Tautology

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

就是离散数学的一个模拟，判断所给的式子是不是永真公式
用栈逆序推还不算难
 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<stack>
 5 using namespace std;
 6 stack<int>sta;
 7 int p,q,r,s,t,a,b;
 8 char str[102];
 9 int f()
10 {
11     int len=strlen(str);
12     for(int i=len-1; i>-1; i--)
13     {
14         if(str[i]=='p')sta.push(p);//是字符直接入栈
15         else if(str[i]=='q')sta.push(q);
16         else if(str[i]=='r')sta.push(r);
17         else if(str[i]=='s')sta.push(s);
18         else if(str[i]=='t')sta.push(t);
19         else if(str[i]=='K')//是运算就进行运算
20         {
21             a=sta.top();
22             sta.pop();
23             b=sta.top();
24             sta.pop();
25             sta.push(a&&b);
26         }
27         else if(str[i]=='A')
28         {
29             a=sta.top();
30             sta.pop();
31             b=sta.top();
32             sta.pop();
33             sta.push(a||b);
34         }
35         else if(str[i]=='C')
36         {
37             a=sta.top();
38             sta.pop();
39             b=sta.top();
40             sta.pop();
41             if(a&&!b)sta.push(0);
42             else sta.push(1);
43         }
44         else if(str[i]=='E')
45         {
46             a=sta.top();
47             sta.pop();
48             b=sta.top();
49             sta.pop();
50             if(a==b)sta.push(1);
51             else sta.push(0);
52         }
53         else if(str[i]=='N')
54         {
55             a=sta.top();
56             sta.pop();
57             sta.push(!a);
58         }
59     }
60     return sta.top();
61 }
62 int sf()//枚举各种情况
63 {
64     for(p=0; p<2; p++)
65         for(q=0; q<2; q++)
66             for(r=0; r<2; r++)
67                 for(s=0; s<2; s++)
68                     for(t=0; t<2; t++)
69                         if(!f())
70                             return 0;
71     return 1;
72 }
73 int main()
74 {
75     while(gets(str))
76     {
77         if(!strcmp(str,"0"))
78             break;
79         else
80         {
81             if(sf())
82                 printf("tautology\n");
83             else
84                 printf("not\n");
85         }
86     }
87     return 0;
88 }
View Code

 

posted @ 2013-08-06 20:10  孔凡凯凯  阅读(457)  评论(0)  编辑  收藏  举报