Tautology
| 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.
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.
| Definitions of K, A, N, C, and E |
| 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
Waterloo Local Contest, 2006.9.30
用栈处理计算过程，每种情况暴力跑一次
| 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\n",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\n",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\n",p,q,ss[1],flag);
100 }
101
102 void solve()
103 {
104 //printf(" sol\n");
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\n");
118 }
119 else{
120 printf("not\n");
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 }