| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9302 | Accepted: 3549 |
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:
|
|
| 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
题意为根据输入的不同的字符串，来求这个逻辑表达式的值。这里用到了栈，和表达式求值思想类似，从后往前扫面输入的字符串，如果是数（题中的p,q,r,s,t）就进栈，如果是操作符就从栈中取数，运算后再进栈。题中的数p,q,r,s,t只有0,1取值，所以5重循环，对获得的表达式求值。遇到0就退出。
代码：
#include <iostream>
#include <string.h>
#include <stack>
using namespace std;
string wff;
int p,q,r,s,t;
bool ok;
int compute(string str)//栈中的操作
{
stack<int>st;
int len=str.length();
for(int i=len-1;i>=0;i--)
{
if(str[i]=='p')
st.push(p);
else if(str[i]=='q')
st.push(q);
else if(str[i]=='r')
st.push(r);
else if(str[i]=='s')
st.push(s);
else if(str[i]=='t')
st.push(t);
else if(str[i]=='K')
{
int x=st.top();
st.pop();
int y=st.top();
st.pop();
st.push(x&y);
}
else if(str[i]=='A')
{
int x=st.top();
st.pop();
int y=st.top();
st.pop();
st.push(x||y);
}
else if(str[i]=='N')
{
int x=st.top();
st.pop();
st.push(!x);
}
else if(str[i]=='C')
{
int x=st.top();
st.pop();
int y=st.top();
st.pop();
st.push(!x||y);
}
else if(str[i]=='E')
{
int x=st.top();
st.pop();
int y=st.top();
st.pop();
st.push(x==y);
}
}
return st.top();
}
int main()
{
while(cin>>wff&&wff!="0")
{
ok=1;
for(p=0;p<2;p++)//枚举结果，遇到0就退出循环
for(q=0;q<2;q++)
for(r=0;r<2;r++)
for(s=0;s<2;s++)
for(t=0;t<2;t++)
{
if(compute(wff)==0)
{
ok=0;
goto label;
}
}
label:
if(ok)
cout<<"tautology"<<endl;
else
cout<<"not"<<endl;
}
return 0;
}
[ACM] POJ 3295 Tautology (构造）,布布扣,bubuko.com
原文：http://blog.csdn.net/sr_19930829/article/details/37764471