Poj 3295 Tautology

1.Link：

http://poj.org/problem?id=3295

2.content:

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9793		Accepted: 3712

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

Source

Waterloo Local Contest, 2006.9.30

3.method:

将p,q,r,s,t分别置入以0，1代入，使用五个for循环实现，这样会出现相同计算多次的情况，不过由于循环次数很少，所以不影响大的效率。

将字符串从后向前置入栈，遇到运算符则进行运算，然后将结果置入栈。

最后结果非1的话利用goto跳出多层循环

4.code:

#include <iostream>
#include <string>
#include <stack>

using namespace std;

int main()
{
    //freopen("D://input.txt","r",stdin);

    string str;
    cin >> str;

    while(str.size() != 1 || str[0] != '0')
    {
        //cout << str << endl;

        int p,q,r,s,t;
        for(p = 0; p < 2; ++p)
        {
            for(q = 0; q < 2; ++q)
            {
                for(r = 0; r < 2; ++r)
                {
                    for(s = 0; s < 2; ++s)
                    {
                        for(t = 0; t < 2; ++t)
                        {
                            stack<int> s_str;
                            int w,x;
                            for(string::reverse_iterator r_iter = str.rbegin(); r_iter != str.rend(); ++r_iter)
                            //for(int str_i = str.size() - 1; str_i >= 0; --str_i)
                            {
                                if(*r_iter == 'p') s_str.push(p);
                                else if(*r_iter == 'q') s_str.push(q);
                                else if(*r_iter == 'r') s_str.push(r);
                                else if(*r_iter == 's') s_str.push(s);
                                else if(*r_iter == 't') s_str.push(t);
                                else
                                {
                                    w = s_str.top();
                                    s_str.pop();
                                    if(*r_iter == 'N') s_str.push(!w);
                                    else
                                    {
                                        x = s_str.top();
                                        s_str.pop();
                                        if(*r_iter == 'K') s_str.push(w&&x);
                                        else if(*r_iter == 'A') s_str.push(w||x);
                                        else if(*r_iter == 'C') {if(w == 1 && x == 0) s_str.push(0); else s_str.push(1);}
                                        else {if(w == x) s_str.push(1); else s_str.push(0);}
                                    }
                                }
                            }
                            if(s_str.top() != 1) goto end;
                        }
                    }
                }
            }
        }

end:
        if(p >= 2) cout << "tautology" << endl;
        else cout << "not" << endl;

        cin >> str;
    }
    return 0;
}

 5.Reference:

http://blog.csdn.net/lyy289065406/article/details/6642766