poj日记（3295）

转载：https://www.cnblogs.com/kuangbin/archive/2012/08/13/2636855.html

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, Nwis 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

 

 

 

 

简单题。看代码就应该可以看懂的

/*POJ 3295构造法p,q,r,s,t枚举所有可能取值用一个堆栈从字符串末尾进行操作AC  G++  684K  0MS*/#include<stdio.h>#include<iostream>#include<string.h>using namespace std;const int MAXN=120;int sta[MAXN];//数组实现堆栈char str[MAXN];int p,q,r,s,t;void  DoIt(){    int top=0;    int len=strlen(str);    for(int i=len-1;i>=0;i--)    {        if(str[i]=='p') sta[top++]=p;        else if(str[i]=='q') sta[top++]=q;        else if(str[i]=='r') sta[top++]=r;        else if(str[i]=='s') sta[top++]=s;        else if(str[i]=='t') sta[top++]=t;        else if(str[i]=='K')        {            int t1=sta[--top];            int t2=sta[--top];            sta[top++]=(t1&&t2);        }        else if(str[i]=='A')        {            int t1=sta[--top];            int t2=sta[--top];            sta[top++]=(t1||t2);        }        else if(str[i]=='N')        {            int t1=sta[--top];            sta[top++]=(!t1);        }        else if(str[i]=='C')        {            int t1=sta[--top];            int t2=sta[--top];            if(t1==1&&t2==0)sta[top++]=0;            else sta[top++]=1;        }        else if(str[i]=='E')        {            int t1=sta[--top];            int t2=sta[--top];            if((t1==1&&t2==1)||(t1==0&&t2==0)) sta[top++]=1;            else sta[top++]=0;        }    }}bool solve(){    for(p=0;p<2;p++)      for(q=0;q<2;q++)        for(r=0;r<2;r++)           for(s=0;s<2;s++)              for(t=0;t<2;t++)              {                  DoIt();                  if(sta[0]==0)return false;              }    return true;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    while(scanf("%s",&str))    {        if(strcmp(str,"0")==0)break;        if(solve())printf("tautology\n");        else printf("not\n");    }    return 0;}