My solution works 5.5 sec on the following test. I believe my computer is faster than yours. So, please add this test and make me lose AC.
10 27
a b
a c
a d
a e
a f
a g
a h
a i
a j
d h
d i
d j
e f
e g
e h
e i
e j
f g
f h
f i
f j
g h
g i
g j
h i
h j
i j
a

It seems someone have suggested to hold a common memory segment for all stacks. Here is my way to solve with some details.

Allocate a large array - a common segment for all stacks, and define the size of a block.
int a[BLOCKS*BLOCK_SIZE];
Each block stores (BLOCK_SIZE-1) cells for the numbers and one cells to link it with the previous block for the stack.

Also it is useful to have a bit map reflecting if a particular block is occupied. Make functions find_free_block and release_block.
Also i was needed a couple of arrays of int[1000].

When the next stack grows above (BLOCK_SIZE-1)*k, the new block is occupied and the stack's memory becomes BLOCK_SIZE*(k+1). On the other hand, when i can release a block after POP i do this for memory reuse.

It was found experimetally at the price of 5 MLE attempts that acceptable parameters are:
BLOCKS = 3200; //number of blocks
BLOCK_SIZE = 50; //integers, not bytes

Apparently, the first 12 tests are complete crap: I passed them using dx instead of dy and hypotenuse length instead of leg length. So the hint is to check your program for stupid mistakes.

Be sure to pass the following testcase:

6
1 10
1 10
1 10
1 10
1 10
6 3
1
4 1

z = fact(z)/(2*pow(log(2),n+1));

if(fmod(z,1)>=0.5) z = floor(z) + 1;
else z = floor(z);


I tried the example test case and got true, but got WA with 1st test case... T_T

WTF test?

Why ??

My best time for most programs is 0.015 at least. I see lots of 0.001s exec time . Can share how to write one ? Even the easiest problem A+B

Can someone give me test 6?

For example:
7 10
1 2
2 3
2 4
2 5
3 5
4 5
5 6
5 7
6 7
1 7
At first the DFS procedure find the loop:
(1,2,3,5,6,7,1),then it'll find (1,2,3,5,7,1).
and next,no loop with unused edge will be found!?!
In fact,the maximum T is 3(it's obvious),but use greedy
method + DFS the number of T only reach 2.
Unless we regulated the order of search
(it means DFS can't be right!),otherwise we have to
delete a found loop to let more loop to have unused
edge,then how to make the regulations?

Hello, i try send my programm but always get some WA. Can you explaint few things.
First:
In END of input ".?!"? When i added check for it i get 1WA.
Second:
"ssss[S]ssss" r there mistake? When i check for it i get 2WA
Third:
When i delete this conditions i get 10WA.
Im confused. I've readed all threads and added conditions for every case which i found. What in 10 test?

var
  a:array[0..14000] of longint;
  b:array[-1..200000] of longint;
  n,i,j,k,max:longint;

begin
  read(n);
  fillchar(a,sizeof(a),0);fillchar(b,sizeof(b),0);    b[-
1]:=10000;max:=-1;
  for i:=0 to n-2 do begin
    read(a[i]);
    if a[i]=-1 then begin writeln(max);halt;end;
    inc(b[a[i]]);
    if a[i]>max then max:=a[i];
  end;
  read(j);
  while j>=0 do begin
    inc(i);
    i:=i mod n;
    dec(b[a[i]]);a[i]:=j;
    inc(b[j]);
    if j>max then max:=j;
    while b[max]<=0 do dec(max);
    writeln(max);
    read(j);
  end;
end.

when will the new problems be add to problemset(-)

I have been thinking about the problem these days, and found that the solution is much related to the 4, 5 or 6-point groups of the N points.

For the former two situations, we could simply calculate that the total number of the figures is
4 * C(n, 4)
and
5 * C(n, 5)
but what have we do when there are 6-point groups? According to the problem statement, the chords must intersect pairwise, so the solution for 6-point groups will be based on the detail position of the points. For example, we must find out if the chords will parallel with each other.

On the other side, I found we have more than one ways to connect 6 points to form the figure. For those 6-point groups with no paralleling chords, we have actually 4 ways to connect them: (1-4, 2-5, 3-6), (1-4, 2-6, 3-5), (2-5, 1-3, 4-6) and (3-6, 1-5, 2-4). Here we numbered the points in CW or CCW order.

Could anybody tell me that why the correct answer could be
4 * C(n, 4) + 5 * C(n, 5) + C(n, 6), (especially the last term)
If my analysis above is right, this formula will be wrong even in N = 7.

Thanks in advance.

why????

some one pls explain sandras biography  plz

We got CE for some reason while our compilers manage to run it.

Для конкретного постамента, направление, куда переместится с него этот несчастный артефакт, заранее зафиксировано или он может с одного и того же постамента в одном и том же сценарии и направо ходить, и налево по своему желанию?