##CC10 牛牛的双链表求和

#include <stdlib.h>
#include<stdio.h>
//单向循环链表
typedef struct node
 {
    int data;
    struct node* next;
} node;
//创建链表b头节点
node* add_head() 
{
    node* Head = (node*)malloc(sizeof(node));
    if (Head == NULL)
        return NULL;
    Head->next = Head;
    return Head;
}
//尾插法
void add_node(node* Head, int data)
 {
    node* new = (node*)malloc(sizeof(node));
    if (new == NULL)
        return;
    //节点成员赋值
    new->data = data;
    new->next = NULL;
    //链接
    node* pT = NULL;
    for (pT = Head; pT->next != Head; pT = pT->next);
    new->next = pT->next;
    pT->next = new;
}
//输出链表
void output(node* Head) 
{
    if (Head->next == Head)
        return;
    node* pT = Head->next;
    while (pT != Head) {
        printf("%d ", pT->data);
        pT = pT->next;
    }
}
int main(void)
 {
    node* Head = add_head();//链表头节点
    int n, i, j;
    scanf("%d", &n);
    int arr[n];
    int brr[n];
    //将键盘键入的数据存放到数组中
    for (i = 0; i < n; i++)
        scanf("%d", &arr[i]);
    for (i = 0; i < n; i++)
        scanf("%d", &brr[i]);
    //将数据插入链表
    for (j = 0; j < n; j++)
        add_node(Head, arr[j] + brr[j]);
    output(Head);
    return 0;
}