POJ 3295 Tautology (栈模拟)

Tautology

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10096 Accepted: 3847

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, andequals 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

ApNpApNq0

Sample Output

tautologynot

Source

Waterloo Local Contest, 2006.9.30

题目链接：http://poj.org/problem?id=3295

题目大意：按题意计算所给表达式的值，相当于一个带栈功能的位运算器

题目分析：暴力枚举所有p,q,r,s,t的可能值，用栈模拟算出结果，若为0则not否则是tautology

#include <cstdio>#include <stack>#include <cstring>using namespace std;int const MAX = 150;stack <int> st;char s[MAX];void cal(int p, int q, int r, int s1, int t){    int len = strlen(s);    for(int i = len - 1; i >= 0; i --)    {        if(s[i] == 'p')            st.push(p);        else if(s[i] == 'q')            st.push(q);        else if(s[i] == 'r')            st.push(r);        else if(s[i] == 's')            st.push(s1);        else if(s[i] == 't')            st.push(t);        else if(s[i] == 'K')        {            int t1 = st.top();            st.pop();            int t2 = st.top();            st.pop();            st.push(t1 && t2);        }        else if(s[i] == 'A')        {            int t1 = st.top();            st.pop();            int t2 = st.top();            st.pop();            st.push(t1 || t2);        }        else if(s[i] == 'N')        {            int t1 = st.top();            st.pop();            st.push(!t1);        }        else if(s[i] == 'C')        {            int t1 = st.top();            st.pop();            int t2 = st.top();            st.pop();            if(t1 == 1 && t2 == 0)                st.push(0);            else                 st.push(1);        }        else if(s[i] == 'E')        {            int t1 = st.top();            st.pop();            int t2 = st.top();            st.pop();            if((t1 == 1 && t2 == 1) || (t1 == 0 && t2 == 0))                st.push(1);            else                st.push(0);        }    }}bool judge(){    for(int p = 0; p < 2; p++)        for(int q = 0; q < 2; q++)            for(int r = 0; r < 2; r++)                 for(int s1 = 0; s1 < 2; s1++)                    for(int t = 0; t < 2; t++)                     {                        cal(p, q, r, s1, t);                        if(st.top() == 0)                            return false;                    }    return true;}int main(){    while(scanf("%s", s) != EOF && !(strlen(s) == 1 && s[0] == '0'))       {        if(judge())            printf("tautology\n");        else            printf("not\n");    } }