POJ 3295 Tautology

题意：逻辑运算题，最多有5个变量，依次枚举就可以AC了

k	A	N	C	E
a && b	a || b	!a	!a || b	a ==b

思路：之前是用递归写的，但是WA或是超时，郁闷。。。后来在网上参考了别人的代码，用栈一次AC。

 

 

 

10962407

NY_lv10

3295

Accepted

208K

16MS

C++

1648B

2012-10-27 20:08:57

 

用栈AC代码

#include <iostream>

#include <stack>

//#include <fstream>

#include <string>

using namespace std;



//ofstream out("1.txt");

string str;

stack<int> stk;

bool v[6];

bool tautology;



bool GetSult(/*int p*/)

{

    int a, b;

    for (int i = str.length()-1; i>=0; i--)

    {

        switch(str[i])

        {

        case 'K':

        {

            a = stk.top();

            stk.pop();

            b = stk.top();

            stk.pop();

            stk.push(a && b);

            //GetSult(p+1) && GetSult(p+2);

            break;

        }

        case 'A':

        {

            a = stk.top();

            stk.pop();

            b = stk.top();

            stk.pop();

            stk.push(a || b);

            break;

            //return GetSult(p+1) || GetSult(p+2);

        }

        case 'N':

        {

            a = stk.top();

            stk.pop();

            stk.push(!a);

            break;

            //return !GetSult(p+1);

        }

            //    return !v[str[p+1]-'p'];

        case 'C':

        {

            a = stk.top();

            stk.pop();

            b = stk.top();

            stk.pop();

            stk.push(!a || b);

            break;

            //return (!GetSult(p+1)) || GetSult(p+2);

        }

        case 'E':

        {

            a = stk.top();

            stk.pop();

            b = stk.top();

            stk.pop();

            stk.push(a == b);

            break;

            //return GetSult(p+1) == GetSult(p+2);

        }

        default:

            stk.push( v[str[i]-'p']);

        }

    }

    return stk.top();

}



int main()

{

    int i, j;

    while (cin>>str && str!="0")

    {

        while (!stk.empty())

            stk.pop();

        tautology = true;

        for (i=0; i<=31; i++)

        {

            for (j=0; j<5; j++)

                v[j] = (i>>j) % 2;

            

            if (!GetSult(/*0*/))        

            {

                tautology = false;

                goto L;

            }

        }

L:

        if (tautology)

        {

        //    out<<"tautology"<<endl;

            cout<<"tautology"<<endl;

        }

        else

        {

        //    out<<"not"<<endl;

            cout<<"not"<<endl;

        }

    }



    return 0;

}

 

递归WA || 超时

#include <iostream>

#include <stack>

#include <fstream>

#include <string>

using namespace std;



ofstream out("1.txt");

string str;

stack<int> stk;

bool v[6];

bool tautology;



bool GetSult(int p)

{

    

    switch(str[p])

    {

    case 'K':

        return GetSult(p+1) && GetSult(p+2);

    case 'A':

        return GetSult(p+1) || GetSult(p+2);

    case 'N':

        return !GetSult(p+1);

        //    return !v[str[p+1]-'p'];

    case 'C':

         return (!GetSult(p+1)) || GetSult(p+2);

    case 'E':

        return GetSult(p+1) == GetSult(p+2);

    default:

        return v[str[p]-'p'];

    }

}



int main()

{

    int i, j;

    while (cin>>str && str!="0")

    {

        tautology = true;

        for (i=0; i<=31; i++)

        {

            for (j=0; j<5; j++)

                v[j] = (i>>j) % 2;

            

            if (!GetSult(0))        

            {

                tautology = false;

                goto L;

            }

        }

L:

        if (tautology)

        {

            out<<"tautology"<<endl;

            cout<<"tautology"<<endl;

        }

        else

        {

            out<<"not"<<endl;

            cout<<"not"<<endl;

        }

    }



    return 0;

}