poj3295 Tautology

Tautology
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 16817   Accepted: 6500

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

思路：把真值表枚举完就行。

下附代码：
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 const int MAXN = 120;
 6 int sta[MAXN];
 7 char str[MAXN];
 8 int p, q, r, s, t;
 9 void judge() {
10     int top = 0;
11     int len = strlen(str);
12     for (int i = len - 1; i >= 0; i--) {
13         if (str[i] == 'p')
14             sta[top++] = p;
15         else if (str[i] == 'q')
16             sta[top++] = q;
17         else if (str[i] == 'r')
18             sta[top++] = r;
19         else if (str[i] == 's')
20             sta[top++] = s;
21         else if (str[i] == 't')
22             sta[top++] = t;
23         else if (str[i] == 'K') {
24             int t1 = sta[--top];
25             int t2 = sta[--top];
26             sta[top++] = (t1 && t2);
27         } else if (str[i] == 'A') {
28             int t1 = sta[--top];
29             int t2 = sta[--top];
30             sta[top++] = (t1 || t2);
31         } else if (str[i] == 'N') {
32             int t1 = sta[--top];
33             sta[top++] = (!t1);
34         } else if (str[i] == 'C') {
35             int t1 = sta[--top];
36             int t2 = sta[--top];
37             if (t1 == 1 && t2 == 0)
38                 sta[top++] = 0;
39             else
40                 sta[top++] = 1;
41         } else if (str[i] == 'E') {
42             int t1 = sta[--top];
43             int t2 = sta[--top];
44             if ((t1 == 1 && t2 == 1) || (t1 == 0 && t2 == 0))
45                 sta[top++] = 1;
46             else
47                 sta[top++] = 0;
48         }
49     }
50 }
51 bool solve() {
52     for (p = 0; p < 2; p++)
53         for (q = 0; q < 2; q++)
54             for (r = 0; r < 2; r++)
55                 for (s = 0; s < 2; s++)
56                     for (t = 0; t < 2; t++) {
57                         judge();
58                         if (sta[0] == 0)
59                             return 0;
60                     }
61     return 1;
62 }
63 int main() {
64     while (scanf("%s", str)) {
65         if (strcmp(str, "0") == 0)
66             break;
67         if (solve())
68             printf("tautology\n");
69         else
70             printf("not\n");
71     }
72     return 0;
73 }
View Code

 

posted @ 2020-10-04 20:45  我是菜狗QAQ  阅读(150)  评论(0)  编辑  收藏  举报