[POJ3295]Tautology

试题描述

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.

输入

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.

输出

For each test case, output a line containing tautology or not as appropriate.

输入示例

ApNp
ApNq
0

输出示例

tautology
not

数据规模及约定

见“输入”

题解

枚举 p, q, r, s, t 的值，然后带进去递归求出这个串的值，如果都为真那么就是“tautology”，否则是“not”。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

#define maxn 110
#define maxal 300
char S[maxn];
int ord[maxal];
bool val[10];

struct Info {
	int p, v;
	Info() {}
	Info(int _, int __): p(_), v(__) {}
} ;
Info check(int l) {
	if(islower(S[l])) return Info(l + 1, val[ord[S[l]]]);
	if(S[l] == 'N') {
		Info t;
		t = check(l + 1);
		return Info(t.p, t.v ^ 1);
	}
	if(S[l] == 'K') {
		Info t, t2;
		t = check(l + 1);
		t2 = check(t.p);
		return Info(t2.p, t.v & t2.v);
	}
	if(S[l] == 'A') {
		Info t, t2;
		t = check(l + 1);
		t2 = check(t.p);
		return Info(t2.p, t.v | t2.v);
	}
	if(S[l] == 'C') {
		Info t, t2;
		t = check(l + 1);
		t2 = check(t.p);
		return Info(t2.p, (t.v && !t2.v) ? 0 : 1);
	}
	if(S[l] == 'E') {
		Info t, t2;
		t = check(l + 1);
		t2 = check(t.p);
		return Info(t2.p, t.v == t2.v);
	}
	return Info(0, 0);
}

int main() {
	ord['p'] = 0;
	ord['q'] = 1;
	ord['r'] = 2;
	ord['s'] = 3;
	ord['t'] = 4;
	while(scanf("%s", S + 1) == 1) {
		int all = (1 << 5) - 1, n = strlen(S + 1);
		if(n == 1 && S[1] == '0') break;
		bool ok = 1;
		for(int i = 0; i <= all; i++) {
			for(int j = 0; j < 5; j++)
				val[j] = (i >> j & 1);
			if(!check(1).v){ ok = 0; break; }
		}
		puts(ok ? "tautology" : "not");
	}
	
	return 0;
}