poj 3295

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

题意如下，

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。

代码偷来的

//这个代码让我懂了先根遍历该咋写- -
//慢慢学吧，不会的还有很多
#include<iostream>
#include<cstring>
#include<stack>
using namespace std;

stack <bool> sta;
bool val[5],flag;

int main()
{
    int i,j;
    string str;
    bool tmp1,tmp2;
    while(cin>>str && str!="0")
    {
        flag=1;
        for(i=0;i<32 && flag;i++)
        {
            memset(val,0,sizeof(val));
            for(j=0;j<5;j++)
            {
                if(i & (1<<j))
                    val[j]=1;
            }
            for(j=str.length()-1;j>=0;j--)
            {
                switch(str[j])
                {
                case 'p':case 'q':case 'r':case 's':case 't':
                    sta.push(val[str[j]-'p']);
                    break;
                case 'N':
                    tmp1=sta.top();
                    sta.pop();
                    sta.push(!tmp1);
                    break;
                case 'E':
                    tmp1=sta.top();
                    sta.pop();
                    tmp2=sta.top();
                    sta.pop();
                    sta.push(tmp1==tmp2);
                    break;
                case 'A':
                    tmp1=sta.top();
                    sta.pop();
                    tmp2=sta.top();
                    sta.pop();
                    sta.push(tmp1||tmp2);
                    break;
                case 'K':
                    tmp1=sta.top();
                    sta.pop();
                    tmp2=sta.top();
                    sta.pop();
                    sta.push(tmp1 && tmp2);
                    break;
                case 'C':
                    tmp1=sta.top();
                    sta.pop();
                    tmp2=sta.top();
                    sta.pop();
                    if(!tmp1||tmp2)
                        tmp1=1;
                    else
                        tmp1=0;
                    sta.push(tmp1);
                    break;
                default:
                    break;
                };
            }
            if(!sta.top())
                flag=0;
        }
        if(flag)
            cout<<"tautology"<<endl;
        else
            cout<<"not"<<endl;
    }
    return 0;
}