进阶递归之Boolean Expressions

时间: 1ms        内存:64M

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.

示例输入:

( V | V ) & F & ( F| V)
!V | V & V & !F & (F | V ) & (!F | F | !V & V)
(F&F|V|!V&!F&!(F|F&V))

示例输出:

Expression 1: F
Expression 2: V
Expression 3: V

提示:

参考答案(内存最优[1100]):

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#include <math.h>
char oprd[50];//操作数
char oprt[50];//操作符
char str[200];
int lend,lent;
void calculate()
{
    if(oprt[lent]=='!')
        oprd[lend]^=16;
    else if(oprt[lent]=='&')
    {
        --lend;
        if(oprd[lend]=='F'||oprd[lend+1]=='F')
            oprd[lend]='F';
    }
    else if(oprt[lent]=='|')
    {
        --lend;
        if(oprd[lend]=='V'||oprd[lend+1]=='V')
            oprd[lend]='V';
    }
    --lent;
}
int cmpPriority(char oper1,char oper2)
{
    if(oper1=='(')
        return 0;
    if(oper1=='|'&&(oper2=='&'||oper2=='!'))
        return 0;
    if(oper1=='&'&&oper2=='!')
        return 0;
    return 1;
}
int main()
{
    int cnt=1;
    while(gets(str))
    {
        int len=strlen(str);
        lend=lent=0;
        for(int i=0;i<=len;i++)
        {
            if(str[i]=='('||str[i]=='!')
               oprt[++lent]=str[i];
            else if(str[i]==')')
            {
                while(oprt[lent]!='(')
                    calculate();
                 --lent;
            }
            else if(str[i]=='&'||str[i]=='|')
            {
                while(cmpPriority(oprt[lent],str[i])&&lent)
                    calculate();
                oprt[++lent]=str[i];
            }
            else if(str[i]=='F'||str[i]=='V')
                oprd[++lend]=str[i];
        }
        while(lent)
            calculate();
        printf("Expression %d: %c\n",cnt++,oprd[1]);
    }
    return 0;
}

参考答案(时间最优[0]):

#include<iostream>
#include<cstdio>
#include<stack>
#include<queue>
#include<string>
using namespace std;

stack<char> s;
queue<char> q;  //存后缀表达式
string c;

int f(char c){  //求表达式的优先级
	if(c=='(') return 4;
	if(c=='!') return 3;
	if(c=='&') return 2;
	if(c=='|') return 1;
}

int f2(char c){  //转换F和V
	if(c=='F') return 0;
	else return 1;
}

void houzui(){   //求表达式对应的后缀表达式
	while(s.size()){
		s.pop();
	}
	for(int i=0;i<c.size();i++){
		if(c[i]!=' '){  //除去空格
			if(c[i]=='F'||c[i]=='V') q.push(c[i]);
			else if(c[i]=='!'&&s.size()&&s.top()=='!'){  //除去两个连续的'!'
				s.pop();
			}
			else if(!s.size()) s.push(c[i]);
			else if(c[i]==')'){           //如果是右括号,则弹出对应左括号前的所有的运算符
				while(s.top()!='('){
					q.push(s.top());
					s.pop();
				}
				s.pop();
				continue;
			}
			else if(f(s.top())==4||(f(c[i])>f(s.top()))) s.push(c[i]);  //左括号优先级最高,入栈后优先级最低
			else if(f(s.top())!=4&&f(c[i])<=f(s.top())){
				q.push(s.top());
				s.pop();
				while(s.size()&&f(s.top())!=4&&f(c[i])<=f(s.top())){      //弹出不低于c[i]优先级的运算
					q.push(s.top());
					s.pop();
				}
				s.push(c[i]);
			}
		}
	}
	while(s.size()){
		q.push(s.top());
		s.pop();
	}
}

void qiuzhi(){     //后缀表达式求值
	bool r=1;
    char x,y;
	while(q.size()){
		if(q.front()=='V'||q.front()=='F') {
		s.push(q.front());
		q.pop();}
		else{
			if(q.front()=='&') {
				x=s.top();
				s.pop();
				y=s.top();
				s.pop();
				r=f2(x)&&f2(y);
				if(r==1)
				s.push('V');
				else
				s.push('F');
			}
			else if(q.front()=='|'){
				x=s.top();
				s.pop();
				y=s.top();
				s.pop();
				r=f2(x)||f2(y);
				if(r==1)
				s.push('V');
				else
				s.push('F');
			}
			else{  //!运算
				x=s.top();
				s.pop();
				if(f2(x)==1)
				s.push('F');
				else
				s.push('V');
			}
			q.pop();
		}
	}
}

int main(){
	int flag=0;        //统计次数
	while(getline(cin,c)){
		flag++;
	houzui();
	qiuzhi();
	cout<<"Expression "<<flag<<": "<<s.top()<<endl;
	}
	return 0;
}

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