poj 3295

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
题意如下,

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

 

解题思路:

p, q, r, s, t不同的取值组合共32种情况,枚举不同取值组合代入逻辑表达式WFF进行计算。

如果对于所有的取值组合,WFF值都为 true, 则结果为 tautology,否则为 not。    

WFF的计算方法:

从字符串WFF的末尾开始依次向前读取字符。

构造一个栈stack,当遇到逻辑变量 p, q, r, s ,t 则将其当前的值压栈;

遇到 N 则取栈顶元素进行非运算,运算结果的值压栈;

遇到K, A, C, E则从栈顶中弹出两个元素进行相应的运算,将结果的值压栈。 

由于输入是合法的,当字符串WFF扫描结束时,栈stack中只剩一个值,该值就是逻辑表达式WFF的值。

代码偷来的
//这个代码让我懂了先根遍历该咋写- -
//慢慢学吧,不会的还有很多
#include<iostream>
#include<cstring>
#include<stack>
using namespace std;

stack <bool> sta;
bool val[5],flag;

int main()
{
    int i,j;
    string str;
    bool tmp1,tmp2;
    while(cin>>str && str!="0")
    {
        flag=1;
        for(i=0;i<32 && flag;i++)
        {
            memset(val,0,sizeof(val));
            for(j=0;j<5;j++)
            {
                if(i & (1<<j))
                    val[j]=1;
            }
            for(j=str.length()-1;j>=0;j--)
            {
                switch(str[j])
                {
                case 'p':case 'q':case 'r':case 's':case 't':
                    sta.push(val[str[j]-'p']);
                    break;
                case 'N':
                    tmp1=sta.top();
                    sta.pop();
                    sta.push(!tmp1);
                    break;
                case 'E':
                    tmp1=sta.top();
                    sta.pop();
                    tmp2=sta.top();
                    sta.pop();
                    sta.push(tmp1==tmp2);
                    break;
                case 'A':
                    tmp1=sta.top();
                    sta.pop();
                    tmp2=sta.top();
                    sta.pop();
                    sta.push(tmp1||tmp2);
                    break;
                case 'K':
                    tmp1=sta.top();
                    sta.pop();
                    tmp2=sta.top();
                    sta.pop();
                    sta.push(tmp1 && tmp2);
                    break;
                case 'C':
                    tmp1=sta.top();
                    sta.pop();
                    tmp2=sta.top();
                    sta.pop();
                    if(!tmp1||tmp2)
                        tmp1=1;
                    else
                        tmp1=0;
                    sta.push(tmp1);
                    break;
                default:
                    break;
                };
            }
            if(!sta.top())
                flag=0;
        }
        if(flag)
            cout<<"tautology"<<endl;
        else
            cout<<"not"<<endl;
    }
    return 0;
}


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