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.
| 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
就是离散数学的一个模拟，判断所给的式子是不是永真公式
用栈逆序推还不算难
1 #include<stdio.h> 2 #include<iostream> 3 #include<string.h> 4 #include<stack> 5 using namespace std; 6 stack<int>sta; 7 int p,q,r,s,t,a,b; 8 char str[102]; 9 int f() 10 { 11 int len=strlen(str); 12 for(int i=len-1; i>-1; i--) 13 { 14 if(str[i]=='p')sta.push(p);//是字符直接入栈 15 else if(str[i]=='q')sta.push(q); 16 else if(str[i]=='r')sta.push(r); 17 else if(str[i]=='s')sta.push(s); 18 else if(str[i]=='t')sta.push(t); 19 else if(str[i]=='K')//是运算就进行运算 20 { 21 a=sta.top(); 22 sta.pop(); 23 b=sta.top(); 24 sta.pop(); 25 sta.push(a&&b); 26 } 27 else if(str[i]=='A') 28 { 29 a=sta.top(); 30 sta.pop(); 31 b=sta.top(); 32 sta.pop(); 33 sta.push(a||b); 34 } 35 else if(str[i]=='C') 36 { 37 a=sta.top(); 38 sta.pop(); 39 b=sta.top(); 40 sta.pop(); 41 if(a&&!b)sta.push(0); 42 else sta.push(1); 43 } 44 else if(str[i]=='E') 45 { 46 a=sta.top(); 47 sta.pop(); 48 b=sta.top(); 49 sta.pop(); 50 if(a==b)sta.push(1); 51 else sta.push(0); 52 } 53 else if(str[i]=='N') 54 { 55 a=sta.top(); 56 sta.pop(); 57 sta.push(!a); 58 } 59 } 60 return sta.top(); 61 } 62 int sf()//枚举各种情况 63 { 64 for(p=0; p<2; p++) 65 for(q=0; q<2; q++) 66 for(r=0; r<2; r++) 67 for(s=0; s<2; s++) 68 for(t=0; t<2; t++) 69 if(!f()) 70 return 0; 71 return 1; 72 } 73 int main() 74 { 75 while(gets(str)) 76 { 77 if(!strcmp(str,"0")) 78 break; 79 else 80 { 81 if(sf()) 82 printf("tautology "); 83 else 84 printf("not "); 85 } 86 } 87 return 0; 88 }