c语言编写输入10个数字，并根据这些数字构造单向链表，之后，再反转链表并打印输出，即A={a1,a2,…,a_n}反转成A={a_n,…,a2,a1} 输入提示: "Please input %d numbers:\n" 输入格式: "%d" 输出格式: "%5d"

好的，以下是 C 语言的实现代码：

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

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode* prev = NULL;
    struct ListNode* curr = head;
    while (curr != NULL) {
        struct ListNode* next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

int main() {
    const int N = 10;
    int nums[N];
    printf("Please input %d numbers:\n", N);
    for (int i = 0; i < N; i++) {
        scanf("%d", &nums[i]);
    }

    struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
    head->val = nums[0];
    head->next = NULL;
    struct ListNode* tail = head;
    for (int i = 1; i < N; i++) {
        struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
        node->val = nums[i];
        node->next = NULL;
        tail->next = node;
        tail = node;
    }

    head = reverseList(head);

    printf("After reverse:\n");
    struct ListNode* p = head;
    while (p != NULL) {
        printf("%5d", p->val);
        p = p->next;
    }
    printf("\n");

    return 0;
}

输出结果：

Please input 10 numbers:
1 2 3 4 5 6 7 8 9 10
After reverse:
   10    9    8    7    6    5    4    3    2    1