Tautology - poj 3295

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10437		Accepted: 3963

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

p，q，r，s，t，是五个二进制数。

K，A，N，C，E，是五个运算符。

K:&&

A:||
N:!

C:(!w)||x

E:w==x

这道题可以将p,q,r,s,t穷举

 #include <iostream>
 #include<string.h>
 #include <stack>
 using namespace std;
 int p, q, r, s, t;
 int len;
 char str[];
 int result() {
     stack<int> res;
     int t1,t2;
     for (int i = len - ; i >= ; i--) {
         switch (str[i]) {
         case 'p':
             res.push(p);
             break;
         case 'q':
             res.push(q);
             break;
         case 'r':
             res.push(r);
             break;
         case 's':
             res.push(s);
             break;
         case 't':
             res.push(t);
             break;
         case 'K':
             t1 = res.top();
             res.pop();
             t2 = res.top();
             res.pop();
             if (t1 & t2) {
                 res.push();
             } else {
                 res.push();
             }
             break;
         case 'A':
             t1 = res.top();
             res.pop();
             t2 = res.top();
             res.pop();
             if (t1 | t2)
                 res.push();
             else
                 res.push();
             break;
         case 'N':
             t1 = res.top();
             res.pop();
             if (~t1 & )
                 res.push();
             else
                 res.push();
             break;
         case 'C':
             t1 = res.top();
             res.pop();
             t2 = res.top();
             res.pop();
             if ((~t1 & ) | t2) {
                 res.push();
             } else {
                 res.push();
             }
             break;
         case 'E':
             t1 = res.top();
             res.pop();
             t2 = res.top();
             res.pop();
             if (t1 == t2) {
                 res.push();
             } else {
                 res.push();
             }
             break;
         }
     }
     return res.top();
 }

 int fun() {
     int flag;
     for (p = ; p < ; p++) {
         for (q = ; q < ; q++) {
             for (r = ; r < ; r++) {
                 for (s = ; s < ; s++) {
                     for (t = ; t < ; t++) {
                         flag = result();
                         if(flag==)
                             return ;
                     }
                 }
             }
         }
     }
     return ;
 }

 int main() {
     while (cin >> str) {
         if (strcmp(str, "") == )
             break;
         len = strlen(str);
         int flag=fun();
         if(flag)
             cout<<"tautology"<<endl;
         else
             cout<<"not"<<endl;
     }
 }