Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 10437   Accepted: 3963

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

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

p,q,r,s,t,是五个二进制数。

K,A,N,C,E,是五个运算符。

K:&&

A:||
N:!

C:(!w)||x

E:w==x

这道题可以将p,q,r,s,t穷举

 #include <iostream>
#include<string.h>
#include <stack>
using namespace std;
int p, q, r, s, t;
int len;
char str[];
int result() {
stack<int> res;
int t1,t2;
for (int i = len - ; i >= ; i--) {
switch (str[i]) {
case 'p':
res.push(p);
break;
case 'q':
res.push(q);
break;
case 'r':
res.push(r);
break;
case 's':
res.push(s);
break;
case 't':
res.push(t);
break;
case 'K':
t1 = res.top();
res.pop();
t2 = res.top();
res.pop();
if (t1 & t2) {
res.push();
} else {
res.push();
}
break;
case 'A':
t1 = res.top();
res.pop();
t2 = res.top();
res.pop();
if (t1 | t2)
res.push();
else
res.push();
break;
case 'N':
t1 = res.top();
res.pop();
if (~t1 & )
res.push();
else
res.push();
break;
case 'C':
t1 = res.top();
res.pop();
t2 = res.top();
res.pop();
if ((~t1 & ) | t2) {
res.push();
} else {
res.push();
}
break;
case 'E':
t1 = res.top();
res.pop();
t2 = res.top();
res.pop();
if (t1 == t2) {
res.push();
} else {
res.push();
}
break;
}
}
return res.top();
} int fun() {
int flag;
for (p = ; p < ; p++) {
for (q = ; q < ; q++) {
for (r = ; r < ; r++) {
for (s = ; s < ; s++) {
for (t = ; t < ; t++) {
flag = result();
if(flag==)
return ;
}
}
}
}
}
return ;
} int main() {
while (cin >> str) {
if (strcmp(str, "") == )
break;
len = strlen(str);
int flag=fun();
if(flag)
cout<<"tautology"<<endl;
else
cout<<"not"<<endl;
}
}

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