POJ 2049Tautology（模拟）

Tautology

Time Limit: 1000MS Memory Limit: 65536KTotal 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 ofp. On the other hand,ApNq is not, because it has the value 0 forp=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

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;}