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[..] of node;
     son:array[..,..] of longint;
     size:array[..] of longint;
     a:array[..] 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:=;
     for i:= downto  do
     begin
       y:=x and ( shl i);
       if y> then y:=;
       if son[p,y]= 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:=;
     ask:=;
     for i:= downto  do
     begin
       y:=x and ( shl i);
       if y> then y:=;
       if size[son[p,y]]>=k then p:=son[p,y]
       else begin
         ask:=ask+ shl i;
         k:=k-size[son[p,y]];
         p:=son[p,-y];
       end;
     end;
   end;

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

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