POJ 3295：Tautology-白红宇

POJ 3295：Tautology

阅读量：6150 次

发布时间：2019-06-21

本文共 3119 字，大约阅读时间需要 10 分钟。

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10482		Accepted: 3982

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

ApNpApNq0

Sample Output

tautologynot

这个题记得是离散数学里面的内容，题意是判断给定的字符串是不是永远为1，就是不管p、q、r、s、t取什么值，其结果都是1。

1AC。反正自从遇到了上一次类似的题目之后，做这种题自己的感受就是两点：

1.构造一个栈

2.从后往前面撸。

代码：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;stack 
          
            o_sta;int p,q,r,s,t;string test;int len,i;void push1(char a){ switch (a) { case 'p': o_sta.push(p); break; case 'q': o_sta.push(q); break; case 'r': o_sta.push(r); break; case 's': o_sta.push(s); break; case 't': o_sta.push(t); break; default: break; }}void cal(char a){ int temp1,temp2; switch (a) { case 'N': temp1=o_sta.top(); o_sta.pop(); temp1=!temp1; o_sta.push(temp1); break; case 'K': temp1=o_sta.top(); o_sta.pop(); temp2=o_sta.top(); o_sta.pop(); temp1=temp1&temp2; o_sta.push(temp1); break; case 'A': temp1=o_sta.top(); o_sta.pop(); temp2=o_sta.top(); o_sta.pop(); temp1=temp1|temp2; o_sta.push(temp1); break; case 'C': temp1=o_sta.top(); o_sta.pop(); temp2=o_sta.top(); o_sta.pop(); temp1=temp1-temp2; if(temp1==1) o_sta.push(0); else o_sta.push(1); break; case 'E': temp1=o_sta.top(); o_sta.pop(); temp2=o_sta.top(); o_sta.pop(); temp1=temp1-temp2; if(temp1==0) o_sta.push(1); else o_sta.push(0); break; default: break; }}bool solve(){ for(p=0;p<=1;p++) { for(q=0;q<=1;q++) { for(r=0;r<=1;r++) { for(s=0;s<=1;s++) { for(t=0;t<=1;t++) { for(i=len-1;i>=0;i--) { if(test[i]=='p'||test[i]=='q'||test[i]=='r'||test[i]=='s'||test[i]=='t') push1(test[i]); else cal(test[i]); } if(o_sta.top()==0) return false; } } } } } return true;}int main(){ while(cin>>test) { if(test=="0") break; len=test.length(); if(solve()) { cout<<"tautology"<

转载于:https://www.cnblogs.com/lightspeedsmallson/p/4785832.html

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