POJ 3295

Tautology

 

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

 

 

总结：

递归题，表达式，永真式

 

做了两个版本

WA了很多次，原来是把题目意思看错了。开始理解为不管任何pqrst值，计算结果都一样的是tautology。

 

AC代码1

#include <iostream>#include <string>#include <cctype>using namespace std;int p, q, r, s, t;int K[2][2]={0, 0, 0, 1}, A[2][2]={0, 1, 1, 1}, N[2]={1, 0},C[2][2]={1, 1, 0, 1}, E[2][2]={1, 0, 0, 1};int getv(char ch){switch (ch){case (int)'p': return p;case (int)'q': return q;case (int)'r': return r;case (int)'s': return s;case (int)'t': return t;}}int dfs(string str, int& len){int a, b, ret, len1=0, len2=0;if (isupper(str[1]))a=dfs( str.substr(1), len1);else{a=getv(str[1]);len1=1;}if (str[0]!='N'){if (isupper(str[len1+1]))b=dfs( str.substr(len1+1), len2 );else{b=getv(str[len1+1]);len2=1;}}len=1+len1+len2;switch(str[0]){case (int)'K': ret=K[a][b]; break;case (int)'A': ret=A[a][b]; break;case (int)'N': ret=N[a]; break;case (int)'C': ret=C[a][b]; break;case (int)'E': ret=E[a][b]; break;}return ret;}int main(){string str;while (cin >> str && str!="0"){p=0, q=0, r=0, s=0, t=0;int len=0;bool done=false;for (p=0;p<=1 && !done;p++)for (q=0;q<=1 && !done; q++)for (r=0;r<=1 && !done; r++)for (s=0;s<=1 && !done; s++)for (t=0;t<=1 && !done; t++)if (dfs(str, len)!=1)done=true;if (done)cout << "not/n";elsecout << "tautology/n";}return 0;} 

 

后来看了discuss的提示，给出了另一个很精简的写法

AC代码2

#include <iostream>#include <string>using namespace std;int p, q, r, s, t;int K[2][2]={0, 0, 0, 1}, A[2][2]={0, 1, 1, 1}, N[2]={1, 0},C[2][2]={1, 1, 0, 1}, E[2][2]={1, 0, 0, 1};string str;int now;int calc(){now++;switch(str[now]){case (int)'K':return calc() & calc();case (int)'A':return calc() | calc();case (int)'N':return !calc();case (int)'C':return C[calc()][calc()];case (int)'E':return E[calc()][calc()];case (int)'p':return p;case (int)'q':return q;case (int)'r':return r;case (int)'s':return s;case (int)'t':return t;}}int main(){while (cin >> str && str!="0"){bool done=false;for (p=0;p<=1 && !done;p++)for (q=0;q<=1 && !done; q++)for (r=0;r<=1 && !done; r++)for (s=0;s<=1 && !done; s++)for (t=0;t<=1 && !done; t++){now=-1;if (calc()!=1)done=true;}if (done)cout << "not/n";elsecout << "tautology/n";}return 0;} 

 

在网上，还发现一个用位运算的更精简的写法，学习了。

http://www.cnblogs.com/asuran/archive/2009/09/28/1575379.html

 

// POJ3295.cpp : Defines the entry point for the console application.//#include <iostream>#include <string>using namespace std;static int pos = -1;bool WFF(const string& formula, int i){ ++pos; switch(formula[pos]) { case 'p': return i & 1; case 'q': return (i >> 1) & 1; case 'r': return (i >> 2) & 1; case 's': return (i >> 3) & 1; case 't': return (i >> 4) & 1; case 'N': return !WFF(formula, i); case 'K': return WFF(formula, i) & WFF(formula, i); case 'A': return WFF(formula, i) | WFF(formula, i); case 'C': return !WFF(formula, i) | WFF(formula, i); case 'E': return WFF(formula, i) == WFF(formula, i); } return false;};bool isTautology(string formula){ for (int i = 0; i < 32; ++i) { pos = -1; if (WFF(formula, i)==false) return false;; } return true;};int main(int argc, char* argv[]){ string ln; while (cin >> ln && ln[0] != '0') { if (isTautology(ln)) cout << "tautology/n"; else cout << "not/n"; } return 0;}