POJ3295——Tautology

Tautology

Description

WFF 'N PROOF is a logic game played with dice. Each die has six faces representing some subset of the possible symbols K, A, N, C, E, p, q, r, s, t. A Well-formed formula (WFF) is any string of these symbols obeying the following rules:

• p, q, r, s, and t are WFFs
• if w is a WFF, Nw is a WFF
• if w and x are WFFs, Kwx, Awx, Cwx, and Ewx are WFFs.

The meaning of a WFF is defined as follows:

• p, q, r, s, and t are logical variables that may take on the value 0 (false) or 1 (true).
• K, A, N, C, E mean and, or, not, implies, and equals as defined in the truth table below.

Definitions of K, A, N, C, and E

w  x	Kwx	Awx	Nw	Cwx	Ewx
1  1	1	1	0	1	1
1  0	0	1	0	0	0
0  1	0	1	1	1	0
0  0	0	0	1	1	1

A tautology is a WFF that has value 1 (true) regardless of the values of its variables. For example, ApNp is a tautology because it is true regardless of the value of p. On the other hand, ApNq is not, because it has the value 0 for p=0, q=1.

You must determine whether or not a WFF is a tautology.

Input

Input consists of several test cases. Each test case is a single line containing a WFF with no more than 100 symbols. A line containing 0 follows the last case.

Output

For each test case, output a line containing tautology or not as appropriate.

Sample Input

ApNp
ApNq
0

Sample Output

tautology
not

题目大意：逻辑表达式求知，K, A, N, C, E为逻辑运算符，p, q, r, s, and t 为真假值。 
　　　　　　Kxy -> x&&y
　　　　　　Axy -> x||y
　　　　　　Nx  -> !x
　　　　　　Cxy -> x||(!y)
　　　　　　Exy -> x==y
　　　　　　判断是否表达式恒为真。
解题思路：  数值变量总共就5个，枚举这五个变量的值，有32种情况。
　　　　　　处理字符串的时候，类似于逆波兰表达式的求值过程。
　　　　　　从(S.length()-1)->0遍历，遇到小写字母将其对应的布尔值存入栈。
　　　　　　遇到大写字母（除N外） 去除栈顶2个元素进行处理，后存入栈。
　　　　　　遇到N，去除栈顶一个元素，取反后存入栈。
　　　　　　遍历完返回S.top()。
Code：

 #include<iostream>
 #include<string>
 #include<stack>
 using namespace std;
 int q,p,s,r,t;
 bool Is_Tau(string S)
 {
     int len=S.length()-,i,t1,t2;
     stack<char> ST;
     for (i=len;i>=;i--)
     {
         if (S[i]=='q') ST.push(q);
         if (S[i]=='p') ST.push(p);
         if (S[i]=='r') ST.push(r);
         if (S[i]=='s') ST.push(s);
         if (S[i]=='t') ST.push(t);
         if (S[i]=='K')
         {
             t1=ST.top();
             ST.pop();
             t2=ST.top();
             ST.pop();
             ST.push(t1&&t2);
         }
         if (S[i]=='A')
         {
             t1=ST.top();
             ST.pop();
             t2=ST.top();
             ST.pop();
             ST.push(t1||t2);
         }
         if (S[i]=='C')
         {
             t1=ST.top();
             ST.pop();
             t2=ST.top();
             ST.pop();
             ST.push(t1||(!t2));
         }
         if (S[i]=='E')
         {
             t1=ST.top();
             ST.pop();
             t2=ST.top();
             ST.pop();
             ST.push(t1==t2);
         }
         if (S[i]=='N')
         {
             t1=ST.top();
             ST.pop();
             ST.push(!t1);
         }
     }
     return ST.top();
 }
 int main()
 {
     string WFF;
     while (cin>>WFF)
     {
         int OK=;
         if (WFF=="") break;
         for (q=; q<=; q++)
             for (p=; p<=; p++)
                 for (r=; r<=; r++)
                     for (s=; s<=; s++)
                         for (t=; t<=; t++)
                             if (!Is_Tau(WFF))
                             {
                                 OK=;
                                 break;
                             }
         if (OK) printf("tautology\n");
         else printf("not\n");
     }
     return ;
 }