H - Tautology

H - Tautology

题目描述：

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

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：

Sample Output：

题目大意：

这道题真的看哭了，这道题给给出了4个逻辑变量p，q，r，s，可以为0和1，还有5个表达式分别为N,C,A,E,K。
N—相当于非
C—相当于（非r||q）
A—相当于或
K—相当于且
E—相当于等于（==）
然后题目会给出字符串，叫你判断是否重定性，这个重定性（就是他这字符串为真），我是这样理解的。

思路分析：

他并没有明确给逻辑变量是1还是0所以有32种搭配方式，你要一个一个的去枚举，然后表达式与逻辑变量，应该从内部一个一个分，比如AqNp，这个可A（q（N（q）））这样分。

代码：

	#include<stdio.h>
	#include<string.h>
	char str[1000];
	int a[32];
	int lemon1[1000];
	int lemon()
	{
		int flag=0;
		int i,j,k;
		for(i=0;i<32;i++)//这里是枚举32种情况 
		{
			int num=0;
			for(j=0;j<5;j++)
			{
				if(i & (1 << j))//这是位运算，如果实在不懂可以百度和自己运行一下 
				{
					a[j]=1;
				}
				else
				a[j]=0;
			}
			for(k=strlen(str)-1;k>=0;k--)//开始从内部来分 
			{
				if(str[k]>='p' && str[k]<='t')
				{
					lemon1[num++]=a[str[k]-'p'];
				}
				else
				{
				switch(str[k])
				{
					case 'K':
						num--;
						lemon1[num-1]=lemon1[num-1] && lemon1[num];
						break;
						case 'A':
							num--;
							lemon1[num-1]=lemon1[num-1] || lemon1[num];
							break;
							case 'N':
								lemon1[num-1]=(!lemon1[num-1]);
								break;
								case 'C':
									num--;
									lemon1[num-1]=((!lemon1[num-1])|| lemon1[num]);
									break;
									case 'E':
										num--;
										lemon1[num-1]=(lemon1[num-1]==lemon1[num]);
										break;
									}
								}
							}
									if(!lemon1[0])
									{
										flag=1;
										break;
									}
				}
		return flag;
	}
	
	void lemon2(int flag)//判断重定式 
	{
		if(flag)
		{
			printf("not\n");
		}
		else
		printf("tautology\n");
	}
	int main()
	{
		while(scanf("%s",str)!=EOF)
		{
			int flag;
			if(str[0]=='0')
			{
				break;
			}
		flag=lemon();
		lemon2(flag);
		}
	}