用一个结构体type0是字符，type1是数字，然后建一个vector。把所有字符数字数字化（比如‘1’,‘2’,‘3’变成123），然后用栈每找到一个括号对就solve那一段。solve直接用insert和erase，把那一段变成数字。先扫一遍^，然后/*然后+-，同样找到一个符号就把左右和符号本身erase和insert成一个数字。

注意细节处理比如负号，EOF，多余括号。好久没打这种暴力模拟了。。。

C++ 代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iostream>
#include<stack> 
//#pragma G++ optimize(2)
using namespace std;
const int maxn = 1e6+5;

struct node{
    bool Type;//0 char    1 int
    int num;
    char ch;

    void print()
    {
        if(Type==0) printf("%c", ch);
        else printf("%d", num);
    }
};

vector <node> s;
stack <char> stk1;
stack <int> stk2;

node addnode(bool Type, int num, char ch)
{
    node tmp;
    tmp.Type = Type;
    tmp.num = num;
    tmp.ch = ch;
    return tmp;
}


inline int poww(int a, int b)//快速幂
{
    int ans = 1;
    while(b)
    {
        if(b&1)
        {
            ans *= a;
            --b;
        }
        else
        {
            a *= a;
            b >>= 1;
        }
    }
    return ans;
}






inline void Solve(int l, int r)//把l和r之间的表达式合并为一个数
{
    for(register int i=l; i<=r; ++i)
    {
        if(s[i].ch == '^')
        {
            int tmp1 = s[i-1].num;
            int tmp2 = s[i+1].num;
            s.erase(s.begin()+i-1, s.begin()+i+2);
            s.insert(s.begin()+i-1, addnode(1, poww(tmp1, tmp2), 0));
            r-=2;
            i-=1;
        }
    }

    for(register int i=l; i<=r; ++i)
    {
        if(s[i].ch == '*')
        {
            int tmp1 = s[i-1].num;
            int tmp2 = s[i+1].num;
            s.erase(s.begin()+i-1, s.begin()+i+2);
            s.insert(s.begin()+i-1, addnode(1, tmp1 * tmp2, 0));
            r-=2;
            i-=1;
        }
        else if(s[i].ch == '/')
        {
            int tmp1 = s[i-1].num;
            int tmp2 = s[i+1].num;
            s.erase(s.begin()+i-1, s.begin()+i+2);
            s.insert(s.begin()+i-1, addnode(1, tmp1 / tmp2, 0));
            r-=2;
            i-=1;
        }
    }

    for(register int i=l; i<=r; ++i)
    {
        if(s[i].ch == '+')
        {
            int tmp1 = s[i-1].num;
            int tmp2 = s[i+1].num;
            s.erase(s.begin()+i-1, s.begin()+i+2);
            s.insert(s.begin()+i-1, addnode(1, tmp1 + tmp2, 0));
            r-=2;
            i-=1;
        }
        else if(s[i].ch == '-')
        {
            int tmp1 = s[i-1].num;
            int tmp2 = s[i+1].num;
            s.erase(s.begin()+i-1, s.begin()+i+2);
            s.insert(s.begin()+i-1, addnode(1, tmp1 - tmp2, 0));
            r-=2;
            i-=1;
        }
    }
}




int main()
{
//  freopen("input.txt", "r", stdin);
    bool isread = 0;
    char ch;
    bool iseof = 0;


    //读入
    char lstch='(';
    while(1)
    {
        if(iseof) break;
        if(!isread) 
        {
            if(scanf("%c", &ch)==EOF) break;
        }
        int num=0;
        if(ch=='\n' || ch=='\r') break;

        if(ch=='-' && lstch=='(')
        {
            s.push_back(addnode(1, 0, 0));
            s.push_back(addnode(0, 0, '-'));
            lstch = ch;
            continue;
        }
        lstch = ch;
        if(isdigit(ch))
        {
            num = num * 10 + ch - '0';
            isread = 1;
            while(isdigit(ch))
            {
                if(scanf("%c", &ch)==EOF)
                {
                    iseof = 1;
                    break;
                }
                if(!isdigit(ch)) break;
                lstch = ch;
                num = num * 10 + ch - '0';
            }
            s.push_back(addnode(1, num, 0));

        }
        else
        {
            isread = 0;
            s.push_back(addnode(0, 0, ch));
        }


    }



    //用栈找括号并solve，容易发现找到的括号里面没有括号，stk2记录左括号在vector中的位置
    for(register unsigned int i=0; i<s.size(); ++i)
    {
        if(s[i].ch == '(')
        {
            stk1.push('(');
            stk2.push(i);
        }
        else if(s[i].ch == ')')
        {
            if(!stk1.empty() && stk1.top()=='(')
            {
                Solve(stk2.top()+1, i-1);
                s.erase(s.begin()+stk2.top());
                s.erase(s.begin()+stk2.top()+1);
                i = stk2.top();
                stk1.pop();    stk2.pop();
            }
            else
            {
                s.erase(s.begin()+i);//多余右括号删掉
                --i;
            }
        }
    }
    //处理多余左括号
    while(!stk1.empty())
    {
        s.erase(s.begin()+stk2.top());
        stk1.pop();
        stk2.pop();
    }
    Solve(0, s.size()-1);
    printf("%d", s[0].num);
    return 0;


}