| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10107 | Accepted: 3850 |
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
Source
| 13990104 | njczy2010 | 3295 | Accepted | 552K | 47MS | G++ | 2592B | 2015-03-21 17:29:21 |
1 #include <cstdio> 2 #include <cstring> 3 #include <stack> 4 #include <vector> 5 #include <algorithm> 6 #include <map> 7 8 #define ll long long 9 int const N = 105; 10 int const M = 205; 11 int const inf = 1000000000; 12 ll const mod = 1000000007; 13 14 using namespace std; 15 16 int l; 17 char ssss[N]; 18 int flag; 19 int ss[N]; 20 21 void ini() 22 { 23 flag=1; 24 l=strlen(ssss); 25 //printf(" l=%d ",l); 26 } 27 28 int fun(int x,int a,int b) 29 { 30 if(x==2){ 31 return a & b; 32 } 33 else if(x==3){ 34 return a | b; 35 } 36 else if(x==4){ 37 if(a==1 && b==0){ 38 return 0; 39 } 40 else return 1; 41 } 42 else if(x==5){ 43 return a==b ? 1 : 0; 44 } 45 return 0; 46 } 47 48 void cal(int p,int q,int r,int s,int t) 49 { 50 // printf(" p=%d q=%d flag=%d ",p,q,flag); 51 int top; 52 map<char,int>mp; 53 mp['p']=p; 54 mp['q']=q; 55 mp['r']=r; 56 mp['s']=s; 57 mp['t']=t; 58 mp['K']=2; 59 mp['A']=3; 60 mp['N']=6; 61 mp['C']=4; 62 mp['E']=5; 63 top=0; 64 int te; 65 int a,b,x; 66 for(int i=0;i<l;i++){ 67 te=mp[ ssss[i] ]; 68 top++; 69 ss[top]=te; 70 if(te>=2){ 71 continue; 72 } 73 else{ 74 while(1) 75 { 76 if(top>=2 && ss[top-1]==6){ 77 te=ss[top]; 78 top--; 79 ss[top]=1-te; 80 } 81 else if(top>=3 && ss[top]<=1 && ss[top-1]<=1 && ss[top-2]>=2 && ss[top-2]<=5){ 82 b=ss[top]; 83 top--; 84 a=ss[top]; 85 top--; 86 x=ss[top]; 87 //top++; 88 ss[top]=fun(x,a,b); 89 // printf(" top=%") 90 } 91 else break; 92 } 93 94 } 95 } 96 if(ss[1]==0){ 97 flag=0; 98 } 99 //printf(" p=%d q=%d ss=%d flag=%d ",p,q,ss[1],flag); 100 } 101 102 void solve() 103 { 104 //printf(" sol "); 105 int p,q,r,s,t; 106 for(p=0;p<=1;p++) 107 for(q=0;q<=1;q++) 108 for(r=0;r<=1;r++) 109 for(s=0;s<=1;s++) 110 for(t=0;t<=1;t++) 111 cal(p,q,r,s,t); 112 } 113 114 void out() 115 { 116 if(flag==1){ 117 printf("tautology "); 118 } 119 else{ 120 printf("not "); 121 } 122 } 123 124 int main() 125 { 126 //freopen("data.in","r",stdin); 127 //scanf("%d",&T); 128 // for(cnt=1;cnt<=T;cnt++) 129 //while(T--) 130 while(scanf("%s",ssss)!=EOF) 131 { 132 if(strcmp(ssss,"0")==0) break; 133 ini(); 134 solve(); 135 out(); 136 } 137 }