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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9936 | Accepted: 3774 |
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:
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| 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：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。
思路：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。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <stack>
using namespace std;
char str[110];
stack<int>Q;
int p,q,r,s,t,len;
int IN_stack()
{
int i;
int x,y;
for(i=strlen(str)-1; i>=0; i--) {
if(str[i] == 'p')
Q.push(p);
else if(str[i] == 'q')
Q.push(q);
else if(str[i] == 'r')
Q.push(r);
else if(str[i] == 's')
Q.push(s);
else if(str[i] == 't')
Q.push(t);
else if(str[i]== 'K') {
x = Q.top();
Q.pop();
y = Q.top();
Q.pop();
Q.push(x && y);
} else if(str[i] == 'A') {
x = Q.top();
Q.pop();
y = Q.top();
Q.pop();
Q.push(x || y);
} else if(str[i] == 'N') {
x = Q.top();
Q.pop();
Q.push(!x);
} else if(str[i] == 'C') {
x = Q.top();
Q.pop();
y = Q.top();
Q.pop();
Q.push((!x)||y);
} else if(str[i] == 'E') {
x = Q.top();
Q.pop();
y = Q.top();
Q.pop();
Q.push(x == y);
}
}
}
int judge()
{
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++) {
IN_stack();
if(Q.top()==0)
return 0;
}
return 1;
}
int main()
{
while(~scanf("%s",str)) {
getchar();
if(str[0] == '0')
break;
if(judge()) {
printf("tautology\n");
} else printf("not\n");
}
return 0;
}
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原文地址：http://blog.csdn.net/u013486414/article/details/42580691
