标签：
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:
| 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 Outputtautology
not
大致题意：
输入由p、q、r、s、t、K、A、N、C、E共10个字母组成的逻辑表达式，
其中p、q、r、s、t的值为1（true）或0（false），即逻辑变量；
K、A、N、C、E为逻辑运算符，
K --> and: x && y
A --> or: x || y
N --> not : !x
C --> implies : (!x)||y
E --> equals : x==y
问这个逻辑表达式是否为永真式。
分析：共 p，q，r，s，t 五个变量，也就是说最多共32种情况，分别进行枚举。
如果表达式的值有结果为零，就输出 not，因为已经构不成永真式
1 #include <iostream> 2 #include <cstring> 3 #include <stack> 4 using namespace std; 5 char str[150]; 6 stack<int>s; 7 int p,q,r,ss,t; //各个逻辑变量的状态 8 int wff; 9 bool fun(char ch) //如果是逻辑变量则压栈 10 { 11 switch(ch) 12 { 13 case ‘p‘: 14 s.push(p); 15 return true; 16 case ‘q‘: 17 s.push(q); 18 return true; 19 case ‘r‘: 20 s.push(r); 21 return true; 22 case ‘s‘: 23 s.push(ss); 24 return true; 25 case ‘t‘: 26 s.push(t); 27 return true; 28 default: 29 return false; 30 } 31 } 32 void operators(char ch) 33 { 34 int a,b; 35 switch(ch) 36 { 37 case ‘K‘: 38 { 39 int a=s.top(); 40 s.pop(); 41 int b=s.top(); 42 s.pop(); 43 s.push(a&&b); 44 break; 45 } 46 case ‘A‘: 47 { 48 a=s.top(); 49 s.pop(); 50 b=s.top(); 51 s.pop(); 52 s.push(a||b); 53 break; 54 } 55 case ‘N‘: 56 { 57 a=s.top(); 58 s.pop(); 59 s.push(!a); 60 break; 61 } 62 case ‘C‘: 63 { 64 a=s.top(); 65 s.pop(); 66 b=s.top(); 67 s.pop(); 68 s.push((!a)||b); 69 break; 70 } 71 case ‘E‘: 72 { 73 a=s.top(); 74 s.pop(); 75 b=s.top(); 76 s.pop(); 77 s.push(a==b); 78 break; 79 } 80 } 81 return ; 82 } 83 int main() 84 { 85 while(cin>>str) 86 { 87 if(str[0]==‘0‘) 88 break; 89 int len=strlen(str)-1; 90 bool flag=false; 91 for(p=0; p<=1; p++) 92 { 93 for(q=0; q<=1; q++) 94 { 95 for(r=0; r<=1; r++) 96 { 97 for(ss=0; ss<=1; ss++) 98 { 99 for(t=0; t<=1; t++) 100 { 101 for(wff=len; wff>=0; wff--) 102 { 103 if(fun(str[wff])==0) 104 operators(str[wff]); 105 } 106 int ans=s.top(); //栈中最后一个元素 107 s.pop(); 108 if(ans==0) 109 { 110 flag=true; 111 } 112 } 113 if(flag) 114 break; 115 } 116 if(flag) 117 break; 118 } 119 if(flag) 120 break; 121 } 122 if(flag) 123 break; 124 } 125 if(flag) 126 cout<<"not"<<endl; 127 else 128 cout<<"tautology"<<endl; 129 } 130 return 0; 131 }
标签：
原文地址：http://www.cnblogs.com/cxbky/p/4915345.html
