Tautology（递推）||（栈（stack））(待整理)

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：

题意很简单，就是数学中的运算符运算；

第一种解法是运用了栈的思想，这样的话比较简单

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<stack>
using namespace std;
stack <bool> st;
int p,q,r,s,t;


void push(char c){//对于运算符压入栈顶
    if(c=='p') st.push(p);
    else if(c=='q') st.push(q);
    else if(c=='r') st.push(r);
    else if(c=='s') st.push(s);
    else if(c=='t') st.push(t);
    return;
}
void f(char c){//对经判断之后的操作符进行运算函数
    bool a,b;
    if(c=='K'){
        a=st.top();
        st.pop();
        b=st.top();
        st.pop();
        st.push(a&&b);
    }
    else if(c=='A'){
        a=st.top();
        st.pop();
        b=st.top();
        st.pop();
        st.push(a||b);
    }
    else if(c=='N'){
        a=st.top();
        st.pop();
        st.push(!a);
    }
    else if(c=='C'){
        a=st.top();
        st.pop();
        b=st.top();
        st.pop();
        st.push((!a)||b);
    }
    else if(c=='E'){
        a=st.top();
        st.pop();
        b=st.top();
        st.pop();
        st.push(a==b);
    }
    return;
}
int main(){
    char wff[110];
    while(scanf("%s",wff)&&wff[0]!='0'){
        int len=strlen(wff);
        bool flag=1;
        int i;
        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(c=='p'||c=='q'||c=='r'||c=='s'||c=='t')
                                    push(wff[i]);//将当前字符压入栈
                                else
                                    f(wff[i]);//对栈顶进行操作，进行运算
                            }
                            if(st.top()==0)//此时的top是最后的结果
                                flag=0;
                            st.pop();//将最后的结果清空，以便下一种情况的遍历
                        }
        }
        if(flag)
            printf("tautology
");
        else
            printf("not
");
    }
}

 

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int state[5];
char s[205];
int l=0;
int ind(){
	char ch=s[l++];
	//printf("");

	switch(ch){
	case 'p':
	case 'q':
	case 'r':
	case 's':
	case 't':	return state[ch-'p'];

	case 'K':	return ind()&ind();
	case 'A':	return ind()|ind();
	case 'N':	return !ind();
	case 'C':	return !ind()|ind();
	case 'E':	return ind()==ind();
	}
}

int main(){
    scanf("%s", s);
    while(s[0]!='0'){
        int len=strlen(s);
        int mark=1;
        for(state[0]=0; state[0]<=1 && mark; state[0]++){
            for(state[1]=0; state[1]<=1 && mark; state[1]++){
                for(state[2]=0; state[2]<=1 && mark; state[2]++){
                    for(state[3]=0; state[3]<=1 && mark; state[3]++){
                        for(state[4]=0; state[4]<=1 && mark; state[4]++){
							l=0;
                            if(ind()==0)
                               mark=0;
                        }
                    }
                 }
            }
        }
        if(mark==1)
            printf("tautology
");
        else
            printf("not
");
        scanf("%s", s);
    }
    return 0;
}