POJ 3295-Tautology（模拟-逻辑表达式）

水深无声 2022-07-11 07:15 117阅读 0赞

Tautology

<table> 
 <tbody> 
 <tr> 
 <td>Time Limit:&nbsp;1000MS</td> 
 <td>&nbsp;</td> 
 <td>Memory Limit:&nbsp;65536K</td> 
 </tr> 
 <tr> 
 <td>Total Submissions:&nbsp;12534</td> 
 <td>&nbsp;</td> 
 <td>Accepted:&nbsp;4776</td> 
 </tr> 
 </tbody> 
</table>

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, N*w* is a WFF
 *  if *w* and *x* are WFFs, K*wx*, A*wx*, C*wx*, and E*wx* 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.

<table> 
 <tbody> 
 <tr> 
 <td> Definitions of K, A, N, C, and E </td> 
 </tr> 
 </tbody> 
</table>

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

## 题目意思： ##

p,q,r,s,t是变量，取值是0或1；

K,A,N,C,E是逻辑运算符，运算规则分别是：

————————————

K and x&&y  
A or x||y  
N not !x  
C implies (!x)||y  
E equals x==y

————————————

给定一个表达式，计算其是否恒为真，如果是，输出“tautology”，否则输出“not”。

## 解题思路： ##

由于这些逻辑运算符都是满足左结合性的，所以读取表达式后，需要从后向前依次进行计算。

而且5个变量一共有2^5=32种情况，所以可使用5重循环来模拟各个情况，分别代入计算判断即可。

#include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 
 int K(int x,int y)//K and x&&y
 {
 return x&&y;
 }
 int A(int x,int y)//A or x||y
 {
 return x||y;
 }
 int N(int x)//N not !x
 {
 return !x;
 }
 int C(int x,int y)//C implies (!x)||y
 {
 return A(N(x),y);
 }
 int E(int x,int y)//E equals x==y
 {
 return x==y;
 }
 
 string str;
 int c[110];
 int p,q,r,s,t;
 int x,y;
 bool solve()
 {
 memset(c,'\0',sizeof(c));
 int j=0;
 for(int i=str.size(); i>=0; --i)
 {
 switch(str[i])
 {
 case 'p':
 c[j++]=p;
 break;
 case 'q':
 c[j++]=q;
 break;
 case 'r':
 c[j++]=r;
 break;
 case 's':
 c[j++]=s;
 break;
 case 't':
 c[j++]=t;
 break;
 
 case 'K':
 x=c[--j];
 y=c[--j];
 c[j++]=K(x,y);
 break;
 case 'A':
 x=c[--j];
 y=c[--j];
 c[j++]=A(x,y);
 break;
 case 'N':
 x=c[--j];
 c[j++]=N(x);
 break;
 case 'C':
 x=c[--j];
 y=c[--j];
 c[j++]=C(x,y);
 break;
 case 'E':
 x=c[--j];
 y=c[--j];
 c[j++]=E(x,y);
 break;
 }
 }
 if(c[0]==0) return false;
 else return true;
 }
 
 int main()
 {
 while(cin>>str&&str!="0")
 {
 bool flag=true;
 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++)
 {
 if(!solve())
 {
 flag=false;
 break;
 }
 }
 if(flag) puts("tautology");
 else puts("not");
 }
 return 0;
 }
 /*
 ApNp
 ApNq
 0
 */

[Waterloo Local Contest]: http://poj.org/searchproblem?field=source&key=Waterloo+Local+Contest