| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8127 | Accepted: 3115 |
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
题目大意是给你一个逻辑判断语句，让你判断是否为用真句，其中小写字母代表逻辑变量，大写字母代表的是逻辑运算符，其中逻辑运算符的运算规则已经在表里没出了，如果不是用真就输出not，如果是用真就输出tautology
解题方法也没简单，建立一个堆栈，然后就类似表达式求值了，但是这里需要说明一下运算符的表示方法
K--------- &
A--------- |
N-------- ！
C（这里让我头疼了好久，数字逻辑没学好啊，没看出来什么关系，看了别人的解析才发现的，比赛中遇到只能写一个函数根据表中的关系得到逻辑关系了）----（！x）| y
E-------- ==
自己写了一个简易的栈，所以代码有点长用STL会短一些，还有就是解题时特意用了一下用一个整形表示了5个数的状态，练习一下位运算了
AC代码 0MS
#include<stdio.h>
#include<string.h>
class bool_stack
{
public:
bool s[200];
int top;
bool_stack()
{
top = 0;
}
void push(bool a)
{
s[top++] = a;
}
bool pop()
{
top--;
return s[top]; }
};
int solve(bool * num, char * str)
{
bool_stack stack;
int len = strlen(str);
int i;
for(i = len - 1; i > -1; i--)
{
bool temp1, temp2;
switch(str[i])
{
case 'K':
temp1 = stack.pop();
temp2 = stack.pop();
stack.push(temp1 & temp2);
break;
case 'A':
temp1 = stack.pop();
temp2 = stack.pop();
stack.push(temp1 | temp2);
break;
case 'N':
temp1 = stack.pop();
stack.push(!temp1);
break;
case 'C':
temp1 = stack.pop();
temp2 = stack.pop();
stack.push((!temp2) | temp1);
break;
case 'E':
temp1 = stack.pop();
temp2 = stack.pop();
stack.push(temp1 == temp2);
break;
default:
stack.push(num[str[i] - 'p']);
break;
}
}
return stack.pop();
}
int main()
{
char str[120];
while(scanf("%s", str), str[0] != '0')
{
int flag = 0;
int i;
for(i = 0; i < 32; i++)
{
bool num[5];
int j;
for(j = 0; j < 5; j++)
{
num[j] = (i >> j) & 1;
}
if(solve(num, str) == 0)
break;
}
if(i == 32)
printf("tautology\n");
else
printf("not\n");
}
return 0;
}