POJ 3295 Tautology(枚举 + 构造)

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5306		Accepted: 1997

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

Source

Waterloo Local Contest, 2006.9.30

解题报告：就是判断所给的式子是否为永真式；其中K表示&&（并）A表示||（或），N表示！（非），C表示a->b(!a||b)，E表示等值(a == b);他们的值有两种1或0，所以用枚举的方法，枚举一遍即可；

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 105;
char str[N];
int flag[6], st;
int Judge()
{
    int p, q;
    switch(str[++st])
    {
        case 'K': p = Judge(); q = Judge(); return p && q;
        case 'A': p = Judge(); q = Judge(); return p || q;
        case 'N': return !Judge();
        case 'C': p = Judge(); q = Judge(); return !p || q;
        case 'E': return Judge() == Judge();
        default: return flag[str[st] - 'p' ];
    }
}
int main()
{
    int i, j, k, l, m, ans;
    while(scanf("%s", str) != EOF && str[0] != '0')
    {
        ans = 1;
        for (i = 0; i <= 1; ++i)//相当于枚举
        {
            flag[0] = i;
            for (j = 0; j <= 1; ++j)
            {
                flag[1] = j;
                for (k = 0; k <= 1; ++k)
                {
                    flag[2] = k;
                    for (l = 0; l <= 1; ++l)
                    {
                        flag[3] = l;
                        for (m = 0; m <= 1; ++m)
                        {
                            flag[4] = m;
                            st = -1;
                            ans = ans && Judge();
                            if (!ans)
                            {
                                break;
                            }
                        }
                    }
                }
            }
        }
        if (ans)
        {
            printf("tautology\n");
        }
        else
        {
            printf("not\n");
        }
    }
    return 0;
}