递归之 Boolean Expressions

关于题目

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
来源
México and Central America 2004

解题思想

主要是应用递归

开始也看了其他的解题方法，可以用栈解决，不过处理起来比较麻烦。布尔表达式本身就是一个递归的定义，项可以由表达式组成也可以由单个字符组成，表达式可以是 项 & 项 的形式，或者是 项 | 项 的形式，或者是 ！项的形式

还有一点需要注意

本题中需注意对空格的处理，所以单独编写了读取函数（函数借鉴于别人）

代码

#include
using namespace std;

bool f1();
bool f2();

char getc_()
{
    char c;
    while((c = cin.get()) == ' ');
    return c;
}
char peekc_()
{
    char c;
    while((c = cin.peek()) == ' ')
    {
        cin.get();
    }
    return c;
}

bool f2()
{
    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();
        if(c == 'V')    return true;
        if(c == 'F')    return false;
    }
}

bool f1()
{
    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 = 0;
    while(cin.peek() != EOF){
        bool f = f1();
        if(cin.peek() == '\n' || cin.peek() == EOF){
            cout<< "Expression " << ++n << ": " << (f?'V':'F') << endl;
            cin.get();
        }
    }
    return 0;
}