Poj 3295

题目大意：略

解：暴力枚举rqst之流的状态就行了，状态数不多，对运算顺序困惑了下，看了看题解是用栈自后往前弄。

//Tautology
const
        ttttt='tautology';
        fffff='not';
        inf='1.txt';
        maxl=111;
var
        ask: string;
        ans: boolean;
        stack: array[0..maxl]of longint;
        tot : longint;

procedure main;
var
        l, tmp, p, q, r, s, t: longint;
        c: char;
begin
  ans := true;
  for p := 0 to 1 do
    for q := 0 to 1 do
      for r := 0 to 1 do
        for s := 0 to 1 do
          for t := 0 to 1 do begin
            l := length(ask);
            tot := 0;
            while l>0 do begin
              c := ask[l];
              case c of
                'q', 'p', 'r', 's', 't': begin
                  inc(tot);
                  case c of
                    'q':stack[tot] := q;
                    'p':stack[tot] := p;
                    'r':stack[tot] := r;
                    's':stack[tot] := s;
                    't':stack[tot] := t;
                  end;
                end;
                'N':stack[tot] := stack[tot] xor 1 ;
                'K':begin
                  dec(tot);
                  stack[tot] := stack[tot] and stack[tot+1];
                end;
                'A':begin
                  dec(tot);
                  stack[tot] := stack[tot] or stack[tot+1];
                end;
                'C':begin
                  dec(tot);
                  stack[tot] := ((stack[tot]xor stack[tot+1])xor 1)or(stack[tot])
                end;
                'E':begin
                  dec(tot);
                  stack[tot] := (stack[tot]xor stack[tot+1])xor 1;
                end;
              end;
              dec(l);
            end;
            if stack[1]=0 then begin
              ans := false; exit;
            end;
          end;
end;

procedure print;
begin
  if ans then writeln(ttttt)
    else writeln(fffff);
end;

begin
  assign(input,inf); reset(input);
  read(ask); readln;
  while ask<>'0' do begin
    main;
    print;
    read(ask); readln;
  end;
end.