POJ 2106 Boolean Expressions
- 总时间限制: 1000ms 内存限制: 65536kB
- 描述
- The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next:
Expression:( V | V ) & F & ( F | V )
- where V is for True, and F is for False. The expressions may include the following operators: ! for not , & for and, | for or , the use of parenthesis for operations grouping is also allowed.
- To perform the evaluation of an expression, it will be considered the priority of the operators, the not having the highest, and the or the lowest. The program must yield V or F , as the result for each expression in the input file.
- 输入
- The expressions are of a variable length, although will never exceed 100 symbols. Symbols may be separated by any number of spaces or no spaces at all, therefore, the total length of an expression, as a number of characters, is unknown.
- The number of expressions in the input file is variable and will never be greater than 20. Each expression is presented in a new line, as shown below.
- 输出
- For each test expression, print "Expression " followed by its sequence number, ": ", and the resulting value of the corresponding test expression. Separate the output for consecutive test expressions with a new line.
- Use the same format as that shown in the sample output shown below.
- 样例输入
( V | V ) & F & ( F| V)
!V | V & V & !F & (F | V ) & (!F | F | !V & V)
(F&F|V|!V&!F&!(F|F&V))
样例输出
Expression : F
Expression : V
Expression : V
解题思路
一开始看到这题就觉得要用栈,毕竟是表达式求值问题,但是自己的中缀表达式求值也没练过,而且这次的课程作业又是练习递归,于是就尝试了递归解法。
但是入门太浅,还是搞不懂递归思路,很难从一个表达式中进行结构的分解,逻辑梳理能力太差。
下面是大佬的代码,真心膜拜,希望我的代码能力能越来越好。
#include<iostream>
using namespace std; bool f1();
bool f2(); char getc_() //读入一个非空格字符
{
char c;
while ((c = cin.get()) == ' ');
return c;
}
char peekc_()//读到第一个非空格字符前停止
{
char c;
while ((c = cin.peek()) == ' ')
{
cin.get();
}
return c;
} bool f2()//计算项
{
char c = peekc_();
if (c == '(')
{
getc_(); //拿走左括号
bool d = f1();
getc_(); //拿走右括号
return d;
}
else
{
getc_();
if (c == '!') return !f2();//运算符作用于下一个对象
if (c == 'V') return true;
if (c == 'F') return false;
}
} bool f1()//计算整个表达式
{
char c = peekc_();
if (c == '\n') return false;
bool a = f2();
while (true)
{
c = peekc_();
if (c == '|')
{
getc_();
bool b = f2();
a = a || b;
}
else if (c == '&')
{
getc_();
bool b = f2();
a = a && b;
}
else break;
}
return a;
}
int main()
{
int n = ;
while (cin.peek() != EOF) {
bool f = f1();
if (cin.peek() == '\n' || cin.peek() == EOF) {
cout << "Expression " << ++n << ": " << (f ? 'V' : 'F') << endl;
cin.get();//清理换行符和结束符
}
}
return ;
}
引用网址:
https://blog.csdn.net/chuxin126/article/details/55105076
POJ 2106 Boolean Expressions的更多相关文章
- [poj 2106] Boolean Expressions 递归
Description The objective of the program you are going to produce is to evaluate boolean expressions ...
- (栈的应用5.2.2)POJ 2106 Boolean Expressions(表达式求值)
/* * POJ_2106.cpp * * Created on: 2013年10月30日 * Author: Administrator */ #include <iostream> # ...
- POJ 2106 Boolean Expressions (布尔表达式求值)
题意:关于!,&,| 的运算,表达式中V代表true,F代表false. 思路:见代码吧,很详细了. 要注意 !!!F,!(...) 的情况. #include <iostream> ...
- poj 2106 Boolean Expressions 课本代码
#include<cstdio> const int maxn=100 +10; int val[maxn],vtop; int op[maxn],otop; void insert(in ...
- Boolean Expressions POJ - 2106 (表达式求值)
The objective of the program you are going to produce is to evaluate boolean expressions as the one ...
- POJ | Boolean Expressions
总时间限制: 1000ms 内存限制: 65536kB 描述The objective of the program you are going to produce is to evaluate ...
- Boolean Expressions
Boolean Expressions Time Limit: 1000MS Memory Limit: 30000K Description The objective of the ...
- shorthand trick with boolean expressions
https://stackoverflow.com/questions/2802055/what-does-the-construct-x-x-y-mean --------------------- ...
- POJ 2106-Boolean Expressions,双栈运用类似表达式求值!
Boolean Expressions 首先声明此题后台可能极水(毕竟这种数据不好造!).昨天写了一天却总是找不到bug,讨论区各种数据都过了,甚至怀疑输入有问题,但看到gets也可以过,难道是思路错 ...
随机推荐
- 在vue项目中使用自己封装的ajax
在 src 目录下新建 vue.extend.js ,内容如下: export default { install(Vue) { Vue.prototype.$http=function(option ...
- [Codeforces Educational Round 71]Div. 2
总结 手速场...像我这种没手速的就直接炸了... 辣鸡 E 题交互,少打了个 ? 调了半个小时... 到最后没时间 G 题题都没看就结束了...结果早上起来被告知是阿狸的打字机...看了看题一毛一样 ...
- PowerDesigner 创建表的时候 没有自增长Id的设置项
今天早上同事创建表的时候,在那个界面没有自增长Id的选项,当时我也纳闷,软件肯定都是一样的,设置的步骤都一样(有些配置好的 我就没改过 然后就忘了还改过些什么步骤了),结果还是没有那个选项 百度了一下 ...
- Hdu 3037 Saving Beans(Lucus定理+乘法逆元)
Saving Beans Time Limit: 3000 MS Memory Limit: 32768 K Problem Description Although winter is far aw ...
- C博客作业02—循环结构
0.展示PTA总分(0----2) 截图展示2次题目集:单循环和嵌套循环题目集,排名分数截图. 1.本章学习总结(2分) 1.1 学习内容总结 整理这两周学习主要知识点,并能对每个知识点介绍简单案例或 ...
- sublime 添加到右键菜单
- mysql 选择所有同学名字
mysql> select * from test; +----+----------+-------+-----------+ | id | name | score | subject | ...
- Java为什么没有指针
为了摒弃指针带来的风险(当然了,也就放弃了指针带来的效率). 1.C/C++为什么有指针? 这个很简单,程序都是在内存中运行的,只要有内存,就有内存地址,有地址,就必然有指针,只是C++对内存地址的访 ...
- 2019 SDN第四次上机作业
作业博客链接:http://edu.cnblogs.com/campus/fzu/fzusdn2019/homework/10017 1. 解压安装OpenDayLight控制器(本次实验统一使用Be ...
- NGINX心跳检测
NGINX心跳检测 upstream springboot { server 10.3.73.223:8080 max_fails=2 fail_timeout=30s; server 10.3.73 ...