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poj 3295 Tautology

2016-07-26 11:41 351 查看

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11717		Accepted: 4438

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

Source

Waterloo Local Contest, 2006.9.30

提示

题意：

有符号K, A, N, C, E, p, q, r, s, t，逻辑命题（WFF）中的符号遵循一下规则：

1.K, A, N, C, E为逻辑运算符，p, q, r, s, t为二进制数（也就是0与1）。

2.逻辑运算符与值作为逻辑命题。（实际上也就是if和else里的判断条件了）

给你一个命题（逻辑表达式）看它是否为恒真命题（真）。

思路：

枚举，但还是去找了下度娘，看着简单做着难。（⊙.⊙）

示例程序

Source Code

Problem: 3295		Code Length: 2860B
Memory: 392K		Time: 0MS
Language: GCC		Result: Accepted
#include <stdio.h>
#include <string.h>
int f(char ch[])
{
int a[100],top,p,q,r,s,t,i;		//借助栈存储数据
for(p=0;p<=1;p++)		//暴力枚举
{
for(q=0;q<=1;q++)
{
for(r=0;r<=1;r++)
{
for(s=0;s<=1;s++)
{
for(t=0;t<=1;t++)
{
top=0;
for(i=strlen(ch)-1;i>=0;i--)
{
if(ch[i]=='p')
{
a[top]=p;
top++;
}
else if(ch[i]=='q')
{
a[top]=q;
top++;
}
else if(ch[i]=='r')
{
a[top]=r;
top++;
}
else if(ch[i]=='s')
{
a[top]=s;
top++;
}
else if(ch[i]=='t')
{
a[top]=t;
top++;
}
else if(ch[i]=='K')
{
top--;
a[top-1]=a[top-1]&&a[top];
}
else if(ch[i]=='A')
{
top--;
a[top-1]=a[top-1]||a[top];
}
else if(ch[i]=='N')
{
a[top-1]=!a[top-1];
}
else if(ch[i]=='C')
{
top--;
a[top-1]=(!a[top-1])||a[top];
}
else if(ch[i]=='E')
{
top--;
a[top-1]=a[top-1]==a[top];
}
}
if(top!=1||a[top-1]==0)		//top不为1说明表达式不合法，后面的则是一有为0的情况就跳出
{
return 0;
}
}
}
}
}
}
return 1;
}
int main()
{
char ch[101];
scanf("%s",ch);
while(ch[0]!='0')
{
if(f(ch)==1)
{
printf("tautology\n");
}
else
{
printf("not\n");
}
scanf("%s",ch);
}
return 0;
}

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