POJ 2049Tautology（模拟）

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12454		Accepted: 4738

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

题意：有p，q，r，s，t五个变量，分别取1或0；有五个关系表达式’K‘（与关系）、’A‘（或关系）、’N‘（非关系）、’C‘（没有定义，只有0,1时为0）、’E‘（同或关系）；输入一行不超过100个字符的字符串，如果五个变量无论什么取值都结果为1输出tautology，否则输出not。

题解：用栈从后往前做运算，当找到一个0时就标记flag=0；结束循环。

#include <iostream>
#include <stdio.h>
#include <stack>
#include <map>
#include <string.h>

using namespace std;

map<char, int>mp;

char str[150];

int judge()
{
    stack<int>p;
    int L = strlen(str);
    for(int i = L-1; i >= 0; i--)
    {
        if(str[i] == 'K'||str[i] == 'A'||str[i] == 'C'||str[i] == 'E')
        {
            int num2 = p.top();
            p.pop();
            int num1 = p.top();
            p.pop();
            if(str[i] == 'K')
            {
                p.push(num1&&num2);
            }
            else if(str[i] == 'A')
            {
                p.push(num1||num2);
            }
            else if(str[i] == 'C')
            {
                if(num1 == 1&&num2 == 0)
                    p.push(0);
                else
                    p.push(1);
            }
            else if(str[i] == 'E')
            {
                if((num1 == 0&&num2==0)||(num1==1&&num2==1))
                {
                    p.push(1);
                }
                else
                    p.push(0);
            }
        }
        else if(str[i] == 'N')
        {
            int num = p.top();
            p.pop();
            p.push(!num);
        }
        else
        {
            p.push(mp[str[i]]);
        }
    }
    int num = p.top();
    while(!p.empty())
        p.pop();
    return num;
}

int main()
{
    int p, q, r, s, t;
    while(~scanf("%s", str) && str[0]!='0')
    {
        int flag = 1;
        for(p = 0; p < 2; p++)
        {
            if(flag == 0)
                break;
            mp['p'] = p;
            for(q = 0; q < 2&&flag; q++)
            {
                if(flag == 0)
                    break;
                mp['q'] = q;
                for(r = 0; r < 2&&flag; r++)
                {
                    if(flag == 0)
                        break;
                    mp['r'] = r;
                    for(s = 0; s < 2&&flag; s++)
                    {
                        if(flag == 0)
                            break;
                        mp['s'] = s;
                        for(t = 0; t < 2&&flag; t++)
                        {
                            if(flag == 0)
                                break;
                            mp['t'] = t;
                            if(judge() == 0)
                                flag = 0;
                        }
                    }
                }
            }
        }
        if(flag == 0)
            printf("not\n");
        else
            printf("tautology\n");
    }
    return 0;
}