bzoj3668

noi水题，直接推出来每一位上取什么值才能取1
直接搞就可以了

 var a,b:array[..] of longint;
     x,n,i,j,k,t,m,ans:longint;
     fl:boolean;
     s:string;
     ch:char;

 procedure doit(x:longint);
   begin
     fillchar(b,sizeof(b),);
     m:=;
     while x<> do
     begin
       inc(m);
       b[m]:=x mod ;
       x:=x shr ;
     end;
   end;

 begin
   readln(n,k);
   t:=;
   for i:= to  do
     a[i]:=;
   for i:= to n do
   begin
     s:='';
     read(ch);
     while ch<>' ' do
     begin
       s:=s+ch;
       read(ch);
     end;
     readln(x);
     doit(x);
     if s='OR' then  //表示这一位取什么值都能使ans这位上取1,-表示去什么都不可以，表示取1才可以，表示取0才可以
     begin
       for j:= to  do
         if b[j]= then a[j]:=;
     end
     else if s='AND' then
     begin
       for j:= to  do
         if b[j]= then a[j]:=-;
     end
     else begin
       for j:= to  do
         if b[j]= then
         begin
           if a[j]=- then a[j]:=
           else if a[j]= then a[j]:=-
           else if a[j]= then a[j]:=
           else if a[j]= then a[j]:=;
         end;
     end;
   end;
   doit(k);
   ans:=;
   fl:=false;  //fl判断高位是否已经严格小于了
   for i:= downto  do
     if fl then
     begin
       if a[i]<>- then
         ans:=ans+ shl (i-);
     end
     else begin
       if (b[i]=) and ((a[i]=) or (a[i]=)) then ans:=ans+ shl (i-);
       if (b[i]=) then
       begin
         if (a[i]=) or (a[i]=) then
         begin
           fl:=true;
           ans:=ans+ shl (i-);
         end
         else if a[i]=- then fl:=true
         else ans:=ans+ shl (i-);
       end;
     end;

   writeln(ans);
 end.