[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…

Tautology Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10061   Accepted: 3826 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…

Tautology Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10428   Accepted: 3959 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…

点击打开链接 Tautology Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8127   Accepted: 3115 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…

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:http://poj.org/problem?id=3295 题意:p,q,r,s,t,是五个二进制数. K,A,N,C,E,是五个运算符. K:&& A:||N:! C:(!w)||x E:w==x 题意是让求如果对于五个数的所有情况一个式子总是恒为1,那么这个式子就是tautology.输出tautology. 否则输出not. 5个数,最多有2^5种情况. 判断式子是不是恒为1,只需要从后往前判断即可. 这题好长时间没看懂,代码也是看网上大神的 #include<iost…

字母:K, A, N, C, E 表示逻辑运算 字母:p, q, r, s, t 表示逻辑变量 0 或 1 给一个字符串代表逻辑表达式,如果是永真式输出tautology 否则输出not 枚举每个逻辑变量的值,5个变量,共2^5种情况,对于每种情况都为真则为永真式. 代码: /*************************************** Problem: 3295 User: Memory: 688K Time: 0MS Language: G++ Result: Accept…

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, a…

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…

题目链接:http://poj.org/problem?id=3295 思路分析:判断逻辑表达式是否为永真式问题.根据该表达式的特点,逻辑词在逻辑变量前,类似于后缀表达式求值问题. 算法中使用两个栈,从表达式的后边开始处理表达式中每个字符:若为逻辑变量,使其入栈SR,否则从栈SR中弹出两个逻辑变量, 进行运算后的结果再入栈SR:直到处理完表达式所有的字符.(PS:使用栈可以很好的处理广义表类似的序列) 代码如下: #include <iostream> #include <stack&g…