从最后一位开始正常的累加，当前位为sum / 10，进位为sum % 10，每一位存的结果存在stack中，最后输出stack即可

#include <iostream>
#include <stack>

using namespace std;

int main() {
    string num1, num2;
    while (cin >> num1 >> num2) {
        int i = num1.length() - 1, j = num2.length() - 1, n1, n2, carry = 0;
        stack<int> st;
        while (i >= 0 || j >= 0 || carry > 0) {
            n1 = i >= 0 ? num1[i] - '0' : 0;
            n2 = j >= 0 ? num2[j] - '0' : 0;
            st.push((n1 + n2 + carry) % 10);
            carry = (n1 + n2 + carry) / 10;
            i--;
            j--;
        }
        while (st.size()) {
            cout << st.top();
            st.pop();
        }
        cout << endl;
    }
}