POJ - 3295 - Tautology （构造）

来源：互联网发布：酷狗铃声截取软件编辑：程序博客网时间：2024/06/06 03:29

题目传送：Tautology

思路：枚举所有变量可能的值（0或1），算出其表达式的值，因为题目是要求是否是永真式，求式子的真值可以用栈来求，栈的话，可以自己构造一个栈，也可以用STL的stack

AC代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <deque>#include <cctype>#define LL long long#define INF 0x7fffffffusing namespace std;char str[105];int p, q, r, s, t;int len;int fun() {stack<int> st;for(int i = len - 1; i >= 0; i --) {if(str[i] == 'p') st.push(p);else if(str[i] == 'q') st.push(q);else if(str[i] == 'r') st.push(r);else if(str[i] == 's') st.push(s);else if(str[i] == 't') st.push(t);else if(str[i] == 'K') {int t1 = st.top(); st.pop();int t2 = st.top(); st.pop();if(t1 == 1 && t2 == 1) st.push(1);else st.push(0);}else if(str[i] == 'A') {int t1 = st.top(); st.pop();int t2 = st.top(); st.pop();if(t1 == 0 && t2 == 0) st.push(0);else st.push(1);}else if(str[i] == 'N') {int t1 = st.top(); st.pop();st.push(!t1);}else if(str[i] == 'C') {int t1 = st.top(); st.pop();int t2 = st.top(); st.pop();if(t1 == 1 && t2 == 0) st.push(0);else st.push(1);}else if(str[i] == 'E') {int t1 = st.top(); st.pop();int t2 = st.top(); st.pop();if((t1 == 1 && t2 == 1) || (t1 == 0 && t2 == 0)) st.push(1);else st.push(0);}}return st.top();}int solve() {for(p = 0; p < 2; p ++)for(q = 0; q < 2; q ++)for(r = 0; r < 2; r ++)for(s = 0; s < 2; s ++)for(t = 0; t < 2; t ++) {int tmp = fun();if(tmp == 0) return 0;}return 1;}int main() {while(scanf("%s", str) != EOF) {if(strcmp(str, "0") == 0) break;len = strlen(str);int flag = solve();if(flag) {printf("tautology\n");}else printf("not\n");}return 0;}

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