Test#4
abaabaaba

ans:
abaabaaba    //for WA4, because S2 must be not empty!
abaabaabaaba //for AC

Edited by author 25.11.2011 17:57

For this task, pay attention to last few sentences.
>>Each subsequent value must be strictly less than the preceding one.<<
and
>>If several answers exist, output any of them. It is guaranteed that there is at least one consistent sequence.<<

Hint: use binary search

Edited by author 24.10.2014 14:27

> > Someone said that it can be solved by Greedy. Just use
net
> > flow to catch the largest number of "+". And then so.
But
> > they all said that it is a wrong arithmetic. Could you
> tell
> > me your way to solve it?

I can poste my bad code, can someone help me?

var
  a, b, n, x, y: int64;

begin
  readln(n, x, y);
  write('King: ');
  if n = 1 then writeln(0) else
  if ((x = 1) and (y = 1)) or ((x = n) and (y = 1)) or ((x = 1) and (y = n)) or ((x = n) and (y = n)) then writeln(3) else
  if (x = 1) or (y = 1) or (x = n) or (y = n) then writeln(5) else writeln(8);
  write('Knight: ');
  if (n = 1) or (n = 2) or ((n = 3) and (x = 2) and (y = 2)) then writeln(0) else
  if (((y = 1) and (x = 1)) or ((y = n) and (x = n)) or ((y = 1) and (x = n)) or ((y = n) and (x = 1))) or (n = 3) and ((x = 2) or (y = 2)) then writeln(2) else
  if ((x = 2) and ((y = 1) or (y = n))) or ((x = n - 1) and ((y = 1) or (y = n)) or ((y = 2) and ((x = 1) or (x = n))) or ((Y = N - 1) and ((x = 1) or (x = n))))  then writeln(3) else
  if (((x >= 3) and (x <= n - 2)) and ((y = 1) or (y = n))) or (((y >= 3) and (y <= n - 2)) and ((x = 1) or (x = n))) or ((x = 2) and ((y = 2) or (y = n - 1)) or ((x = n - 1) and ((y = 2) or (y = n - 1)))) then  writeln(4)  else
  if (((x >= 3) and (x <= n - 2)) and ((y = 2) or (y = n - 1))) or (((y >= 3) and (y <= n - 2)) and ((x = 1) or (x = n)))  then writeln(6) else writeln(8);

  if n = 1 then a := 0 else
  if (n - x >= x) and (n - x >= n - y) and (n - x >= y) then a := (n - 1) + 2 * (x - 1) else
  if (x >= n - x) and (x >= n - y)   and (x >= y)   then a := (n - 1) + 2 * (n - x) else
  if (y >= x)   and (y >= n - y)   and (y >= n - x) then a := (n - 1) + 2 * (n - y) else
  if (n - y >= x) and (n - y >= n - x) and (n - y >= y) then a := (n - 1) + 2 * (y - 1);
  if n = 1 then b := 0 else b := (n - 1) * 2;
  writeln('Bishop: ', a);
  writeln('Rook: ', b);
  writeln('Queen: ', a + b);
end.

Edited by author 01.06.2016 11:32

I had OLE #4, and it turned out to be a minor slip in my code. That is, for commands of type «if smth1 <operator> smth2 goto label, i was searching smth1 and smth2 indices in my variable indices array; label i should have searched in my goto indices array, but i accidentally was searching a label index in variable array too; as a result, it returned 0 (which it returned in case if the needed element was the first one, and if the element wasn't found, which shouldn't be happening). So it jumped to the first label and continued from there.

Example:

label1:
A = 10
print 100
label2: goto label2

Due to a mistake, "label2" has been looked for in variables array, which only had "A" in there. It was not found, it returned 0, it jumped to label1, tons of print commands were executed. After mistake is fixed, "label2" is looked for inside labels array, and correctly returns 1.

I can't say for sure you have the very same mistake, but still it is likely just a minor slip like this one.