Tautology

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

  1 #include <iostream>
  2 #include<string.h>
  3 #include <stack>
  4 using namespace std;
  5 int p, q, r, s, t;
  6 int len;
  7 char str[106];
  8 int result() {
  9     stack<int> res;
 10     int t1,t2;
 11     for (int i = len - 1; i >= 0; i--) {
 12         switch (str[i]) {
 13         case 'p':
 14             res.push(p);
 15             break;
 16         case 'q':
 17             res.push(q);
 18             break;
 19         case 'r':
 20             res.push(r);
 21             break;
 22         case 's':
 23             res.push(s);
 24             break;
 25         case 't':
 26             res.push(t);
 27             break;
 28         case 'K':
 29             t1 = res.top();
 30             res.pop();
 31             t2 = res.top();
 32             res.pop();
 33             if (t1 & t2) {
 34                 res.push(1);
 35             } else {
 36                 res.push(0);
 37             }
 38             break;
 39         case 'A':
 40             t1 = res.top();
 41             res.pop();
 42             t2 = res.top();
 43             res.pop();
 44             if (t1 | t2)
 45                 res.push(1);
 46             else
 47                 res.push(0);
 48             break;
 49         case 'N':
 50             t1 = res.top();
 51             res.pop();
 52             if (~t1 & 1)
 53                 res.push(1);
 54             else
 55                 res.push(0);
 56             break;
 57         case 'C':
 58             t1 = res.top();
 59             res.pop();
 60             t2 = res.top();
 61             res.pop();
 62             if ((~t1 & 1) | t2) {
 63                 res.push(1);
 64             } else {
 65                 res.push(0);
 66             }
 67             break;
 68         case 'E':
 69             t1 = res.top();
 70             res.pop();
 71             t2 = res.top();
 72             res.pop();
 73             if (t1 == t2) {
 74                 res.push(1);
 75             } else {
 76                 res.push(0);
 77             }
 78             break;
 79         }
 80     }
 81     return res.top();
 82 }
 83 
 84 int fun() {
 85     int flag;
 86     for (p = 0; p < 2; p++) {
 87         for (q = 0; q < 2; q++) {
 88             for (r = 0; r < 2; r++) {
 89                 for (s = 0; s < 2; s++) {
 90                     for (t = 0; t < 2; t++) {
 91                         flag = result();
 92                         if(flag==0)
 93                             return 0;
 94                     }
 95                 }
 96             }
 97         }
 98     }
 99     return 1;
100 }
101 
102 int main() {
103     while (cin >> str) {
104         if (strcmp(str, "0") == 0)
105             break;
106         len = strlen(str);
107         int flag=fun();
108         if(flag)
109             cout<<"tautology"<<endl;
110         else
111             cout<<"not"<<endl;
112     }
113 }

相关阅读:
c# 委托
 datagrid加下拉列表dropdownlist
MySQL存储过程的基本函数(三)
MySQL存储过程详解 mysql 存储过程(二)
MySQL存储过程（一）
SFTP 文件上传下载引用代码
 JAVA SFTP文件上传、下载及批量下载
 java中遍历MAP的几种方法
 POSTGRESQL 导入导出
 db2move 数据导出整理
原文地址：https://www.cnblogs.com/sdxk/p/4615640.html