构造 + 离散数学、重言式 - POJ 3295 Tautology

Tautology

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


【题目来源】
Waterloo Local Contest, 2006.9.30
http://poj.org/problem?id=3295
【题目大意】
给你一个合式公式，让你判断是否为重言式。
【题目分析】
这题考了一些离散数学的概念在里面，题目并不难，就是构造加模拟。
使用一个栈来从后往前计算。

AC代码：

#include<iostream>
#include<cstring>
#define MAX 110
using namespace std;
int a [ MAX ];
char str [ MAX ];
void calc( int p , int q , int r , int s , int t)
{
    int top = 0;
    int t1 , t2;
    int len = strlen( str);
    for( int i = len - 1; i >= 0; i --) //所有的都判断一遍，直到a[0]
    {
        if( str [ i ] == 'p') a [ top ++ ] =p;
  else if( str [ i ] == 'q') a [ top ++ ] = q;
  else if( str [ i ] == 'r') a [ top ++ ] = r;
  else if( str [ i ] == 's') a [ top ++ ] =s;
  else if( str [ i ] == 't') a [ top ++ ] = t;

        if( str [ i ] == 'K')
        {
          t1 = a [ -- top ];
          t2 = a [ -- top ];
          a [ top ++ ] = t1 && t2;
        }
        else if( str [ i ] == 'A')
        {
          t1 = a [ -- top ];
          t2 = a [ -- top ];
          a [ top ++ ] = t1|| t2;
        }
      else if( str [ i ] == 'N')
        {
          t1 = a [ -- top ];
          a [ top ++ ] =! t1;
        }
      else if( str [ i ] == 'C')
        {
          t1 = a [ -- top ];
          t2 = a [ -- top ];
          if( t1 == 1 && t2 == 0)
              a [ top ++ ] = 0;
          else a [ top ++ ] = 1;
        }
        else if( str [ i ] == 'E')
        {
          t1 = a [ -- top ];
          t2 = a [ -- top ];
          if( t1 == t2)
            a [ top ++ ] = 1;
          else a [ top ++ ] = 0;
        }
    }
}
bool judge()
{
    int p , q , r ,s , t;
    for(p = 0;p < 2;p ++)                       //枚举所有的情况
    for( q = 0; q < 2; q ++)
      for( r = 0; r < 2; r ++)
      for(s = 0;s < 2;s ++)
        for( t = 0; t < 2; t ++)
        {
            calc(p , q , r ,s , t);
            if( a [ 0 ] == 0)
            {
                return false;
            }
        }
    return   true;
}
int main()
{
    while( cin >> str)
    {
        if( strcmp( str , "0") == 0) break;
        if( judge())
            cout << "tautology" << endl;
        else cout << "not" << endl;
    }
}

 

 

转载于:https://www.cnblogs.com/crazyacking/p/3731215.html