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Discussion of Problem 1306. Sequence Median

var
   a : array[1..250001]of longword;
   n : longint;
   t:extended;
Procedure QS(m,n : longint);
var
 i,j,w,x : longint;
BEgin
   i:=m;
   j:=n;
   x:=a[(i+j)div 2];
   while i<=j do
   if a[i]<x then inc(i)
   else if a[j]>x then dec(j)
   else
   begin
   w:=a[i];
   a[i]:=a[j];
   a[j]:=w;
   inc(i);
   dec(j);
   end;
   if m<j then qs(m,j);
   if i<n then qs(i,n);

end;
PRocedure Run;
var
i : longint;
Begin
  read(n);
  for i:=1  to n do
  read(a[i]);
  qs(1,n);
  if n mod 2 = 1 then
  begin
  t:=a[(n+1) div 2];
  writeln(t:0:1)
  end
  else
  begin
     t:=(a[n div 2+1]+a[(n div 2)]) /2;
     writeln(t:0:1);
  end;
end;
BEgin
  run;
end.
or can it be so
var
n,i,j : longint;
t,d,f : double;
begin
  read(n);
  d:=0;
  for i:=1 to n do
    begin
     read(t);
     f:=t/n;
     d:=d+f;
    end;
    writeln(d:0:1);
end.
In first one you can't take array [1..25001] of longword.
Try to do it with less memory but with similar method.
Second program is wrong.