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.
- 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.
|
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
题意如下,
输入由p、q、r、s、t、K、A、N、C、E共10个字母组成的逻辑表达式,
其中p、q、r、s、t的值为1(true)或0(false),即逻辑变量;
K、A、N、C、E为逻辑运算符,
K --> and: x && y
A --> or: x || y
N --> not : !x
C --> implies : (!x)||y
E --> equals : x==y
问这个逻辑表达式是否为永真式。
PS:输入格式保证是合法的
解题思路:
p, q, r, s, t不同的取值组合共32种情况,枚举不同取值组合代入逻辑表达式WFF进行计算。
如果对于所有的取值组合,WFF值都为 true, 则结果为 tautology,否则为 not。
WFF的计算方法:
从字符串WFF的末尾开始依次向前读取字符。
构造一个栈stack,当遇到逻辑变量 p, q, r, s ,t 则将其当前的值压栈;
遇到 N 则取栈顶元素进行非运算,运算结果的值压栈;
遇到K, A, C, E则从栈顶中弹出两个元素进行相应的运算,将结果的值压栈。
由于输入是合法的,当字符串WFF扫描结束时,栈stack中只剩一个值,该值就是逻辑表达式WFF的值。
代码偷来的
//这个代码让我懂了先根遍历该咋写- - //慢慢学吧,不会的还有很多 #include<iostream> #include<cstring> #include<stack> using namespace std; stack <bool> sta; bool val[5],flag; int main() { int i,j; string str; bool tmp1,tmp2; while(cin>>str && str!="0") { flag=1; for(i=0;i<32 && flag;i++) { memset(val,0,sizeof(val)); for(j=0;j<5;j++) { if(i & (1<<j)) val[j]=1; } for(j=str.length()-1;j>=0;j--) { switch(str[j]) { case 'p':case 'q':case 'r':case 's':case 't': sta.push(val[str[j]-'p']); break; case 'N': tmp1=sta.top(); sta.pop(); sta.push(!tmp1); break; case 'E': tmp1=sta.top(); sta.pop(); tmp2=sta.top(); sta.pop(); sta.push(tmp1==tmp2); break; case 'A': tmp1=sta.top(); sta.pop(); tmp2=sta.top(); sta.pop(); sta.push(tmp1||tmp2); break; case 'K': tmp1=sta.top(); sta.pop(); tmp2=sta.top(); sta.pop(); sta.push(tmp1 && tmp2); break; case 'C': tmp1=sta.top(); sta.pop(); tmp2=sta.top(); sta.pop(); if(!tmp1||tmp2) tmp1=1; else tmp1=0; sta.push(tmp1); break; default: break; }; } if(!sta.top()) flag=0; } if(flag) cout<<"tautology"<<endl; else cout<<"not"<<endl; } return 0; }