构造 + 离散数学、重言式 - POJ 3295 Tautology
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 【题目来源】
Waterloo Local Contest, 2006.9.30
http://poj.org/problem?id=3295
【题目大意】
给你一个合式公式,让你判断是否为重言式。
【题目分析】
这题考了一些离散数学的概念在里面,题目并不难,就是构造加模拟。
使用一个栈来从后往前计算。 AC代码:
#include<cstring>
#define MAX 110
using namespace std;
int a[MAX];
char str[MAX];
void calc(int p,int q,int r,int s,int t)
{
int top=;
int t1,t2;
int len=strlen(str);
for(int i=len-;i>=;i--) //所有的都判断一遍,直到a[0]
{
if(str[i]=='p') a[top++]=p;
else if(str[i]=='q') a[top++]=q;
else if(str[i]=='r') a[top++]=r;
else if(str[i]=='s') a[top++]=s;
else if(str[i]=='t') a[top++]=t;
if(str[i]=='K')
{
t1=a[--top];
t2=a[--top];
a[top++]=t1&&t2;
}
else if(str[i]=='A')
{
t1=a[--top];
t2=a[--top];
a[top++]=t1||t2;
}
else if(str[i]=='N')
{
t1=a[--top];
a[top++]=!t1;
}
else if(str[i]=='C')
{
t1=a[--top];
t2=a[--top];
if(t1==&&t2==)
a[top++]=;
else a[top++]=;
}
else if(str[i]=='E')
{
t1=a[--top];
t2=a[--top];
if(t1==t2)
a[top++]=;
else a[top++]=;
}
}
}
bool judge()
{
int p,q,r,s,t;
for(p=;p<;p++) //枚举所有的情况
for(q=;q<;q++)
for(r=;r<;r++)
for(s=;s<;s++)
for(t=;t<;t++)
{
calc(p,q,r,s,t);
if(a[]==)
{
return false;
}
}
return true;
}
int main()
{
while(cin>>str)
{
if(strcmp(str,"0")==) break;
if(judge())
cout<<"tautology"<<endl;
else cout<<"not"<<endl;
}
}
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