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时间:2016-07-06 11:43:41      阅读:184      评论:0      收藏:0      [点我收藏+]

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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

 

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


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;
}

 

poj3295

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原文地址:http://www.cnblogs.com/shenben/p/5646335.html

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