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
做这题的时候正好学了离散- -！。。一开始想的是从前面开始模拟，发现不容易实现，因为比如要实现Apq，告诉了p，但是后面部分还是个算式的话就不容易
算了。所以想到了从后往前不断递推。。这样的话最后栈内的结果就是最后的答案
View Code
1 #include <cstdio> 2 #include <cstring> 3 #include <stack> 4 using namespace std; 5 int p,q,r,s,t; 6 int charge(char c) 7 { 8 if(c=='p') return p; 9 if(c=='q') return q; 10 if(c=='r') return r; 11 if(c=='s') return s; 12 if(c=='t') return t; 13 } 14 int main() 15 { 16 char str[110]; 17 int i,x,y; 18 while(scanf("%s",str)&&str[0]!='0') 19 { 20 int flag=1; 21 stack<int>st; 22 for(p=0;p<=1&&flag;p++) 23 for(q=0;q<=1&&flag;q++) 24 for(r=0;r<=1&&flag;r++) 25 for(s=0;s<=1&&flag;s++) 26 for(t=0;t<=1&&flag;t++) 27 { 28 for(i=strlen(str)-1;i>=0;i--) 29 { 30 if(str[i]=='p'||str[i]=='q'||str[i]=='r' 31 ||str[i]=='s'||str[i]=='t') 32 st.push(charge(str[i])); 33 else if(str[i]=='K') 34 { 35 x=st.top(); 36 st.pop(); 37 y=st.top(); 38 st.pop(); 39 st.push(x&&y); 40 } 41 else if(str[i]=='A') 42 { 43 x=st.top(); 44 st.pop(); 45 y=st.top(); 46 st.pop(); 47 st.push(x||y); 48 } 49 else if(str[i]=='N') 50 { 51 x=st.top(); 52 st.pop(); 53 st.push(!x); 54 } 55 else if(str[i]=='C') 56 { 57 x=st.top(); 58 st.pop(); 59 y=st.top(); 60 st.pop(); 61 if(x==1&&y==0) st.push(0); 62 else st.push(1); 63 } 64 else if(str[i]=='E') 65 { 66 x=st.top(); 67 st.pop(); 68 y=st.top(); 69 st.pop(); 70 st.push((x==y)); 71 } 72 else 73 flag=0; 74 } 75 if(flag) 76 { 77 flag=st.top(); 78 st.pop(); 79 } 80 } 81 if(flag) 82 printf("tautology\n"); 83 else 84 printf("not\n"); 85 } 86 return 0; 87 }