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;
   }
}