[POJ3295]Tautology
[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
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;
}
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原文地址: http://www.cnblogs.com/cresuccess/p/4874330.html