POJ 3295：Tautology

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10482		Accepted: 3982

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

这个题记得是离散数学里面的内容，题意是判断给定的字符串是不是永远为1，就是不管p、q、r、s、t取什么值，其结果都是1。

1AC。反正自从遇到了上一次类似的题目之后，做这种题自己的感受就是两点：

1.构造一个栈

2.从后往前面撸。

代码：

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <cmath>
using namespace std;

stack <int> o_sta;
int p,q,r,s,t;
string test;
int len,i;

void push1(char a)
{
switch (a)
{
case 'p':
o_sta.push(p);
break;
case 'q':
o_sta.push(q);
break;
case 'r':
o_sta.push(r);
break;
case 's':
o_sta.push(s);
break;
case 't':
o_sta.push(t);
break;
default:
break;
}
}

void cal(char a)
{
int temp1,temp2;
switch (a)
{
case 'N':
temp1=o_sta.top();
o_sta.pop();
temp1=!temp1;
o_sta.push(temp1);
break;
case 'K':
temp1=o_sta.top();
o_sta.pop();
temp2=o_sta.top();
o_sta.pop();
temp1=temp1&temp2;
o_sta.push(temp1);
break;
case 'A':
temp1=o_sta.top();
o_sta.pop();
temp2=o_sta.top();
o_sta.pop();
temp1=temp1|temp2;
o_sta.push(temp1);
break;
case 'C':
temp1=o_sta.top();
o_sta.pop();
temp2=o_sta.top();
o_sta.pop();
temp1=temp1-temp2;
if(temp1==1)
o_sta.push(0);
else
o_sta.push(1);
break;
case 'E':
temp1=o_sta.top();
o_sta.pop();
temp2=o_sta.top();
o_sta.pop();
temp1=temp1-temp2;
if(temp1==0)
o_sta.push(1);
else
o_sta.push(0);
break;
default:
break;
}
}

bool solve()
{
for(p=0;p<=1;p++)
{
for(q=0;q<=1;q++)
{
for(r=0;r<=1;r++)
{
for(s=0;s<=1;s++)
{
for(t=0;t<=1;t++)
{
for(i=len-1;i>=0;i--)
{
if(test[i]=='p'||test[i]=='q'||test[i]=='r'||test[i]=='s'||test[i]=='t')
push1(test[i]);
else
cal(test[i]);
}
if(o_sta.top()==0)
return false;
}
}
}
}
}
return true;
}
int main()
{
while(cin>>test)
{
if(test=="0")
break;
len=test.length();

if(solve())
{
cout<<"tautology"<<endl;
}
else
{
cout<<"not"<<endl;
}
}
//system("pause");
return 0;
}