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[ACM] POJ 3295 Tautology (构造)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total 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.
- 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
Source
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; }