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.
- 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
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：逻辑表达式求知，K, A, N, C, E为逻辑运算符，p, q, r, s, and t 为真假值。
Kxy -> x&&y
Axy -> x||y
Nx -> !x
Cxy -> x||(!y)
Exy -> x==y
判断是否表达式恒为真。
解题思路： 数值变量总共就5个，枚举这五个变量的值，有32种情况。
处理字符串的时候，类似于逆波兰表达式的求值过程。
从(S.length()-1)->0遍历，遇到小写字母将其对应的布尔值存入栈。
遇到大写字母（除N外） 去除栈顶2个元素进行处理，后存入栈。
遇到N，去除栈顶一个元素，取反后存入栈。
遍历完返回S.top()。
Code：
1 #include<iostream> 2 #include<string> 3 #include<stack> 4 using namespace std; 5 int q,p,s,r,t; 6 bool Is_Tau(string S) 7 { 8 int len=S.length()-1,i,t1,t2; 9 stack<char> ST; 10 for (i=len;i>=0;i--) 11 { 12 if (S[i]=='q') ST.push(q); 13 if (S[i]=='p') ST.push(p); 14 if (S[i]=='r') ST.push(r); 15 if (S[i]=='s') ST.push(s); 16 if (S[i]=='t') ST.push(t); 17 if (S[i]=='K') 18 { 19 t1=ST.top(); 20 ST.pop(); 21 t2=ST.top(); 22 ST.pop(); 23 ST.push(t1&&t2); 24 } 25 if (S[i]=='A') 26 { 27 t1=ST.top(); 28 ST.pop(); 29 t2=ST.top(); 30 ST.pop(); 31 ST.push(t1||t2); 32 } 33 if (S[i]=='C') 34 { 35 t1=ST.top(); 36 ST.pop(); 37 t2=ST.top(); 38 ST.pop(); 39 ST.push(t1||(!t2)); 40 } 41 if (S[i]=='E') 42 { 43 t1=ST.top(); 44 ST.pop(); 45 t2=ST.top(); 46 ST.pop(); 47 ST.push(t1==t2); 48 } 49 if (S[i]=='N') 50 { 51 t1=ST.top(); 52 ST.pop(); 53 ST.push(!t1); 54 } 55 } 56 return ST.top(); 57 } 58 int main() 59 { 60 string WFF; 61 while (cin>>WFF) 62 { 63 int OK=1; 64 if (WFF=="0") break; 65 for (q=0; q<=1; q++) 66 for (p=0; p<=1; p++) 67 for (r=0; r<=1; r++) 68 for (s=0; s<=1; s++) 69 for (t=0; t<=1; t++) 70 if (!Is_Tau(WFF)) 71 { 72 OK=0; 73 break; 74 } 75 if (OK) printf("tautology "); 76 else printf("not "); 77 } 78 return 0; 79 }