POJ 3295 Tautology（构造法）

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6088		Accepted: 2315

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, Nwis 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

Waterloo Local Contest, 2006.9.30

 

 

 

 

简单题。看代码就应该可以看懂的

/*

POJ 3295

构造法

p,q,r,s,t枚举所有可能取值

用一个堆栈从字符串末尾进行操作



AC  G++  684K  0MS

*/



#include<stdio.h>

#include<iostream>

#include<string.h>

using namespace std;

const int MAXN=120;

int sta[MAXN];//数组实现堆栈

char str[MAXN];

int p,q,r,s,t;

void  DoIt()

{

    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++)

              {

                  DoIt();

                  if(sta[0]==0)return false;

              }

    return true;

}



int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    while(scanf("%s",&str))

    {

        if(strcmp(str,"0")==0)break;

        if(solve())printf("tautology\n");

        else printf("not\n");

    }

    return 0;

}