Poj 3295 tautology (simulation)

Question link: http://poj.org/problem? Id = 3295

Given a logical expression, you can determine whether the formula is permanent.

This is a simple question for children with a little ACM Foundation.

But I spent two hours wa several times.

Cause 1:

The test data mistaken for the question is to ensure that the values will be met before the meeting, and there will be a logical symbol waiting for them after the two values (N). It turns out that I am wrong, my Wa is sad.

Later I thought of this, but I was not sure. I tested the condition on default and used a while (1) for testing. If OJ returns timeout, it indicates that my conjecture is correct, return timeout;

Cause 2:

This error is detected quickly, but it is not obvious, that is, the switch statement. If it is not executed after the case statement, the following s -- that is, oh, no!

Switch (name [Len]) {s --; // This statement will never run case 'K ': ch = my_map [Rec [s] & my_map [Rec [s-1]? '1': '0'; s --; break; Case 'A': CH = my_map [Rec [s] | my_map [Rec [s-1]? '1': '0'; s --; break; Case 'E': CH = my_map [Rec [s] = my_map [Rec [s-1]? '1': '0'; s --; break; Case 'C': CH = (! My_map [Rec [s]) | my_map [Rec [s-1]? '1': '0'; s --; break; Case 'N': CH = my_map [Rec [s]? '0': '1'; break; // default: printf ("s: % d \ n", S );}

Then the problem is solved. We can simulate the stack according to the normal situation!

Here I use map to correspond to the true value of pqr. This is more convenient!

# Include <iostream> # include <stdio. h> # include <map> # include <string. h> using namespace STD; char STR [200]; char temp [200]; char rec [200]; Map <char, bool> my_map; // when this map is to be inserted, note that bool is_ OK (char * Name, int Len) {char ch; int S = 0; Len -- is also inserted at the time of 0 and 1 --; while (LEN> = 0) {If (name [Len]> = 'A' & name [Len] <= 'Z ') {rec [s ++] = Name [Len]; Len --;} else {s --; Switch (name [Len]) {Case 'K ': ch = my_map [Rec [s] & my_map [Rec [s-1]? '1': '0'; s --; break; Case 'A': CH = my_map [Rec [s] | my_map [Rec [s-1]? '1': '0'; s --; break; Case 'E': CH = my_map [Rec [s] = my_map [Rec [s-1]? '1': '0'; s --; break; Case 'C': CH = (! My_map [Rec [s]) | my_map [Rec [s-1]? '1': '0'; s --; break; Case 'N': CH = my_map [Rec [s]? '0': '1'; break;} rec [s ++] = CH; Len -- ;}return my_map [Rec [0];} int main () {int P, Q, R, S, T, Len; bool flag; while (scanf ("% s", STR) {flag = true; len = strlen (STR); If (strcmp (STR, "0") = 0) return 0; For (P = 0; P <2 & flag; P ++) for (q = 0; q <2 & flag; q ++) for (r = 0; r <2 & flag; r ++) for (S = 0; S <2 & flag; s ++) for (t = 0; t <2 & flag; t ++) {my_map ['0'] = false; my_map ['1'] = true; strcpy (temp, STR); my_map ['P'] = P; my_map ['q '] = Q; my_map ['R'] = r; my_map ['s'] = s; my_map ['T'] = T; If (! Is_ OK (temp, Len) Flag = false; my_map.clear ();} If (FLAG) printf ("tautology \ n"); else printf ("not \ n ");} return 0 ;}