poj 3295 Tautology

题目链接：点击打开链接

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：判断是否为永真

输入由p、q、r、s、t、K、A、N、C、E共10个字母组成的逻辑表达式，
K --> and:  x && y
A --> or:  x || y
N --> not :  !x
C --> implies :  (!x)||y
E --> equals :  x==y

其中p q r s t 的真假要假设，输入字符串，判断其是否为永真式

基本思路：将5个字母用5个for循环假设其真假值，将字符串逆序判断，存入栈中，最后会只剩下一个值，判断是否为真．

注　每一个假设都使字符串为真时才是永真式

<span style="font-size:18px;">#include <iostream>
#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
int p,q,r,s,t;
stack<int > h;
char str[1000];
char a[1000];
bool ch(char c)
{
switch(c)
{
case 'p':
h.push(p);
return true;
case 'q':
h.push(q);
return true;
case 'r':
h.push(r);
return true;
case 's':
h.push(s);
return true;
case 't':
h.push(t);
return true;
}
return false;
}
void luoji(char c)
{
switch(c)
{
case 'K' :
{
int x=h.top();
h.pop();
int y=h.top();
h.pop();
h.push(x&&y);
return ;
}
case 'A' :
{
int x=h.top();
h.pop();
int y=h.top();
h.pop();
h.push(x||y);
return ;
}
case 'N' :
{
int x=h.top();
h.pop();
h.push(!x);
return ;
}
case 'C' :
{
int x=h.top();
h.pop();
int y=h.top();
h.pop();
h.push((!x)||y);
return ;
}
case 'E' :
{
int x=h.top();
h.pop();
int y=h.top();
h.pop();
h.push(x==y);
return ;
}
}
}
int  f()
{
int len=strlen(str);
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(int i=len-1; i>=0; i--)
{
if(!ch(str[i]))
{
luoji(str[i]);
}
}
int ans=h.top();
if(!ans)
{
return 0;
}
}
}
}
}
}
return 1;
}
int main()
{

while(cin>>str&&str[0]!='0')
{
int d=f();
if(d==0)printf("not\n");
else printf("tautology\n");
}
return 0;
}</span>