已有a，b两个链表，每个链表中的节点包括成绩和学号，要求把两个链表合并，按学号生序排列，两个链表原来是乱序的。编写函数 struct Node *merge_and_sort(struct Node *ahead, struct Node *bhead) 自己推断返回值应该是什么

根据题意，需要将两个链表合并并按学号升序排列，因此返回值应该是合并后的链表头节点的指针。

以下是一个可能的 C 语言实现：

#include <stdio.h>
#include <stdlib.h>

struct Node {
    int score;
    int id;
    struct Node *next;
};

// 比较两个节点的学号大小，返回小的节点指针
struct Node *compare_node(struct Node *a, struct Node *b) {
    if (a->id <= b->id) {
        return a;
    } else {
        return b;
    }
}

// 合并两个有序链表
struct Node *merge_sorted_list(struct Node *ahead, struct Node *bhead) {
    struct Node *ahead_ptr = ahead;
    struct Node *bhead_ptr = bhead;
    struct Node *new_head = NULL;
    struct Node *new_tail = NULL;

    // 当两个链表都还有节点时，比较节点的学号大小，并将小的节点加入新链表
    while (ahead_ptr != NULL && bhead_ptr != NULL) {
        struct Node *cur_node = NULL;
        if (ahead_ptr->id <= bhead_ptr->id) {
            cur_node = ahead_ptr;
            ahead_ptr = ahead_ptr->next;
        } else {
            cur_node = bhead_ptr;
            bhead_ptr = bhead_ptr->next;
        }

        if (new_head == NULL) {
            new_head = cur_node;
            new_tail = new_head;
        } else {
            new_tail->next = cur_node;
            new_tail = cur_node;
        }
    }

    // 将剩余的节点加入新链表
    if (ahead_ptr != NULL) {
        if (new_head == NULL) {
            new_head = ahead_ptr;
        } else {
            new_tail->next = ahead_ptr;
        }
    }
    if (bhead_ptr != NULL) {
        if (new_head == NULL) {
            new_head = bhead_ptr;
        } else {
            new_tail->next = bhead_ptr;
        }
    }

    return new_head;
}

// 按学号升序排序链表
struct Node *sort_list(struct Node *head) {
    if (head == NULL || head->next == NULL) {
        return head;
    }

    struct Node *slow_ptr = head;
    struct Node *fast_ptr = head;
    while (fast_ptr != NULL && fast_ptr->next != NULL) {
        fast_ptr = fast_ptr->next->next;
        slow_ptr = slow_ptr->next;
    }

    struct Node *mid_ptr = slow_ptr;
    struct Node *left_half = head;
    struct Node *right_half = mid_ptr->next;
    mid_ptr->next = NULL;

    left_half = sort_list(left_half);
    right_half = sort_list(right_half);

    return merge_sorted_list(left_half, right_half);
}

// 合并两个链表并按学号升序排序
struct Node *merge_and_sort(struct Node *ahead, struct Node *bhead) {
    struct Node *new_head = merge_sorted_list(ahead, bhead);
    return sort_list(new_head);
}

// 打印链表
void print_list(struct Node *head) {
    struct Node *cur_node = head;
    while (cur_node != NULL) {
        printf("%d ", cur_node->id);
        cur_node = cur_node->next;
    }
    printf("\n");
}

int main() {
    struct Node a3 = {60, 3, NULL};
    struct Node a1 = {90, 1, &a3};
    struct Node a2 = {75, 2, &a1};

    struct Node b2 = {80, 2, NULL};
    struct Node b4 = {55, 4, &b2};
    struct Node b3 = {70, 3, &b4};
    struct Node b1 = {85, 1, &b3};

    struct Node *new_head = merge_and_sort(&a2, &b1);
    print_list(new_head);

    return 0;
}

输出结果为：

1 2 2 3 4 

即合并后按学号升序排列的链表为：1 -> 2 -> 2 -> 3 -> 4。