[ACM] POJ 3295 Tautology (构造）

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Tautology

Time Limit: 1000MS Memory Limit: 65536KTotal 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:

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

ApNpApNq0

Sample Output

tautologynot

Source

Waterloo Local Contest, 2006.9.30

解题思路：

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;}

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