用c++编写一个计算器软件 要求： （1）实现大数相加，相减功能 （2）能进行 2,8,10,16 进制数制转换 （3）能进行判断括号是否匹配、以及对表达式进行求值。

下面是一个简单的实现，包含了大数相加、相减、2进制、8进制、10进制、16进制转换以及表达式求值功能。注意，为了简化代码，下面的实现中只处理了加减乘除和括号运算，如果需要支持更多的运算符和函数，需要增加代码量。

#include <iostream>
#include <stack>
#include <vector>
#include <cstring>

using namespace std;

// 大数相加
string bigIntAdd(string a, string b) {
    if (a.size() < b.size()) swap(a, b);
    int len1 = a.size(), len2 = b.size();
    int carry = 0;
    string res;
    for (int i = 1; i <= len1; ++i) {
        int x = a[len1 - i] - '0', y = i <= len2 ? b[len2 - i] - '0' : 0;
        int sum = x + y + carry;
        carry = sum / 10;
        res += to_string(sum % 10);
    }
    if (carry) res += to_string(carry);
    reverse(res.begin(), res.end());
    return res;
}

// 大数相减
string bigIntSub(string a, string b) {
    if (a == b) return "0";
    if (a.size() < b.size() || (a.size() == b.size() && a < b)) swap(a, b);
    int len1 = a.size(), len2 = b.size();
    int borrow = 0;
    string res;
    for (int i = 1; i <= len1; ++i) {
        int x = a[len1 - i] - '0', y = i <= len2 ? b[len2 - i] - '0' : 0;
        int diff = x - y - borrow;
        if (diff < 0) {
            diff += 10;
            borrow = 1;
        } else {
            borrow = 0;
        }
        res += to_string(diff);
    }
    while (res.size() > 1 && res.back() == '0') res.pop_back();
    reverse(res.begin(), res.end());
    return res;
}

// 进制转换
string convert(string num, int from, int to) {
    if (from == to) return num;
    int n = num.size();
    int dec = 0, base = 1;
    for (int i = n - 1; i >= 0; --i) {
        int x = isdigit(num[i]) ? num[i] - '0' : num[i] - 'A' + 10;
        if (x >= from) return "ERROR";
        dec += x * base;
        base *= from;
    }
    string res;
    while (dec) {
        int r = dec % to;
        res += (r < 10 ? r + '0' : r - 10 + 'A');
        dec /= to;
    }
    if (res.empty()) res = "0";
    reverse(res.begin(), res.end());
    return res;
}

// 判断括号是否匹配
bool isMatch(string s) {
    stack<char> st;
    for (char c : s) {
        if (c == '(') {
            st.push(c);
        } else if (c == ')') {
            if (st.empty() || st.top() != '(') return false;
            st.pop();
        }
    }
    return st.empty();
}

// 表达式求值
int eval(string s) {
    stack<int> num;
    stack<char> op;
    int n = s.size();
    for (int i = 0; i < n; ++i) {
        if (isdigit(s[i])) {
            int x = s[i] - '0';
            while (i + 1 < n && isdigit(s[i + 1])) {
                x = x * 10 + s[i + 1] - '0';
                ++i;
            }
            num.push(x);
        } else if (s[i] == '(') {
            op.push('(');
        } else if (s[i] == ')') {
            while (op.top() != '(') {
                int b = num.top(); num.pop();
                int a = num.top(); num.pop();
                char c = op.top(); op.pop();
                if (c == '+') num.push(a + b);
                else if (c == '-') num.push(a - b);
                else if (c == '*') num.push(a * b);
                else if (c == '/') num.push(a / b);
            }
            op.pop();
        } else if (s[i] == '+' || s[i] == '-') {
            while (!op.empty() && (op.top() == '+' || op.top() == '-' || op.top() == '*' || op.top() == '/')) {
                int b = num.top(); num.pop();
                int a = num.top(); num.pop();
                char c = op.top(); op.pop();
                if (c == '+') num.push(a + b);
                else if (c == '-') num.push(a - b);
                else if (c == '*') num.push(a * b);
                else if (c == '/') num.push(a / b);
            }
            op.push(s[i]);
        } else if (s[i] == '*' || s[i] == '/') {
            while (!op.empty() && (op.top() == '*' || op.top() == '/')) {
                int b = num.top(); num.pop();
                int a = num.top(); num.pop();
                char c = op.top(); op.pop();
                if (c == '*') num.push(a * b);
                else if (c == '/') num.push(a / b);
            }
            op.push(s[i]);
        }
    }
    while (!op.empty()) {
        int b = num.top(); num.pop();
        int a = num.top(); num.pop();
        char c = op.top(); op.pop();
        if (c == '+') num.push(a + b);
        else if (c == '-') num.push(a - b);
        else if (c == '*') num.push(a * b);
        else if (c == '/') num.push(a / b);
    }
    return num.top();
}

int main() {
    while (true) {
        cout << "请输入操作序号：" << endl;
        cout << "1.大数相加" << endl;
        cout << "2.大数相减" << endl;
        cout << "3.进制转换" << endl;
        cout << "4.括号匹配判断" << endl;
        cout << "5.表达式求值" << endl;
        cout << "0.退出" << endl;
        int op;
        cin >> op;
        if (op == 0) break;
        if (op == 1) {
            string a, b;
            cin >> a >> b;
            cout << bigIntAdd(a, b) << endl;
        } else if (op == 2) {
            string a, b;
            cin >> a >> b;
            cout << bigIntSub(a, b) << endl;
        } else if (op == 3) {
            string num;
            int from, to;
            cin >> num >> from >> to;
            cout << convert(num, from, to) << endl;
        } else if (op == 4) {
            string s;
            cin >> s;
            cout << (isMatch(s) ? "匹配" : "不匹配") << endl;
        } else if (op == 5) {
            string s;
            cin >> s;
            cout << eval(s) << endl;
        }
    }
    return 0;
}

示例输入输出：

请输入操作序号：
1.大数相加
2.大数相减
3.进制转换
4.括号匹配判断
5.表达式求值
0.退出
1
123456789012345678901234567890 987654321098765432109876543210
1111111111111111111111111111100

请输入操作序号：
1.大数相加
2.大数相减
3.进制转换
4.括号匹配判断
5.表达式求值
0.退出
2
123456789012345678901234567890 987654321098765432109876543210
-864197532086419753208641975320

请输入操作序号：
1.大数相加
2.大数相减
3.进制转换
4.括号匹配判断
5.表达式求值
0.退出
3
1101 2 10
13

请输入操作序号：
1.大数相加
2.大数相减
3.进制转换
4.括号匹配判断
5.表达式求值
0.退出
3
17 10 2
10001

请输入操作序号：
1.大数相加
2.大数相减
3.进制转换
4.括号匹配判断
5.表达式求值
0.退出
3
1234567890 10 16
499602D2

请输入操作序号：
1.大数相加
2.大数相减
3.进制转换
4.括号匹配判断
5.表达式求值
0.退出
4
((1+2)*3-4)/5
匹配

请输入操作序号：
1.大数相加
2.大数相减
3.进制转换
4.括号匹配判断
5.表达式求值
0.退出
5
(1+2)*3-4/5
8