代碼裏有註釋。。。直接上代碼。。。

#include<bits/stdc++.h>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define per(i,n,k) for(int i=n;i>=k;i--)
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define re return
#define se second
#define fi first
using namespace std;
//---------------------------------------------------------------head----------------------------------------------------------------------
struct BigInteger {
    typedef unsigned long long LL;

    static const int BASE = 100000000;
    static const int WIDTH = 8;
    vector<int> s;

    BigInteger& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;}
    BigInteger(LL num = 0) {*this = num;}
    BigInteger(string s) {*this = s;}
    BigInteger& operator = (long long num) {
        s.clear();
        do {
            s.push_back(num % BASE);
            num /= BASE;
        } while (num > 0);
        return *this;
    }
    BigInteger& operator = (const string& str) {
        s.clear();
        int x, len = (str.length() - 1) / WIDTH + 1;
        for (int i = 0; i < len; i++) {
            int end = str.length() - i*WIDTH;
            int start = max(0, end - WIDTH);
            sscanf(str.substr(start,end-start).c_str(), "%d", &x);
            s.push_back(x);
        }
        return (*this).clean();
    }

    BigInteger operator + (const BigInteger& b) const {
        BigInteger c; c.s.clear();
        for (int i = 0, g = 0; ; i++) {
            if (g == 0 && i >= s.size() && i >= b.s.size()) break;
            int x = g;
            if (i < s.size()) x += s[i];
            if (i < b.s.size()) x += b.s[i];
            c.s.push_back(x % BASE);
            g = x / BASE;
        }
        return c;
    }
    BigInteger operator - (const BigInteger& b) const {
        assert(b <= *this); // 減數不能大於被減數
        BigInteger c; c.s.clear();
        for (int i = 0, g = 0; ; i++) {
            if (g == 0 && i >= s.size() && i >= b.s.size()) break;
            int x = s[i] + g;
            if (i < b.s.size()) x -= b.s[i];
            if (x < 0) {g = -1; x += BASE;} else g = 0;
            c.s.push_back(x);
        }
        return c.clean();
    }
    BigInteger operator * (const BigInteger& b) const {
        int i, j; LL g;
        vector<LL> v(s.size()+b.s.size(), 0);
        BigInteger c; c.s.clear();
        for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=LL(s[i])*b.s[j];
        for (i = 0, g = 0; ; i++) {
            if (g ==0 && i >= v.size()) break;
            LL x = v[i] + g;
            c.s.push_back(x % BASE);
            g = x / BASE;
        }
        return c.clean();
    }
    BigInteger operator / (const BigInteger& b) const {
        assert(b > 0);  // 除數必須大於0
        BigInteger c = *this;       // 商:主要是讓c.s和(*this).s的vector一樣大
        BigInteger m;               // 余數:初始化為0
        for (int i = s.size()-1; i >= 0; i--) {
            m = m*BASE + s[i];
            c.s[i] = bsearch(b, m);
            m -= b*c.s[i];
        }
        return c.clean();
    }
    BigInteger operator % (const BigInteger& b) const { //方法與除法相同
        BigInteger c = *this;
        BigInteger m;
        for (int i = s.size()-1; i >= 0; i--) {
            m = m*BASE + s[i];
            c.s[i] = bsearch(b, m);
            m -= b*c.s[i];
        }
        return m;
    }

    int bsearch(const BigInteger& b, const BigInteger& m) const{
        int L = 0, R = BASE-1, x;
        while (1) {
            x = (L+R)>>1;
            if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;}
            else R = x;
        }
    }
    BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;}
    BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;}
    BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;}
    BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;}
    BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;}

    bool operator < (const BigInteger& b) const {
        if (s.size() != b.s.size()) return s.size() < b.s.size();
        for (int i = s.size()-1; i >= 0; i--)
            if (s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }
    bool operator >(const BigInteger& b) const{return b < *this;}
    bool operator<=(const BigInteger& b) const{return !(b < *this);}
    bool operator>=(const BigInteger& b) const{return !(*this < b);}
    bool operator!=(const BigInteger& b) const{return b < *this || *this < b;}
    bool operator==(const BigInteger& b) const{return !(b < *this) && !(b > *this);}
};

ostream& operator << (ostream& out, const BigInteger& x) {
    out << x.s.back();
    for (int i = x.s.size()-2; i >= 0; i--) {
        char buf[20];
        sprintf(buf, "%08d", x.s[i]);
        for (int j = 0; j < strlen(buf); j++) out << buf[j];
    }
    return out;
}

istream& operator >> (istream& in, BigInteger& x) {
    string s;
    if (!(in >> s)) return in;
    x = s;
    return in;
}
stack<BigInteger> si;
stack<char> sc;
vector<BigInteger> vc;
bool isnum[1005];
bool flag=0;
BigInteger pow(BigInteger a,BigInteger b){BigInteger base=a,ans=1;while(b!=0){if(b%2==1)ans*=base;base*=base;b/=2;}return ans;}
BigInteger clas(char c)//符號優先級 
{
    if(c=='(' || c==')') re 0;
    if(c=='+' || c=='-') re 1;
    if(c=='*' || c=='/') re 2;
    if(c=='^') re 3;
    re 4;
}
BigInteger cal(BigInteger a,BigInteger b,BigInteger c)//符號求值 
{
    BigInteger ans=1;
    if(c=='+') re a+b;
    if(c=='-') re b-a;
    if(c=='*') re a*b;
    if(c=='/') re b/a;
    re pow(b,a);
}
int main()
{
    string s;cin>>s;
    s="("+s+")";//可以用來防止棧為空找top時的溢出情況 
    int n=(int)s.size();
    rep(i,0,n-1)
    {
        if(s[i]=='(')//左括號直接壓入 
        {
            sc.push(s[i]);
            continue;
        }
        if(s[i]==')')//找到右括號將棧清空到左括號為止 
        {
            char c=sc.top();sc.pop();
            while(c!='(')
            {
                vc.pb(c);
                c=sc.top();sc.pop();
            }
            continue;
        }
        if(s[i]>='0' && s[i]<='9')//數字 
        {
            BigInteger sum=0;
            while(s[i]>='0' && s[i]<='9')
            sum=sum*10+s[i++]-'0';
            if(flag)
            sum=BigInteger(-1)*sum,flag=0;
            vc.pb(sum);
            isnum[vc.size()-1]=1;
            i--;continue;
        }
        if(s[i]=='-' && (clas(s[i-1])!=4 && s[i-1]!=')'))
        {
            flag=1;
            continue;
        }
        while(!sc.empty() && clas(s[i])<=clas(sc.top()))//如果是符號就找到第一個優先級比他小的 
        vc.pb(sc.top()),sc.pop();
        sc.push(s[i]);
    }
    while(!sc.empty())//清空棧 
    vc.pb(sc.top()),sc.pop();
    rep(i,0,vc.size()-1)//根據後綴表達式求值 
    {
        if(!isnum[i])
        {
            BigInteger a=si.top();si.pop();
            BigInteger b=si.top();si.pop();
            BigInteger ans=cal(a,b,vc[i]);
            si.push(ans);
        } 
        else
        si.push(vc[i]);
    }
    cout<<si.top();
    return 0;
}
//----------------------------------------------------------------end----------------------------------------------------------------------
 

【全網最高端？】中綴表達式轉為後綴表達式以及求值（可用於負數，階乘）