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[..] 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 ];
       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-;
         end;
       until i>j;
       if l<j then sort(l,j);
       if i<r then sort(i,r);
     end;

   begin
     sort(,n);
   end;

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