

To Burunduk1: You are wrong. During the contest my O(N^2*logN) solution got accepted (with 0.25 sec). Magic "i++ -> ++i" and "inline" saved me. :)

Shaman :)

Edited by author 10.08.2006 10:09