Tautology
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9580 | Accepted: 3640 |
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.
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.
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
这道题的意思:用字符串的形式给你一个逻辑表达式,判断是否为永真式,逻辑变量只有那五个小写字母
这是我的代码,没什么价值
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const char ch[5]={‘p‘,‘q‘,‘r‘,‘s‘,‘t‘};
char a[120];
int rep[255];
int f[120];
int len;
void printrep(){
for(int i=0;i<5;i++){
printf("%c%d ",ch[i],rep[ch[i]]);
}
puts("");
}
bool isnum(char ch1){
for(int i=0;i<5;i++)if(ch1==ch[i])return true;
return false;
}
bool dfs(int ind){
if(ind<5){if(!dfs(ind+1))return false;rep[ch[ind]]=1;if(!dfs(ind+1))return false;rep[ch[ind]]=0;return true;}
memset(f,0,sizeof(f));
int index=0;
for(int i=len-1;i>=0;i--){
if(isnum(a[i])){
f[index++]=rep[a[i]];
}
else {
if(a[i]==‘K‘){
f[index-2]=f[index-2]&f[index-1];
index--;
}
if(a[i]==‘A‘){
f[index-2]=f[index-2]|f[index-1];
index--;
}
if(a[i]==‘N‘){
f[index-1]=1^f[index-1];
}
if(a[i]==‘C‘){
f[index-2]=((1^f[index-1])|f[index-2]);
index--;
}
if(a[i]==‘E‘){
f[index-2]=1^(f[index-2]^f[index-1]);
index--;
}
}
}
if(f[index-1]==0)return false;
return true;
}
int main()
{
while(scanf("%s",a)==1&&strcmp(a,"0")){
memset(rep,0,sizeof(rep));
len=strlen(a);
if(dfs(0)){
puts("tautology");
}
else {
puts("not");
}
}
return 0;
}
int ind()//这位大神这么做的看起来清洁干净
{
char ch=s[l++];//把整个堆栈过程和变量分开看了
printf("");
switch(ch)
{
case ‘p‘:
case ‘q‘:
case ‘r‘:
case ‘s‘:
case ‘t‘:
return state[ch];
break;
case ‘K‘:
return ind()&ind();
break;
case ‘A‘:
return ind()|ind();
break;
case ‘N‘:
return !ind();
break;
case ‘C‘:
return !ind()|ind();
break;
case ‘E‘:
return ind()==ind();
break;
}
}
还可以字符替换,不贴了,可以增强直观,反正数据不大
讨论区看到的
WA来自那些递归下降求解的代码.
第一种情况，使用|| 和 &&:
例如s为所给串
int getval()
{
switch(s[c_s++])
{
case ‘p‘: return (value & (1 << 0))? 1:0;
case ‘q‘: return (value & (1 << 1))? 1:0;
case ‘r‘: return (value & (1 << 2))? 1:0;
case ‘s‘: return (value & (1 << 3))? 1:0;
case ‘t‘: return (value & (1 << 4))? 1:0;
case ‘K‘: return getval() && getval();
case ‘A‘: return getval() || getval();
case ‘N‘: return !getval();
case ‘C‘: return !getval() || getval();
case ‘E‘: return getval() == getval();
}
}
这种情况简单，大家知道短路求值吧，先对 ||左边的表达式求值，如果非0，则不会对右边的表达式求值，&&同理，如果左边为0，不会对右边求值，这样下标就不会按照事先设想的增加
第二种情况，使用&和|:
int getval()
{
switch(s[c_s++])
{
case ‘p‘: return (value & (1 << 0))? 1:0;
case ‘q‘: return (value & (1 << 1))? 1:0;
case ‘r‘: return (value & (1 << 2))? 1:0;
case ‘s‘: return (value & (1 << 3))? 1:0;
case ‘t‘: return (value & (1 << 4))? 1:0;
case ‘K‘: return getval() & getval();
case ‘A‘: return getval() | getval();
case ‘N‘: return !getval();
case ‘C‘: return !getval() | getval();
case ‘E‘: return getval() == getval();
}
}
首先这段代码用G++会AC，C++会WA，说明此段代码依赖编译器，是未定义的代码
错误原因: C语言对表达式的求值顺序不是明确规定的，而G++是从左从左向右求值，C++则正好相反,比如: getval() | getval() G++先对左边的getval()求值，而C++则先对右边的getval()求值，也就会导致对s的访问顺序不会按预先的步骤进行，所以用G++会ac，C++会WA掉
正确的写法,去掉那些跟依赖求值顺序的代码（好习惯）,分别求值，用两个临时变量保存，这样G++和C++都AC了:
int getval()
{ int temp1, temp2;
switch(s[c_s++])
{
case ‘p‘: return (value & (1 << 0))? 1:0;
case ‘q‘: return (value & (1 << 1))? 1:0;
case ‘r‘: return (value & (1 << 2))? 1:0;
case ‘s‘: return (value & (1 << 3))? 1:0;
case ‘t‘: return (value & (1 << 4))? 1:0;
case ‘K‘: temp1 = getval(); temp2 = getval(); return temp1 & temp2;
case ‘A‘: temp1 = getval(); temp2 = getval(); return temp1 | temp2;
case ‘N‘: return !getval();
case ‘C‘: temp1 = !getval(); temp2 = getval(); return temp1 | temp2;
case ‘E‘: temp1 = getval(); temp2 = getval(); return temp1 == temp2;
}
}