这里估计大部分都是写了c++有一定基础的，那么我就用链表实现a+b问题吧

#include <iostream>
struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(nullptr) {}
};
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
{
    ListNode *dummy = new ListNode(0);
    ListNode *current = dummy;
    int carry = 0;
    while (l1 || l2 || carry)
    {
        int sum = carry;
        if (l1)
        {
            sum += l1->val;
            l1 = l1->next;
        }
        if (l2)
        {
            sum += l2->val;
            l2 = l2->next;
        }
        carry = sum / 10;
        sum = sum % 10;
        current->next = new ListNode(sum);
        current = current->next;
    }
    return dummy->next;
}
ListNode *createLinkedList(int arr[], int n)
{
    if (n == 0)
    {
        return nullptr;
    }
    ListNode *head = new ListNode(arr[0]);
    ListNode *current = head;
    for (int i = 1; i < n; i++)
    {
        current->next = new ListNode(arr[i]);
        current = current->next;
    }
    return head;
}
void printLinkedList(ListNode *head)
{
    ListNode *current = head;
    while (current)
    {
        std::cout << current->val << " ";
        current = current->next;
    }
    std::cout << std::endl;
}
int main()
{
    int arr1[1];
    int arr2[1];
    std::cin >> arr1[0];
    std::cin >> arr2[0];
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    ListNode *l1 = createLinkedList(arr1, n1);
    ListNode *l2 = createLinkedList(arr2, n2);
    ListNode *sum = addTwoNumbers(l1, l2);
    printLinkedList(sum);
    return 0;
}