poj3295

Posted 2020-08-03 神犇(shenben)

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11579		Accepted: 4392

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

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。
PS:输入格式保证是合法的
思路：
对于逻辑变量的每个取值，都依次枚举 判断对于每一种逻辑变量的取值情况表达式是否为真

#include<iostream>
using namespace std;
int cnt;
char str[101];
bool step(char str[101],int tk){
    cnt++;
    switch(str[cnt]){
        case ‘p‘:return tk&1;
        case ‘q‘:return(tk>>1)&1;
        case ‘r‘:return(tk>>2)&1;
        case ‘s‘:return(tk>>3)&1;
        case ‘t‘:return(tk>>4)&1;
        case ‘N‘:return !step(str,tk);
        case ‘K‘:return step(str,tk)&step(str,tk);
        case ‘A‘:return step(str,tk)|step(str,tk);
        case ‘C‘:return !step(str,tk)|step(str,tk);
        case ‘E‘:return step(str,tk)==step(str,tk);
    }
}
bool judge(char str[101]){
    for(int i=0;i<32;i++){
        cnt=-1;
        if(!step(str,i)) return 0;    
    }
    return 1;
}
int main(){
    while(cin>>str){   
        if(str[0]==‘0‘) break;
        if(judge(str))
            cout<<"tautology"<<endl;
        else
            cout<<"not"<<endl;
    }
    return 0;
}