[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 follow…

https://vjudge.net/problem/POJ-3295 题意 有五种运算符和五个参数,现在给你一个不超过100字符的算式,问最后结果是否恒为1? 分析 首先明确各运算符的意义,K(&&),A(||),N(!),E(==),C特殊判断一下.计算的话肯定是从后往前的,遇到变量进栈,遇到运算符则出栈做运算.现在的问题是各个变量的数值未知,由于只有5个变量,总共只有32总可能,所以暴力跑一遍即可,每种取值都试一次. #include<iostream> #include…

题目链接:http://poj.org/problem?id=3295 题意: 输入由p.q.r.s.t.K.A.N.C.E共10个字母组成的逻辑表达式, 其中p.q.r.s.t的值为1(true)或0(false),即逻辑变量: K.A.N.C.E为逻辑运算符, K --> and:  x && y A --> or:  x || y N --> not :  !x C --> implies :  (!x)||y E --> equals :  x==y…

Tautology 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 followin…

题目链接. 分析: 最多有五个变量,所以枚举所有的真假值,从后向前借助于栈验证是否为永真式. #include <iostream> #include <cstring> #include <cstdio> #include <cstdlib> #include <stack> using namespace std; + ; stack<bool> S; char s[maxn]; ], ans; void check(char p…

给你一个表达式,其包括一些0,1变量和一些逻辑运算法,让你推断其是否为永真式. 计算表达式的经常使用两种方法:1.递归: 2.利用栈. code(递归实现) #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> using namespace std; char str[2000]; int pos; bool ca…

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…

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…

直接上分析: 首先 弄清各种大写字母的操作的实质 K 明显 是 and   & A 是 or      | N 是 not   ! C  由表格注意到 当 w<=x 时 值为1 E  当 w==x 时 值 为1; 弄清了这些就把大写字母当做一个运算符,字符串作为一个前缀表达式来做 #include <iostream> #include <string> #include <stack> #define Det 112 using namespace st…

Tautology 思路很简单,对于p.q.r.s.t暴力枚举是0还是1,判断即可.判断时像写表达式求值那样用栈.为了方便可以从后往前,因为最后一个肯定不是运算.那几个奇奇怪怪的函数可以找规律然后转为位运算,简洁又快. #include<cstdio> #include<stack> #include<cstring> using namespace std; const int MAXN=105; char a[MAXN]; int n,p,q,r,s,t; bool…