qq_33312212 于 2016.03.06 11:57 提问

poj3295 运行输入之后就崩溃了 求大神看看 英汉题意如下

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

K、A、N、C、E为逻辑运算符，
K --> and: x && y
A --> or: x || y
N --> not : !x
C --> implies : (!x)||y
E --> equals : x==y

PS:输入格式保证是合法的

`````` #include <iostream>
#include <vector>
#include <string>
#include <stack>

using namespace std;

bool c(bool a,bool b)
{
if(a&&!b)
{
return false;
}
else
{
return true;
}
}

bool e(bool a,bool b)
{
if(a&&b||!a&&!b)
{
return true;
}
else
{
return false;
}
}

bool solve(string str,int p,int q,int r,int s,int t)
{
stack <bool> ele;
unsigned int i;
for(i=str.size()-1;i>=0;i--)
{
switch(str[i])
{
case 'p':ele.push((bool)p);break;
case 'q':ele.push((bool)q);break;
case 'r':ele.push((bool)r);break;
case 's':ele.push((bool)s);break;
case 't':ele.push((bool)t);break;
case 'K':
{
bool a = ele.top();
ele.pop();
bool b = ele.top();
ele.pop();
ele.push(a&&b);
}
break;
case 'A':
{
bool a = ele.top();
ele.pop();
bool b = ele.top();
ele.pop();
ele.push(a||b);
}
break;
{
case 'N':
bool a = ele.top();
ele.pop();
ele.push(!a);
}
break;
case 'C':
{
bool f = ele.top();
ele.pop();
bool g = ele.top();
ele.pop();
ele.push(c(f,g));
}
break;
case 'E':
{
bool h = ele.top();
ele.pop();
bool j = ele.top();
ele.pop();
ele.push(e(h,j));
break;
}
default:break;
}
}
return ele.top();
}

int main()
{
string str;
bool flag = true;
int i = 0;
vector <string> ans;
while(cin>>str&&str!="0")
{
int p,q,r,s,t;
for(p=0;p<=1;p++)
{
for(q=0;q<=1;q++)
{
for(r=0;r<=1;r++)
{
for(s=0;s<=1;s++)
{
for(t=0;t<=0;t++)
{
flag = solve(str,p,q,r,s,t);
if(!flag)
break;
}
if(!flag)
break;
}
if(!flag)
break;
}
if(!flag)
break;
}
if(!flag)
break;
}
if(flag)
{
ans.push_back("tautology");
}
else
{
ans.push_back("not");
flag = true;
}
}
for(i=0;i!=ans.size();i++)
{
cout<<ans[i]<<endl;
}

return 0;
}
``````

1个回答

caozhy      2016.03.06 12:21
qq_33312212 大神。。这个我看过 但是我想自己写一个 求你帮俺一看呗。。