POJ 3295 Tautology 構造數列及STL中棧的運用

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8184		Accepted: 3139

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

由於本人ACM還是很水，這道題剛開始看的時候木有半點思路，題目是一個運算題，我先手動打表，把這張運算表打出來，之後開始進行運算，由於5個數字，每個數字只有0和1，所以直接循環，5個數字分別從0循環到1進行計算，每次循環將5個數帶入judge函數進行運算，運算具體方法是先建立一個棧，將數列從後往前讀入字母，如果是數字，則將數字入棧，如果是運算符，則根據不同運算符分別令數字出棧後運算，將運算結果再次入棧操作，一直運算至數列字母讀取完畢，此時，棧中剩下的唯一數字即爲運算結果，判斷爲0還是1，如果爲0，則不符合tautology，直接跳出循環，輸出not，如果是1，則繼續運算，當返回所有結果均爲1，則輸出tautology，這樣，此題就水過了，此題解法中主頁君使用了STL中的棧，其實此知識點並不難，只需要掌握棧中的pop函數，top函數以及push函數其實就可以熟練使用STL的棧了，難者不會，會者不難，就是這個道理。。。

本題目AC代碼：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<stack>
using namespace std;
char ch[105];
int num[105];
int a[5][2][2];
int judge(int p,int q,int r,int s,int t)
{
	stack<int> st;
	int l1=strlen(ch),p1,p2,i;
	for(i=l1-1;i>=0;i--)
	{
		if(ch[i]=='p')
			st.push(p);
		else if(ch[i]=='q')
			st.push(q);
		else if(ch[i]=='r')
			st.push(r);
		else if(ch[i]=='s')
			st.push(s);
		else if(ch[i]=='t')
			st.push(t);
		else if(ch[i]=='K')
		{
			p1=st.top();
			st.pop();
			p2=st.top();
			st.pop();
			st.push(a[0][p1][p2]);
		}
		else if(ch[i]=='A')
		{
			p1=st.top();
			st.pop();
			p2=st.top();
			st.pop();
			st.push(a[1][p1][p2]);
		}
		else if(ch[i]=='C')
		{
			p1=st.top();
			st.pop();
			p2=st.top();
			st.pop();
			st.push(a[2][p1][p2]);
		}
		else if(ch[i]=='E')
		{
			p1=st.top();
			st.pop();
			p2=st.top();
			st.pop();
			st.push(a[3][p1][p2]);
		}
		else if(ch[i]=='N')
		{
			p1=st.top();
			st.pop();
			if(p1==1)
			    st.push(0);
			else
				st.push(1);
		}
	}
	if(st.top()==1)
		return 1;
	else
		return 0;
}
int main()
{
	a[0][1][1]=1;   //k
	a[0][1][0]=0;
	a[0][0][1]=0;
	a[0][0][0]=0;
	a[1][1][1]=1;   //a
	a[1][1][0]=1;
	a[1][0][1]=1;
	a[1][0][0]=0;
	a[2][1][1]=1;   //c
	a[2][1][0]=0;
	a[2][0][1]=1;
	a[2][0][0]=1;
	a[3][1][1]=1;   //e
	a[3][1][0]=0;
	a[3][0][1]=0;
	a[3][0][0]=1;
	int p, q, r, s, t,flag;
	while(1)
	{
		flag=1;
		scanf("%s",ch);
		if(strcmp(ch,"0")==0)
			break;
	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++)
					{
						flag=judge(p,q,r,s,t);
						if(flag==0)
							break;
					}
					if(flag==0)
						break;
				}
				if(flag==0)
					break;
			}
			if(flag==0)
				break;
		}
		if(flag==0)
			break;
	}
	if(flag==0)
		printf("not\n");
	else
		printf("tautology\n");
	}
	return 0;
}