OpenJudge 005:Boolean Expressions

OpenJudge 005: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 1: F
Expression 2: V
Expression 3: V

思想：
说一些个人理解：
根据优先级有：！>&>|，
所以可以划分为：｜……｜……｜…… …… …… …… …… …… …… （1）
上一行省略号中的内容：&……&……&………… …… …… …… …… （2）
上一行省略号中的内容：！……！……！………… …… …… …… …（3）
而上一行省略号的内容呢又可以继续划分为上面的（1）（2）（3）
特别有‘（’和‘）’时，就又回到（1）中情形如此形成递归，其中是优先级的关系，如：
要求（1）中的值，就要求省略号，就要往下求，如此一直到最终读到‘V’或者’F’。

代码：

#include<iostream>
using namespace std;

void get_space() {
   
   
    while (cin.peek() == ' ')
        cin