Tautology POJ - 3295

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

题意转自http://user.qzone.qq.com/289065406/blog/1309062835

 

大致题意：

输入由p、q、r、s、t、K、A、N、C、E共10个字母组成的逻辑表达式，

其中p、q、r、s、t的值为1（true）或0（false），即逻辑变量；

K、A、N、C、E为逻辑运算符，

K --> and:  x && y

A --> or:  x || y

N --> not :  !x

C --> implies :  (!x)||y

E --> equals :  x==y

问这个逻辑表达式是否为永真式。

PS:输入格式保证是合法的

 

解题思路：

p, q, r, s, t不同的取值组合共32种情况，枚举不同取值组合代入逻辑表达式WFF进行计算。

如果对于所有的取值组合，WFF值都为 true, 则结果为 tautology，否则为 not。
  

WFF的计算方法：

从字符串WFF的末尾开始依次向前读取字符。

构造一个栈stack，当遇到逻辑变量 p, q, r, s ,t 则将其当前的值压栈；

遇到 N 则取栈顶元素进行非运算，运算结果的值压栈；

遇到K, A, C, E则从栈顶中弹出两个元素进行相应的运算，将结果的值压栈。 

由于输入是合法的，当字符串WFF扫描结束时，栈stack中只剩一个值，该值就是逻辑表达式WFF的值。

 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

char str[110];
int ans[110];

int put()
{
    int pp,qq,ss,tt,rr,i,j;
    for(pp=0;pp<=1;pp++)
    {
        for(qq=0;qq<=1;qq++)
        {
            for(ss=0;ss<=1;ss++)
            {
                for(tt=0;tt<=1;tt++)
                {
                    for(rr=0;rr<=1;rr++)
                    {
                        j=0;
                        for(i=strlen(str)-1;i>=0;i--)
                        {
                            if(str[i]=='p'){ans[j]=pp;j++;}
                            if(str[i]=='s'){ans[j]=ss;j++;}
                            if(str[i]=='t'){ans[j]=tt;j++;}
                            if(str[i]=='r'){ans[j]=rr;j++;}
                            if(str[i]=='q'){ans[j]=qq;j++;}
                            if(str[i]=='K'){j--;ans[j-1]=ans[j]&&ans[j-1];}
                            if(str[i]=='A'){j--;ans[j-1]=ans[j]||ans[j-1];}
                            if(str[i]=='C'){j--;ans[j-1]=(!ans[j-1])||ans[j];}
                            if(str[i]=='N'){ans[j-1]=!ans[j-1];}
                            if(str[i]=='E'){j--;ans[j-1]=ans[j]==ans[j-1];}
                        }
                        if(j!=1||ans[j-1]==0)
                            return 0;
                    }
                }
            }
        }
    }
    return 1;
}

int main()
{
    while(~scanf("%s",str))
    {
        if(str[0]=='0')
            break;
        else
        {
            if(put()==1)
                printf("tautology\n");
            else
                printf("not\n");
        }
    }
    return 0;
}