Tautology POJ 3295

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本文介绍了一种名为WFF'nProof的逻辑游戏，玩家通过骰子上的符号组合成逻辑公式，并判断这些公式是否为恒真式。文章详细解释了游戏规则及逻辑运算符的定义，并提供了一个算法示例来解决这一问题。

题目:

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

#include<iostream>
#include<stdio.h>
#include<stack>
#include<string.h>
using namespace std;
stack<int> digit;
int len;
char temp[110];
// p, q, r, s, t
int d[6];
bool isOpe(char c){  //K, A, N, C, E
    if('K'==c||'A'==c||'N'==c||'C'==c||'E'==c) return true;
    return false;
}
bool solve(){
    int i;
    int op1,op2;
    for(i=len-1;i>=0;i--){
//        cout<<"i="<<i<<" ->"<<temp[i]<<endl;
        if(isOpe(temp[i])){ //如果是操作符，进行运算
            switch(temp[i]){
            case 'K':{
                op2=digit.top();
                digit.pop();
                op1=digit.top();
                digit.pop();
                if(op1&&op2) digit.push(1);
                else digit.push(0);
                break;
            }
            case 'A':{
                op2=digit.top();
                digit.pop();
                op1=digit.top();
                digit.pop();
                if(op1||op2) digit.push(1);
                else digit.push(0);
                break;
            }
            case 'N':{
                op1=digit.top();
                digit.pop();
                if(op1) digit.push(0);
                else digit.push(1);
                break;
            }
            case 'C':{
                op1=digit.top();
                digit.pop();
                op2=digit.top();
                digit.pop();
                if(op1==1&&op2==0) digit.push(0);
                else digit.push(1);
                break;
            }
            case 'E':{
                op2=digit.top();
                digit.pop();
                op1=digit.top();
                digit.pop();
                if(op1==op2) digit.push(1);
                else digit.push(0);
                break;
            }
        }
        }else{
            switch(temp[i]){
            case 'p':
                digit.push(d[0]);
                break;
            case 'q':
                digit.push(d[1]);
                break;
            case 'r':
                digit.push(d[2]);
                break;
            case 's':
                digit.push(d[3]);
                break;
            case 't':
                digit.push(d[4]);
                break;
            }
        }
    }
    int ans=digit.top();
    if(ans) return true;
    return false;
}
bool change(){
    d[0]++;
    int i;
    i=0;
    while(i<5){
        if(d[i]>=2){
            d[i+1]++;
            d[i]-=2;
        }
        i++;
    }
    if(d[5]) return true;
    return false;
}
int main(){
    while(scanf("%s",temp),temp[0]!='0'){
        len=strlen(temp);
//        printf("%s %d\n",temp,len);
        while(!digit.empty()){
            digit.pop();
        }
        memset(d,0,sizeof(d));
        bool res;
        while(1){
            res=solve();
            if(!res) break;
            if(change()) break;
        }
        if(res) printf("tautology\n");
        else printf("not\n");
    }
    return 0;
}

实现时使用堆栈，简化操作。注意每个运算符操作的是右边离他最近的2个数，不要把操作数和操作符分开存储，之前写的代码是把操作数和操作符分开，导致代码出错。