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…

题目网址:http://poj.org/problem?id=3295 题目: Tautology Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13231   Accepted: 5050 Description WFF 'N PROOF is a logic game played with dice. Each die has six faces representing some subset of the po…

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,…

Tautology Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9302   Accepted: 3549 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 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…

http://poj.org/problem?id=3295 题意: 判断表达式是否为永真式. 思路: 把每种情况都枚举一下. #include<iostream> #include<string> #include<cstring> using namespace std; ; int sta[MAXN]; char str[MAXN]; int p, q, r, s, t; void judge() { ; int len = strlen(str); ; i &g…

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…

Tautology Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7716   Accepted: 2935 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,…

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…

题目链接: http://poj.org/problem?id=3295 题目描述: 给一个字符串,字符串所表示的表达式中p, q, r, s, t表示变量,取值可以为1或0.K, A, N, C, E 分别表示且,或,非,真蕴含,等值.问表达式是不是永真的,如果是输出“tautology”,否则输出“not”. 解题思路: 这里借用到了递归的本质,也就是对栈的模拟,用递归进行压栈,求表达式的值,再加上对变量状态压缩进行枚举. #include <cstdio>//本代码用G++交就ac,c+…