Tautology

                                              Tautology

                                                             Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Problem 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

 

这里的题意涉及到 离散数学 上的一点知识，即 蕴含式 和 永真式 。蕴含式“C”是“0”的情况 当且仅当 w=1 && x=0 ；永真式 即无论怎么取值，结果永为真。

 

思路：这里用的是从后往前依次判定，用到了数据结构中 栈 的一点东西。

 

我的代码：（虽然有点长，但很好理解，不会的看看也应该就能懂了）

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=121;
int a[maxn];
char str[maxn];
void DO(int p,int q,int r,int s,int t)
{
    int top=0;
    int len=strlen(str);
    for(int i=len-1;i>=0;i--)
    {
        int t1,t2;
        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);
        }
        if(str[i]=='A')
        {
           t1=a[--top];
           t2=a[--top];
           a[top++]=(t1||t2);
        }
        if(str[i]=='N')
        {
           t1=a[--top];
           a[top++]=(!t1);
        }
        if(str[i]=='C')
        {
           t1=a[--top];
           t2=a[--top];
           if(t1==1&&t2==0)
               a[top++]=0;
           else sta[top++]=1;
        }
        if(str[i]=='E')
        {
           t1=a[--top];
           t2=a[--top];
           if(t1==1&&t2==1 ||t1==0&&t2==0)
            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++)
        {
            DO(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;
    }
}