[ACM] poj 3295 tautology (constructor)

Tautology

Time limit:1000 ms		Memory limit:65536 K
Total submissions:9302		Accepted:3549

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
• IfWIs a WFF, nWIs a WFF
• IfWAndXAre wffs, KWX,WX, CWX, And EWXAre 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 meanAnd, or, not, implies,AndEqualsAs 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

ATautologyIs a WFF that has value 1 (true) regardless of the values of its variables. For example,ApnpIs a tautology because it is true regardless of the valueP. On the other hand,ApnqIs not, because it has the value 0P = 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 containingTautologyOrNotAs appropriate.

Sample Input

ApNpApNq0

Sample output

tautologynot

Source

Waterloo local contest, 2006.9.30

Solution:

Calculate the value of the logical expression based on different input strings. The stack is used here, which is similar to the expression evaluate idea. The string entered from the back to the front is written into the stack if it is a number (P, Q, R, S, T in the question, if it is an operator, the number is obtained from the stack, and then the operation is performed on the stack. The numbers P, Q, R, S, and T in the question are only 0, 1 values. Therefore, the result is a 5-round loop and the obtained expression is evaluated. Exit when 0 is encountered.

Code:

# Include <iostream> # include <string. h ># include <stack> using namespace STD; string WFF; int P, Q, R, S, T; bool OK; int compute (string Str) // stack operation {stack <int> st; int Len = Str. length (); For (INT I = len-1; I> = 0; I --) {If (STR [I] = 'P') ST. push (p); else if (STR [I] = 'q') ST. push (Q); else if (STR [I] = 'R') ST. push (r); else if (STR [I] = 's') ST. push (s); else if (STR [I] = 'T') ST. push (t); else if (STR [I] = 'k') {int x = ST. top (); St. pop (); int y = ST. top (); ST. pop (); ST. push (X & Y);} else if (STR [I] = 'A') {int x = ST. top (); ST. pop (); int y = ST. top (); ST. pop (); ST. push (X | Y);} else if (STR [I] = 'n') {int x = ST. top (); ST. pop (); ST. push (! X);} else if (STR [I] = 'C') {int x = ST. top (); ST. pop (); int y = ST. top (); ST. pop (); ST. push (! X | Y);} else if (STR [I] = 'E') {int x = ST. top (); ST. pop (); int y = ST. top (); ST. pop (); ST. push (x = y) ;}} return St. top () ;}int main () {While (CIN >>> WFF & WFF! = "0") {OK = 1; for (P = 0; P <2; p ++) // The enumerated result. If it is 0, the loop 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 (compute (WFF) = 0) {OK = 0; goto label ;}} label: if (OK) cout <"tautology" <Endl; else cout <"not" <Endl;} return 0 ;}