bzoj3721

不是说好的20s吗，怎么我19s都超时……逗我
最后还得写成c++才能过……
首先不难发现询问肯定是O(1）的复杂度
我们先把奇数和偶数分开排序，不难发现几个性质
1. 奇数的个数一定是奇数
2. 奇数选取随k成单调增
然后就能在O(n)的时间预处理了……

type arr=array[0..1000010] of int64;
var a,b,f:arr;
    j,i,n1,n2,n,m,x:longint;
    ch:boolean;

procedure qsort(var a:arr;n:longint);
  procedure sort(l,r: longint);
    var i,j: longint;
        x,y:int64;
    begin
      i:=l;
      j:=r;
      x:=a[(l+r) div 2];
      repeat
        while a[i]>x do inc(i);
        while x>a[j] do dec(j);
        if not(i>j) then
        begin
          y:=a[i];
          a[i]:=a[j];
          a[j]:=y;
          inc(i);
          j:=j-1;
        end;
      until i>j;
      if l<j then sort(l,j);
      if i<r then sort(i,r);
    end;

  begin
    sort(1,n);
  end;

begin
  readln(n);
  for i:=1 to n do
  begin
    read(x);
    if x mod 2=1 then
    begin
      inc(n1);
      a[n1]:=x;
    end
    else begin
      inc(n2);
      b[n2]:=x;
    end;
  end;
  qsort(a,n1);
  qsort(b,n2);
  for i:=1 to n1 do
    a[i]:=a[i-1]+a[i];
  for i:=1 to n2 do
    b[i]:=b[i-1]+b[i];
  if n1=0 then
    f[1]:=-1
  else f[1]:=a[1];
  i:=1;
  j:=1;
  while i<n do
  begin
    inc(i);
    f[i]:=-1;
    while (i-j>n2) and (j<=n1) do j:=j+2;
    ch:=false;
    while (j<=n1) and (i-j>=0) and (f[i]<a[j]+b[i-j]) do
    begin
      f[i]:=a[j]+b[i-j];
      j:=j+2;
      ch:=true;
    end;
    if ch then j:=j-2;
  end;
  readln(m);
  for i:=1 to m do
  begin
    readln(x);
    writeln(f[x]);
  end;
end.