POJ 3295 Tautology 构造难度:1

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

#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;
}

View Code

int ind()
{
    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

POJ 3295 Tautology 构造 难度:1

POJ 3295 Tautology 构造难度:1