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，我可以帮您解决这个问题。以下是 C++ 代码实现：

#include <iostream>
using namespace std;

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

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

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

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

    head = reverseList(head);

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

    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