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 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 题目大意：逻辑表达式求知，K, A, N, C, E为逻辑运算符，p, q, r, s, and t 为真假值。
Kxy -> x&&y
Axy -> x||y
Nx -> !x
Cxy -> x||(!y)
Exy -> x==y
判断是否表达式恒为真。
解题思路： 数值变量总共就5个，枚举这五个变量的值，有32种情况。
处理字符串的时候，类似于逆波兰表达式的求值过程。
从(S.length()-1)->0遍历，遇到小写字母将其对应的布尔值存入栈。
遇到大写字母（除N外） 去除栈顶2个元素进行处理，后存入栈。
遇到N，去除栈顶一个元素，取反后存入栈。
遍历完返回S.top()。
Code：
#include<iostream>
#include<string>
#include<stack>
using namespace std;
int q,p,s,r,t;
bool Is_Tau(string S)
{
int len=S.length()-,i,t1,t2;
stack<char> ST;
for (i=len;i>=;i--)
{
if (S[i]=='q') ST.push(q);
if (S[i]=='p') ST.push(p);
if (S[i]=='r') ST.push(r);
if (S[i]=='s') ST.push(s);
if (S[i]=='t') ST.push(t);
if (S[i]=='K')
{
t1=ST.top();
ST.pop();
t2=ST.top();
ST.pop();
ST.push(t1&&t2);
}
if (S[i]=='A')
{
t1=ST.top();
ST.pop();
t2=ST.top();
ST.pop();
ST.push(t1||t2);
}
if (S[i]=='C')
{
t1=ST.top();
ST.pop();
t2=ST.top();
ST.pop();
ST.push(t1||(!t2));
}
if (S[i]=='E')
{
t1=ST.top();
ST.pop();
t2=ST.top();
ST.pop();
ST.push(t1==t2);
}
if (S[i]=='N')
{
t1=ST.top();
ST.pop();
ST.push(!t1);
}
}
return ST.top();
}
int main()
{
string WFF;
while (cin>>WFF)
{
int OK=;
if (WFF=="") break;
for (q=; q<=; q++)
for (p=; p<=; p++)
for (r=; r<=; r++)
for (s=; s<=; s++)
for (t=; t<=; t++)
if (!Is_Tau(WFF))
{
OK=;
break;
}
if (OK) printf("tautology\n");
else printf("not\n");
}
return ;
}