2013-07-12 13:22:46Morris
[UVA][前序] 11108 - Tautology
Problem D: Tautology
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.
- 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.
|
|
| 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 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. For each test case, output a line containing tautology or not as appropriate.
Sample Input
ApNp ApNq 0
Possible Output for Sample Input
tautology not
Gordon V. Cormack
題目給定一個前序式，小寫表示運算元(0/1)，大寫表示運算子。
tautology 的定義是，不管變數值為何，運算出來的值恆為真(true 1)。
窮舉所有變數的分配方式 2^5 = 32 種，進行前序運算。
#include <stdio.h>
#include <string.h>
#include <stack>
using namespace std;
int main() {
char s[1005];
int i, j;
while(scanf("%s", s) && s[0] != '0') {
int ret = 1, len = strlen(s);
int a, b;
for(i = 0; i < 32; i++) {
stack<int> stk;
for(j = len-1; j >= 0; j--) {
if(s[j] >= 'p') {
int val = (i>>(s[j]-'p'))&1;
stk.push(val);
} else {
a = stk.top(), stk.pop();
if(s[j] != 'N') {
b = stk.top(), stk.pop();
}
if(s[j] == 'K')
stk.push(a&b);
if(s[j] == 'A')
stk.push(a|b);
if(s[j] == 'N')
stk.push(!a);
if(s[j] == 'C')
stk.push(!(a&(!b)));
if(s[j] == 'E')
stk.push(a == b);
}
}
ret &= stk.top();
}
puts(ret ? "tautology" : "not");
}
return 0;
}