bzoj3689

这题做法很多
可以通过类似noi超级钢琴那道题目的做法用可持久化+trie来做
还可以直接在trie树上维护size域然后类似查找k大的做法做
总之还是比较水的

type node=record
       kth,num,ans:longint;
     end;

var heap:array[0..100010] of node;
    son:array[0..3000010,0..1] of longint;
    size:array[0..3000010] of longint;
    a:array[0..100010] of longint;
    t,i,n,m:longint;

procedure swap(var a,b:node);
  var c:node;
  begin
    c:=a;
    a:=b;
    b:=c;
  end;

procedure add(x:longint);
  var i,p,y:longint;
  begin
    p:=1;
    for i:=30 downto 0 do
    begin
      y:=x and (1 shl i);
      if y>1 then y:=1;
      if son[p,y]=0 then
      begin
        inc(t);
        son[p,y]:=t;
      end;
      p:=son[p,y];
      inc(size[p]);
    end;
  end;

function ask(x,k:longint):longint;
  var p,i,y:longint;
  begin
    p:=1;
    ask:=0;
    for i:=30 downto 0 do
    begin
      y:=x and (1 shl i);
      if y>1 then y:=1;
      if size[son[p,y]]>=k then p:=son[p,y]
      else begin
        ask:=ask+1 shl i;
        k:=k-size[son[p,y]];
        p:=son[p,1-y];
      end;
    end;
  end;

procedure sift(i:longint);
  var j:longint;
  begin
    j:=i shl 1;
    while j<=t do
    begin
      if (j+1<=t) and (heap[j].ans>heap[j+1].ans) then inc(j);
      if heap[i].ans>heap[j].ans then
      begin
        swap(heap[i],heap[j]);
        i:=j;
        j:=i shl 1;
      end
      else break;
    end;
  end;

begin
  readln(n,m);
  t:=1;
  for i:=1 to n do
  begin
    readln(a[i]);
    add(a[i]);
  end;
  for i:=1 to n do
  begin
    heap[i].num:=a[i];
    heap[i].kth:=2;
    heap[i].ans:=ask(a[i],2);
  end;
  t:=n;
  for i:=n div 2 downto 1 do
    sift(i);
  for i:=1 to 2*m do
  begin
    if i mod 2=1 then write(heap[1].ans,' ');  //注意会被重复计算
    if heap[1].kth=n then
    begin
      swap(heap[1],heap[t]);
      dec(t);
    end
    else begin
      inc(heap[1].kth);
      heap[1].ans:=ask(heap[1].num,heap[1].kth);
    end;
    sift(1);
  end;
end.