POJ-2106 Boolean Expressions-！||&计算

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本文探讨如何通过编程解决包含复杂布尔运算符的表达式评估问题，详细介绍了输入格式、解析过程及输出规则，旨在提供一种高效的方法来处理此类逻辑问题。

Boolean Expressions

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3679		Accepted: 1107

Description

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.

Input

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.

Output

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.

Sample Input

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

Sample Output

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

Source

México and Central America 2004

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=100+10;
int val[maxn],vtop;
int op[maxn],otop;
void insert(int t)
{
    while(otop&&op[otop-1]==3)
    {
        t=!t;
        --otop;
    }
    val[vtop++]=t;
}
void calc()
{
    int b=val[--vtop];
    int a=val[--vtop];
    int o=op[--otop];
    int c=(a&b);
    if(o==1)c=(a|b);
    insert(c);
}
int main()
{
    int loop=1;
    char c;
    while((c=getchar())!=EOF)
    {
        vtop=otop=0;
        do
        {
            if(c=='(')
            {
                op[otop++]=0;
            }
            else if(c==')')
            {
                while(otop&&op[otop-1]!=0)
                    calc();
                --otop;
                insert(val[--vtop]);
            }
            else if(c=='!')
            {
                op[otop++]=3;
            }
            else if(c=='&')
            {
                while(otop&&op[otop-1]>=2)
                    calc();
                op[otop++]=2;
            }
            else if(c=='|')
            {
                while(otop&&op[otop-1]>=1)
                    calc();
                op[otop++]=1;
            }
            else if(c=='V'||c=='F')
            {
                insert(c=='V'?1:0);
            }
        }
        while((c=getchar())!='\n'&&c!=EOF);
        while(otop)calc();
        printf("Expression %d: %c\n",loop++,(val[0]?'V':'F'));
    }
    return 0;
}