bzoj3878

当初只会暴力，现在差不多觉得水了
显然离线处理，对输入的数排序然后会发现不管怎么修改都是结果总是单调不降的
对于每次处理，我们只要找到那段越界的即可
显然上线段树，话说jsoi这么喜欢线段树？
下面在bzoj上过不去，因为pascal编译器处理比较严格，可能某处爆了int64，我也懒得查了，本地是能过的

 type node=record
        mi,mx,att,mul,an:int64;
      end;

 var tree:array[..*] of node;
     ans,b,c:array[..] of longint;
     a:array[..] of int64;
     ch:array[..] of char;
     m,f,t,i,n:longint;

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

 procedure sort(l,r:longint);
   var i,j:longint;
       x,y:int64;

   begin
     i:=l;
     j:=r;
     x:=a[(l+r) shr ];
     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;
         swap(c[i],c[j]);
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sort(l,j);
     if i<r then sort(i,r);
   end;

 procedure cal(i,x,y:longint;x1,x2,x3:int64);
   begin
     with tree[i] do
     begin
       mi:=mi*x1+x2*a[x]+x3;
       mx:=mx*x1+x2*a[y]+x3;
       mul:=mul*x1;
       att:=att*x1;
       an:=an*x1;
       att:=att+x2;
       an:=an+x3;
     end;
   end;

 procedure build(i,l,r:longint);
   var m:longint;
   begin
     tree[i].mi:=a[l];
     tree[i].mx:=a[r];
     tree[i].mul:=;
     if l<>r then
     begin
       m:=(l+r) shr ;
       build(i*,l,m);
       build(i*+,m+,r);
     end;
   end;

 procedure push(i,l,r:longint);
   var m:longint;
   begin
     m:=(l+r) shr ;
     with tree[i] do
     begin
       cal(i*,l,m,mul,att,an);
       cal(i*+,m+,r,mul,att,an);
       mul:=;
       att:=;
       an:=;
     end;
   end;

 procedure max(i,l,r:longint);
   var m:longint;
   begin
     if l=r then cal(i,l,l,,,t)
     else begin
       m:=(l+r) shr ;
       push(i,l,r);
       if tree[i*].mx>t then
       begin
         cal(i*+,m+,r,,,t);
         max(i*,l,m);
       end
       else max(i*+,m+,r);
       tree[i].mx:=tree[i*+].mx;
       tree[i].mi:=tree[i*].mi;
     end;
   end;

 procedure min(i,l,r:longint);
   var m:longint;
   begin
     if l=r then cal(i,l,l,,,f)
     else begin
       m:=(l+r) shr ;
       push(i,l,r);
       if tree[i*+].mi<f then
       begin
         cal(i*,l,m,,,f);
         min(i*+,m+,r);
       end
       else min(i*,l,m);
       tree[i].mx:=tree[i*+].mx;
       tree[i].mi:=tree[i*].mi;
     end;
   end;

 procedure ask(i,l,r:longint);
   var m:longint;
   begin
     if l=r then
       ans[c[l]]:=tree[i].mx
     else begin
       m:=(l+r) shr ;
       push(i,l,r);
       ask(i*,l,m);
       ask(i*+,m+,r);
     end;
   end;

 begin
   readln(m,f,t);
   for i:= to m do
     readln(ch[i],b[i]);
   readln(n);
   for i:= to n do
   begin
     readln(a[i]);
     if a[i]>t then a[i]:=t;
     if a[i]<f then a[i]:=f;
     c[i]:=i;
   end;
   sort(,n);
   build(,,n);
   for i:= to m do
     if ch[i]='+' then
     begin
       cal(,,n,,,b[i]);
       if tree[].mx>t then max(,,n);
     end
     else if ch[i]='-' then
     begin
       cal(,,n,,,-b[i]);
       if tree[].mi<f then min(,,n);
     end
     else if ch[i]='*' then
     begin
       cal(,,n,b[i],,);
       if tree[].mx>t then max(,,n);
     end
     else begin
       cal(,,n,,b[i],);
       if tree[].mx>t then max(,,n);
     end;
   ask(,,n);
   for i:= to n do
     writeln(ans[i]);
 end.