poj3295 Tautology

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16817		Accepted: 6500

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

思路：把真值表枚举完就行。

下附代码：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int MAXN = 120;
int sta[MAXN];
char str[MAXN];
int p, q, r, s, t;
void judge() {
    int top = 0;
    int len = strlen(str);
    for (int i = len - 1; i >= 0; i--) {
        if (str[i] == 'p')
            sta[top++] = p;
        else if (str[i] == 'q')
            sta[top++] = q;
        else if (str[i] == 'r')
            sta[top++] = r;
        else if (str[i] == 's')
            sta[top++] = s;
        else if (str[i] == 't')
            sta[top++] = t;
        else if (str[i] == 'K') {
            int t1 = sta[--top];
            int t2 = sta[--top];
            sta[top++] = (t1 && t2);
        } else if (str[i] == 'A') {
            int t1 = sta[--top];
            int t2 = sta[--top];
            sta[top++] = (t1 || t2);
        } else if (str[i] == 'N') {
            int t1 = sta[--top];
            sta[top++] = (!t1);
        } else if (str[i] == 'C') {
            int t1 = sta[--top];
            int t2 = sta[--top];
            if (t1 == 1 && t2 == 0)
                sta[top++] = 0;
            else
                sta[top++] = 1;
        } else if (str[i] == 'E') {
            int t1 = sta[--top];
            int t2 = sta[--top];
            if ((t1 == 1 && t2 == 1) || (t1 == 0 && t2 == 0))
                sta[top++] = 1;
            else
                sta[top++] = 0;
        }
    }
}
bool solve() {
    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++) {
                        judge();
                        if (sta[0] == 0)
                            return 0;
                    }
    return 1;
}
int main() {
    while (scanf("%s", str)) {
        if (strcmp(str, "0") == 0)
            break;
        if (solve())
            printf("tautology
");
        else
            printf("not
");
    }
    return 0;
}