A customer is planning to buy a two enclosure Power 770, and will implement 4 LPARs with the following requirements:Each LPAR must have a private, dedicated connection to the network.The adapter for this network connection must be hot-swappable.The customer wants to minimize the number of I/O slots used. Which of the following fulfills the customer requirements?（）A、Two dual-port PCIe adaptersB、A four-port PCIe Ethernet adapterC、Four separate PCIe Ethernet adaptersD、Two embedded Integrated Virtual Ethernet adapters (IVE)
A customer is planning to buy a two enclosure Power 770, and will implement 4 LPARs with the following requirements:Each LPAR must have a private, dedicated connection to the network.The adapter for this network connection must be hot-swappable.The customer wants to minimize the number of I/O slots used. Which of the following fulfills the customer requirements?（）
- A、Two dual-port PCIe adapters
- B、A four-port PCIe Ethernet adapter
- C、Four separate PCIe Ethernet adapters
- D、Two embedded Integrated Virtual Ethernet adapters (IVE)
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阅读以下说明和c++代码，将应填入(n)处的字句写在对应栏内。【说明】本程序将两个从小到大的有序链表合成一个新的从小到大的有序链表。链表的每一项由类 Node描述，而链表由List描述，类List的成员函数有以下几个：creatList(): 创建从小到大的有序链表。multiplyList(List L1, Llst L2): 将链表L1和链表L2合并。print(): 打印链表。【C++代码】include ＜iostream＞using namespace std;class List;class Node{friend class List;public:Node(int data){(1);}private:int data;Node *next;};class List{public：List(){list=NULL;}void multiplyList(List L1, List L2);void creatList();void print();private:Node *list;};void List::creatList(){Node *p, *u, *pre;int dara;list=NULL;wbile(1){cout＜＜"输入链表的一项: (小于零，结束链表) "＜＜endl;cin＞＞data;if(dara＜0)break;//小于零，结束输入p=list;while(p !=NULL dara＞p-＞data){//查找插入点pre=p;p=p-＞next;}u=(2);if(p==list)list=u;else pre-＞next=u;(3);}}void List::multiplyList(List L1, List L2){Node *pL1, *pL2, *pL, *u;list = NULL;pL1 = L1.list;pL2 = L2.11st;while(pL1 != NULL pL2 != NULL){if(pL1-＞data ＜ pL2-＞data){u = new Node(pL1-＞data);pL1 = pL1-＞next;}else{u = new Node(pL2-＞data);pL2 = pL2-＞next;}if(list == NULL){list =(4);}else{pL-＞next=u;pL=u;}}pL1 = (pL1 != NULL)? pL1:pL2;while(pL1 != NULL){u=(5);pL1 = pL1-＞next;if(list == NULL){list=pL=u;}else{pL-＞next=u;pL=u;}}}void List::print(){Node *p;p = list;while(p !=NULL){cout＜＜p-＞data＜＜"\t";p=p-＞next;&
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