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[0..40] 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),0);
    m:=0;
    while x<>0 do
    begin
      inc(m);
      b[m]:=x mod 2;
      x:=x shr 1;
    end;
  end;

begin
  readln(n,k);
  t:=31;
  for i:=1 to 31 do
    a[i]:=1;
  for i:=1 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  //2表示这一位取什么值都能使ans这位上取1,-1表示去什么都不可以，1表示取1才可以，0表示取0才可以
    begin
      for j:=1 to 31 do
        if b[j]=1 then a[j]:=2;  
    end
    else if s='AND' then
    begin
      for j:=1 to 31 do
        if b[j]=0 then a[j]:=-1;
    end
    else begin
      for j:=1 to 31 do
        if b[j]=1 then
        begin
          if a[j]=-1 then a[j]:=2
          else if a[j]=2 then a[j]:=-1
          else if a[j]=1 then a[j]:=0
          else if a[j]=0 then a[j]:=1;
        end;
    end;
  end;
  doit(k);
  ans:=0;
  fl:=false;  //fl判断高位是否已经严格小于了
  for i:=31 downto 1 do
    if fl then
    begin
      if a[i]<>-1 then
        ans:=ans+1 shl (i-1);
    end
    else begin
      if (b[i]=0) and ((a[i]=2) or (a[i]=0)) then ans:=ans+1 shl (i-1);
      if (b[i]=1) then
      begin
        if (a[i]=2) or (a[i]=0) then
        begin
          fl:=true;
          ans:=ans+1 shl (i-1);
        end
        else if a[i]=-1 then fl:=true
        else ans:=ans+1 shl (i-1);
      end;
    end;

  writeln(ans);
end.