POJ 3295：Tautology

Tautology

Time Limit: 1000MS Memory Limit: 65536KTotal 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 <iostream>#include <string>#include <cstring>#include <algorithm>#include <stack>#include <cmath>using namespace std;stack <int> 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);elseo_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);elseo_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]);elsecal(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"<<endl;}else{cout<<"not"<<endl;}}//system("pause");return 0;}