Poj 3295 Tautology

1.Link：

http://poj.org/problem?id=3295

2.content:

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9793		Accepted: 3712

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

Source

Waterloo Local Contest, 2006.9.30

3.method:

将p,q,r,s,t分别置入以0，1代入，使用五个for循环实现，这样会出现相同计算多次的情况，不过由于循环次数很少，所以不影响大的效率。

将字符串从后向前置入栈，遇到运算符则进行运算，然后将结果置入栈。

最后结果非1的话利用goto跳出多层循环

4.code:

 1 #include <iostream>

 2 #include <string>

 3 #include <stack>

 4 

 5 using namespace std;

 6 

 7 int main()

 8 {

 9     //freopen("D://input.txt","r",stdin);

10 

11     string str;

12     cin >> str;

13 

14     while(str.size() != 1 || str[0] != '0')

15     {

16         //cout << str << endl;

17 

18         int p,q,r,s,t;

19         for(p = 0; p < 2; ++p)

20         {

21             for(q = 0; q < 2; ++q)

22             {

23                 for(r = 0; r < 2; ++r)

24                 {

25                     for(s = 0; s < 2; ++s)

26                     {

27                         for(t = 0; t < 2; ++t)

28                         {

29                             stack<int> s_str;

30                             int w,x;

31                             for(string::reverse_iterator r_iter = str.rbegin(); r_iter != str.rend(); ++r_iter)

32                             //for(int str_i = str.size() - 1; str_i >= 0; --str_i)

33                             {

34                                 if(*r_iter == 'p') s_str.push(p);

35                                 else if(*r_iter == 'q') s_str.push(q);

36                                 else if(*r_iter == 'r') s_str.push(r);

37                                 else if(*r_iter == 's') s_str.push(s);

38                                 else if(*r_iter == 't') s_str.push(t);

39                                 else

40                                 {

41                                     w = s_str.top();

42                                     s_str.pop();

43                                     if(*r_iter == 'N') s_str.push(!w);

44                                     else

45                                     {

46                                         x = s_str.top();

47                                         s_str.pop();

48                                         if(*r_iter == 'K') s_str.push(w&&x);

49                                         else if(*r_iter == 'A') s_str.push(w||x);

50                                         else if(*r_iter == 'C') {if(w == 1 && x == 0) s_str.push(0); else s_str.push(1);}

51                                         else {if(w == x) s_str.push(1); else s_str.push(0);}

52                                     }

53                                 }

54                             }

55                             if(s_str.top() != 1) goto end;

56                         }

57                     }

58                 }

59             }

60         }

61 

62 end:

63         if(p >= 2) cout << "tautology" << endl;

64         else cout << "not" << endl;

65 

66         cin >> str;

67     }

68     return 0;

69 }

 5.Reference:

http://blog.csdn.net/lyy289065406/article/details/6642766