Tautology 分类: POJ 2015-06-28 18:40 10人阅读 评论(0) 收藏

Tautology
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 10428   Accepted: 3959

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
实在是读不懂什么意思,后来找篇博客,才知道什么意思,英语好渣
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <string>
#include <stack>
#include <algorithm>
using namespace std;

const int Max=1100000;

int Arr[6];
int top;
string str;
stack<int>a;
void Bit(int num)
{
    int top=0;
    memset(Arr,0,sizeof(Arr));
    while(num)
    {
        Arr[top++]=num%2;
        num/=2;
    }
}
bool Jud(char s)
{
    switch(s)
    {
        case 'p':a.push(Arr[0]);break;
        case 'q':a.push(Arr[1]);break;
        case 'r':a.push(Arr[2]);break;
        case 's':a.push(Arr[3]);break;
        case 't':a.push(Arr[4]);break;
        default :return 0;
    }
    return 1;
}
void Count(char s)
{
    switch(s)
    {
    case 'K':
        {
            int x=a.top();a.pop();
            int y=a.top();a.pop();
            a.push(x&&y);
        }
        break;
    case 'A':
        {
            int x=a.top();a.pop();
            int y=a.top();a.pop();
            a.push(x||y);
        }
        break;
    case 'N':
        {
             int x=a.top();a.pop();
             a.push(!x);
        }
        break;
    case 'C':
        {
            int x=a.top();a.pop();
            int y=a.top();a.pop();
            a.push((!x)||y);
        }
        break;
    case 'E':
        {
            int x=a.top();a.pop();
            int y=a.top();a.pop();
            a.push(x==y);
        }
        break;
    }
}
bool Cal()
{
    int len=str.length();
    for(int i=0;i<=31;i++)
    {
        Bit(i);
        for(int j=len-1;j>=0;j--)
        {
            if(!Jud(str[j]))
            {
                Count(str[j]);
            }
        }
        int x=a.top();
        a.pop();
        if(!x)
            return false;
    }
    return true;
}
int main()
{
    while(cin>>str)
    {
        if(str=="0")
            break;
        if(Cal())
        {
            cout<<"tautology"<<endl;
        }
        else
        {
            cout<<"not"<<endl;
        }
    }

}


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posted @ 2015-06-28 18:40  一骑绝尘去  阅读(200)  评论(0编辑  收藏  举报