用双向链表实现超长整数加减法

好的，我可以为您解答这个问题。

首先，我们需要定义一个双向链表结构体，包含三个成员变量：前驱节点指针、后继节点指针和当前节点的值。代码如下：

struct Node {
    Node *prev, *next;
    int val;
};

接下来，我们可以定义一个类来实现超长整数的加减法。在这个类中，我们需要定义两个双向链表，分别存储两个超长整数的每一位。加法操作需要从链表的尾部开始遍历，同时在遍历的过程中需要考虑进位。减法操作需要先比较两个超长整数的大小，然后从链表的尾部开始遍历，同时在遍历的过程中需要考虑借位。

下面是一个示例代码，可以实现超长整数的加减法：

class LongInt {
public:
    LongInt() {
        head1 = new Node;
        head2 = new Node;
        tail1 = head1;
        tail2 = head2;
        head1->prev = NULL;
        head2->prev = NULL;
        head1->next = tail1;
        head2->next = tail2;
        tail1->prev = head1;
        tail2->prev = head2;
        tail1->next = NULL;
        tail2->next = NULL;
    }

    void add(int x) {
        Node *node = new Node;
        node->val = x;
        node->prev = tail1;
        node->next = tail1->next;
        tail1->next = node;
        tail1 = node;
    }

    void sub(int x) {
        Node *node = new Node;
        node->val = x;
        node->prev = tail2;
        node->next = tail2->next;
        tail2->next = node;
        tail2 = node;
    }

    LongInt operator+(const LongInt& other) const {
        LongInt res;
        Node *p = tail1->prev;
        Node *q = other.tail1->prev;
        int carry = 0;
        while (p != head1->prev || q != other.head1->prev || carry != 0) {
            int x = (p != head1->prev ? p->val : 0);
            int y = (q != other.head1->prev ? q->val : 0);
            int sum = x + y + carry;
            carry = sum / 10;
            sum %= 10;
            res.add(sum);
            if (p != head1->prev) p = p->prev;
            if (q != other.head1->prev) q = q->prev;
        }
        return res;
    }

    LongInt operator-(const LongInt& other) const {
        LongInt res;
        Node *p = tail1->prev;
        Node *q = other.tail1->prev;
        bool borrow = false;
        if (compare(other) < 0) {
            borrow = true;
            std::swap(p, q);
        }
        while (p != head1->prev || q != other.head1->prev) {
            int x = (p != head1->prev ? p->val : 0);
            int y = (q != other.head1->prev ? q->val : 0);
            if (borrow) x--;
            if (x < y) {
                x += 10;
                borrow = true;
            } else {
                borrow = false;
            }
            int diff = x - y;
            res.sub(diff);
            if (p != head1->prev) p = p->prev;
            if (q != other.head1->prev) q = q->prev;
        }
        return res;
    }

    int compare(const LongInt& other) const {
        Node *p = tail1->prev;
        Node *q = other.tail1->prev;
        while (p != head1->prev || q != other.head1->prev) {
            int x = (p != head1->prev ? p->val : 0);
            int y = (q != other.head1->prev ? q->val : 0);
            if (x > y) {
                return 1;
            } else if (x < y) {
                return -1;
            }
            if (p != head1->prev) p = p->prev;
            if (q != other.head1->prev) q = q->prev;
        }
        return 0;
    }

private:
    Node *head1, *head2, *tail1, *tail2;
};

使用示例：

LongInt a, b;
a.add(9);
a.add(9);
a.add(9);
b.add(1);
b.add(0);
LongInt c = a + b;
LongInt d = a - b;

在这个示例中，我们创建了两个超长整数 a 和 b，然后分别添加了三个数字和两个数字。接着，我们通过重载运算符实现了加法和减法操作，并将结果存储在变量 c 和 d 中。