Tautology(栈或者用数组模拟)

原题：

Tautology

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

Total Submission(s) : 32   Accepted Submission(s) : 22

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 ofp. 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

 

 

 

 

 

知识背景：需要离散数学的知识（顿时感觉到了数学的强大）。

题目大意：

K, A, N, C, E   分别代表“且”，“或”，“非”，“蕴含”，“相等”五种运算，对p, q, r, s, t进行上述五种运算，若表达式为永真式，输出“tautology”，否则输出“not”。

解题思路：

枚举p, q, r, s, t的取值，共有2的5次方种情况。

 

 

解法一：栈

 

 #include <cstdio>
 #include <cstring>
 #include <stack>
  using namespace std;
  int p,q,r,s,t;
  int charge(char c)
  {
      if(c=='p') return p;
      if(c=='q') return q;
     if(c=='r') return r;
     if(c=='s') return s;
     if(c=='t') return t;
 }
 int main()
 {
     char str[110];
     int i,x,y;
     while(scanf("%s",str)&&str[0]!='0')
     {
         int flag=1;
         stack<int>st;
         for(p=0;p<=1&&flag;p++)
             for(q=0;q<=1&&flag;q++)
                 for(r=0;r<=1&&flag;r++)
                     for(s=0;s<=1&&flag;s++)
                         for(t=0;t<=1&&flag;t++)
                         {
                             for(i=strlen(str)-1;i>=0;i--)
                             {
                                 if(str[i]=='p'||str[i]=='q'||str[i]=='r'
                                    ||str[i]=='s'||str[i]=='t')
                                    st.push(charge(str[i]));
                                 else if(str[i]=='K')
                                 {
                                     x=st.top();
                                     st.pop();
                                     y=st.top();
                                     st.pop();
                                     st.push(x&&y);
                                 }
                                 else if(str[i]=='A')
                                 {
                                     x=st.top();
                                     st.pop();
                                     y=st.top();
                                     st.pop();
                                     st.push(x||y);
                                }
                                 else if(str[i]=='N')
                                {
                                     x=st.top();
                                     st.pop();
                                     st.push(!x);
                                 }
                                 else if(str[i]=='C')
                                 {
                                     x=st.top();
                                     st.pop();
                                     y=st.top();
                                     st.pop();
                                     if(x==1&&y==0) st.push(0);
                                    else st.push(1);
                                 }
                                 else if(str[i]=='E')
                                 {
                                    x=st.top();
                                    st.pop();
                                     y=st.top();
                                     st.pop();
                                     st.push((x==y));
                                 }
                                 else
                                    flag=0;                             }
                             if(flag)
                             {
                                 flag=st.top();
                                 st.pop();
                             }
                         }
         if(flag)
            printf("tautology\n");
         else
            printf("not\n");
     }
     return 0;
 }

 

 

 

 

解法二：用数组模拟栈

 

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int stack[1000];
char str[1000];
int put()
{
    int pp,qq,rr,ss,tt,n,top,i;
    for(pp=0;pp<=1;pp++)
    {
        for(qq=0;qq<=1;qq++)
        {
            for(rr=0;rr<=1;rr++)
            {
                for(ss=0;ss<=1;ss++)
                {
                    for(tt=0;tt<=1;tt++)
                    {
                        top=0;
                        n=strlen(str);
                        for(i=n-1;i>=0;i--)
                        {
                            if(str[i]=='q')stack[top++]=qq;
                            if(str[i]=='p')stack[top++]=pp;
                            if(str[i]=='r')stack[top++]=rr;
                            if(str[i]=='t')stack[top++]=tt;
                            if(str[i]=='s')stack[top++]=ss;
                            if(str[i]=='K')top--,stack[top-1]=(stack[top-1]&&stack[top]);
                            if(str[i]=='A')top--,stack[top-1]=(stack[top-1]||stack[top]);
                            if(str[i]=='N')stack[top-1]=!stack[top-1];
                            if(str[i]=='C')top--,stack[top-1]=((!stack[top-1])||stack[top]);
                            if(str[i]=='E')top--,stack[top-1]=((stack[top-1])==stack[top]);
                        }
                        if(top!=1||stack[top-1]!=1)
                        return 0;
                    }
                }
            }
        }
    }
    return 1;
}
int main()
{
    while(gets(str)&&str[0]!='0')
    {
        if(put()==1)printf("tautology\n");
        else
        printf("not\n");
    }
    return 0;
}