POJ3295--Tautology(模拟)

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.

w x Kwx Awx Nw Cwx Ewx 1 1 1 1 0 1 1 1 0 0 1 0 0 0 0 1 0 0 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

有K , C , A, E, N 五种运算符， p,q,r,s,t五种运算数，运算得到重言式输出tautology， 不是重言式输出not，
重言式是指永真式,即无论输入什么，结果都为1.

思路

运算数p , q , r, s, t都只有0，1两种输入，枚举所有情况即可。
map映射
只有32种情况，int型变量的二进制数就可以代表所以情况（状态压缩）
K 相当于 p && q
A 相当于 p || q
N 相当于 !p
C 相当于 p→q
E 相当于 p↔q

代码

#include <cstdio>#include <iostream>#include <stack>#include <map>#include <cstring>using namespace std;char str[110];stack<int > s;map<char, int> mp;int wwf;int len, n;int tmp1 , tmp2;void getbit(int i){    mp['s'] = (i >> 0) & 1;    mp['p'] = (i >> 1) & 1;    mp['q'] = (i >> 2) & 1;    mp['r'] = (i >> 3) & 1;    mp['t'] = (i >> 4) & 1;}bool gao(){    for(int i = 0; i < 32; i ++)    {        getbit(i);        for(int j = len-1; j >= 0; j --)        {            switch(str[j])            {                case 'K':                tmp1 = s.top(); s.pop();                tmp2 = s.top(); s.pop();                s.push(tmp1 && tmp2); //与                break;                case 'A':                tmp1 = s.top(); s.pop();                tmp2 = s.top(); s.pop();                s.push(tmp1 || tmp2);//或                break;                case 'C':                tmp1 = s.top(); s.pop();                tmp2 = s.top(); s.pop();                s.push(!tmp1 || tmp2);//→                break;                case 'N':                tmp1 = s.top(); s.pop();                s.push(!tmp1);//非                break;                case 'E':                tmp1 = s.top(); s.pop();                tmp2 = s.top(); s.pop();                s.push((!tmp1 || tmp2)&&(tmp1 || !tmp2));//↔                break;                default :                s.push(mp[str[j]]);            }        }        if(s.top() == 0)            return false;    }    return true;}int main(){    while(cin >> str && str[0] != '0')    {        len = strlen(str);        if(gao())            cout << "tautology" << endl;        else            cout << "not" << endl;    }}

位运算大法好