Tautology-状压

Tautology

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11716 Accepted: 4437

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

题目链接：http://poj.org/problem?id=3295

题目大意：

此题是让你判断式子最后的答案是否永真，离散好的大神应该看图就能看懂题，根据上表，判断输入的字符串是否为永真式，poj水题，本想不懂脑子用多重for循环直接水过，但是恰巧金桔路过，深深地鄙视了我的for循环，我。。。。。。最终决定，上状压，当然，金桔就是金桔，代码力求简洁。

题解：

共五个变元，p,q,r,s,t，每一个变元有0和1两种状态，共有1<<5种状态，对应每一个变元的值，直接上状压！

本渣的代码：

#include <cstdio>#include <cstring>#include <iostream>#include <stack>using namespace std;bool f(int h1,int h2,char c){    switch(c)    {    case 'K':        return h1&&h2;    case 'A':        return h1||h2;    case 'C':        return (!h1&&h2)||h1==h2;    case 'E':        return h1==h2;    default :        break;    }    return 0;}bool judge(char s[]){    int len=strlen(s);    bool v[10];    stack<bool> Q;    int i,j;    for(i=0;i<1<<5;i++)    {        int t=i;        for(j=0;j<5;j++)        {            v[j]=t%2;            t>>=1;        }        while(!Q.empty())            Q.pop();        for(j=len-1;j>=0;j--)        {            if(s[j]>='p'&&s[j]<='t')                Q.push(v[s[j]-'p']);            else if(s[j]=='N')            {                int h=Q.top();                Q.pop();                h=!h;                Q.push(h);            }            else            {                int h1=Q.top();                Q.pop();                int h2=Q.top();                Q.pop();                int h=f(h1,h2,s[j]);                Q.push(h);            }        }        if(!Q.top())            return 0;    }    if(Q.top())        return 1;    return 0;}int main(){    char s[1000];    while(~scanf("%s",s)&&*s!='0')    {        printf(judge(s)?"tautology\n":"not\n");    }    return 0;}

好久没有写点东西了，颓废了好长时间了，也落后了别人好多了，现在只想认真努力，借用前两天看的电影中的一句话，我们学习不是为了分数和比赛的成绩，而是为了完善我们自己，加油吧！