poj 3295 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, 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.
Definitions of K, A, N, C, and E |
w x | Kwx | Awx | Nw | Cwx | Ewx |
1 1 | 1 | 1 | 0 | 1 | 1 |
1 0 | 0 | 1 | 0 | 0 | 0 |
0 1 | 0 | 1 | 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
题目大意是给你一个逻辑判断语句,让你判断是否为用真句,其中小写字母代表逻辑变量,大写字母代表的是逻辑运算符,其中逻辑运算符的运算规则已经在表里没出了,如果不是用真就输出not,如果是用真就输出tautology
解题方法也没简单,建立一个堆栈,然后就类似表达式求值了,但是这里需要说明一下运算符的表示方法
K--------- &
A--------- |
N-------- !
C(这里让我头疼了好久,数字逻辑没学好啊,没看出来什么关系,看了别人的解析才发现的,比赛中遇到只能写一个函数根据表中的关系得到逻辑关系了)----(!x)| y
E-------- ==
自己写了一个简易的栈,所以代码有点长用STL会短一些,还有就是解题时特意用了一下用一个整形表示了5个数的状态,练习一下位运算了
AC代码 0MS
- #include<stdio.h>
- #include<string.h>
- class bool_stack
- {
- public:
- bool s[200];
- int top;
- bool_stack()
- {
- top = 0;
- }
- void push(bool a)
- {
- s[top++] = a;
- }
- bool pop()
- {
- top--;
- return s[top];
- }
- };
- int solve(bool * num, char * str)
- {
- bool_stack stack;
- int len = strlen(str);
- int i;
- for(i = len - 1; i > -1; i--)
- {
- bool temp1, temp2;
- switch(str[i])
- {
- case 'K':
- temp1 = stack.pop();
- temp2 = stack.pop();
- stack.push(temp1 & temp2);
- break;
- case 'A':
- temp1 = stack.pop();
- temp2 = stack.pop();
- stack.push(temp1 | temp2);
- break;
- case 'N':
- temp1 = stack.pop();
- stack.push(!temp1);
- break;
- case 'C':
- temp1 = stack.pop();
- temp2 = stack.pop();
- stack.push((!temp2) | temp1);
- break;
- case 'E':
- temp1 = stack.pop();
- temp2 = stack.pop();
- stack.push(temp1 == temp2);
- break;
- default:
- stack.push(num[str[i] - 'p']);
- break;
- }
- }
- return stack.pop();
- }
- int main()
- {
- char str[120];
- while(scanf("%s", str), str[0] != '0')
- {
- int flag = 0;
- int i;
- for(i = 0; i < 32; i++)
- {
- bool num[5];
- int j;
- for(j = 0; j < 5; j++)
- {
- num[j] = (i >> j) & 1;
- }
- if(solve(num, str) == 0)
- break;
- }
- if(i == 32)
- printf("tautology\n");
- else
- printf("not\n");
- }
- return 0;
- }
poj 3295 Tautology的更多相关文章
- poj 3295 Tautology (构造)
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 ...
- poj 3295 Tautology(栈)
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, ...
- POJ 3295 Tautology(构造法)
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< ...
- poj 3295 Tautology 伪递归
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, ...
- POJ 3295 Tautology(构造法)
题目网址:http://poj.org/problem?id=3295 题目: Tautology Time Limit: 1000MS Memory Limit: 65536K Total Su ...
- 构造 + 离散数学、重言式 - POJ 3295 Tautology
Tautology Description WFF 'N PROOF is a logic game played with dice. Each die has six faces represen ...
- POJ 3295 Tautology (构造题)
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 ...
- POJ 3295 Tautology 构造 难度:1
Tautology Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9580 Accepted: 3640 Descrip ...
- POJ 3295 Tautology (构造法)
Tautology Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7716 Accepted: 2935 Descrip ...
随机推荐
- SMO序列最小最优化算法
SMO例子: 1 from numpy import * 2 import matplotlib 3 import matplotlib.pyplot as plt 4 5 def loadDataS ...
- Innodb 表修复(转)
摘要: 突然收到MySQL报警,从库的数据库挂了,一直在不停的重启,打开错误日志,发现有张表坏了.innodb表损坏不能通过repair table 等修复myisam的命令操作.现在记录下 ...
- 用Navicat连接Oracle数据库时报错ORA-28547:connection to server failed,probable Oracle Net admin error
用Navicat连接Oracle数据库时出现如下错误 上网一查原来是oci.dll版本不对.因为Navicat是通过Oracle客户端连接Oracle服务器的,Oracle的客户端分为两种,一种是标准 ...
- mysql 获取设置环境变量
mysql 获取环境变量 show global variables; 获取指定环境变量 show global variables like '%timeout'; 设置环境变量 set globa ...
- vue通过判断写样式(v-bind)
v-bind:style="$index % 2 > 0?'background-color:#FFF;':'background-color:#D4EAFA;'"
- 【转】JDBC为什么要使用PreparedStatement而不是Statement
http://www.importnew.com/5006.html PreparedStatement是用来执行SQL查询语句的API之一,Java提供了 Statement.PreparedSta ...
- bzoj 3920: Yuuna的礼物
Description 转眼就要到Karin的生日了!Yuuna她们想为她准备生日礼物!现在有许多礼物被排列成了一个一维序列,每个礼物都有一个价值.Yuuna对这个序列十分感兴趣.因此,你需要多次回答 ...
- 【Android学习日记】
(一) Android 开发基础 1 Android平台的特性 1) 应用程序框架支持组建的重用和替换,包括打电话应用程序.文件管理器等. 2) Dalvik虚拟机专门为移动设备做了优化,Dalv ...
- HDMI接口与协议
深入了解HDMI接口 一.HDMI接口的工作原理这张图是HDMI接口的架构示意图.从左边的信号源中你可以看到,HDMI接口的信源可以是任何支持HDMI输出的设备,而接入端也可以是任何带有HDMI输 入 ...
- iphone dev 入门实例3:Delete a Row from UITableView
How To Delete a Row from UITableView I hope you have a better understanding about Model-View-Control ...