有符号一位数中缀式的计算。HOJ2189 Single Digit Adder。

用stl stack实现中缀转后缀及后缀运算。

转换之前将缺少的操作数补0。

Single Digit Adder

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Time limit:	1sec.	Submitted:	197
Memory limit:	32M	Accepted:	45

Source: The 2005 ACM Pacific Northwest Programming Contest

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Write a program that can evaluate expressions from the following roughly BNF (Backus Naur Form) grammar:

expr ::= term | expr ‘+’ term | expr ‘-’ term
unary_op ::= ‘+’ term | ‘-’ term
term ::= ‘(’ expr ‘)’ | ‘(’ unary_op ‘)’ | literal
literal ::= [0-9]

There will be no whitespace within an expression. All expressions will consist solely of the characters (, ), +, -, and the digits 0 through 9. You may assume that all input is well-formed.

Input

The input will consist of one expression per line followed by a newline. There will be no blank lines in the file.

Output

For each expression, output its integer value, followed by a single newline.

Sample Input

1
(-2)+3
(1-(2+3))
(1-2+3)
(1-(+(2-3)))

Sample Output

1
1
-4
2
2

说明： 
给出表达式。 由有符号一位数和运算符 + - 及 括号 组成。
计算结果。
为了将中缀表达为后缀。 将单操作数运算符前补零。
如 (-2)+3 先转换为 (0-2)+3。
然后中缀转后缀。 后缀运算。 
stl stack 实现。 
代码如下：
#include<iostream>
#include<stack>
using namespace std;

bool isd(char a) {
    if (a >= '0' && a <= '9')return true;
    return false;
}

int main() {
    string ex;
    while (cin >> ex) {
        stack<char> a;
        string tmp = "";
        for (int i = 0; i < ex.length(); i++) {
            if ((ex[i] == '+' || ex[i] == '-') && (i - 1 < 0 || (!isd(ex[i - 1]) && ex[i - 1] != ')'))) {
                ex.insert(i, "0");
                i--;
            } else if (isd(ex[i]))tmp += ex[i];
            else if (ex[i] == '(') a.push(ex[i]);
            else if (ex[i] == ')') {
                while (a.top() != '(') {
                    tmp += a.top();
                    a.pop();
                }
                a.pop();
            } else {
                if (a.empty() || a.top() == '(')a.push(ex[i]);
                else {
                    tmp += a.top();
                    a.pop();
                    a.push(ex[i]);
                }
            }
        }
        while (!a.empty()) {
            tmp += a.top();
            a.pop();
        }
        stack<int> b;
        for (int i = 0; i < tmp.length(); i++) {
            if (isd(tmp[i]))b.push(tmp[i] - '0');
            else {
                int tmp1, tmp2;
                tmp2 = b.top();
                b.pop();
                tmp1 = b.top();
                b.pop();
                if (tmp[i] == '+')b.push(tmp1 + tmp2);
                else b.push(tmp1 - tmp2);
            }
        }
        cout << b.top() << endl;
    }
    return 0;
}