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;
}