A、With reference to
B、Refer
C、Referring
D、With refer to
第1题:
在循环双链表的p结点之后插入s结点的操作是______。
A.p→next=s; p→next→prior=s; S→prior=p; S→next=p→next;
B.s→next=p; s→next=p→next; p→next=s; p→next→prior=s;
C.p→next=s; s→prior=p; p→next→prior=s; s→next=p→next;
D.s→prior=p; s→next=p→next; p→next→prior=s; p→next=S;
第2题:
在循环双链表的p节点之后插入s节点的操作是______。
A.p→next=s; p→next→prior=s; s→prior=p; S→next=p→next;
B.s→next=p; s→next=p→next; p→next=S; p→next→prior=s;
C.p→next=s; s→prior=p; p→next→prior=s; s→next=p→next;
D.s→prior=p; s→next=p→next; p→next→prior=s; p→next=s;
第3题:
在循环双链表的p结点之后插入s结点的操作是______。
A.p→next=s;p→next→prior=s;s→prior=p;s→next=p→next;
B.s→!next=p;s→next=p→next;p→next=s;p→next→prior=s;
C.p→next=s;s→prior=p;p→next→prior=s;s→next=p→next;
D.s→prior=p;s→next=p→next;p→next→prior=s;p→next=s;
第4题:
16、在单链表结点p之后插入结点s,正确的操作是()。
A.p.next=s;s.next=p.next;
B.s.next=p.next;p.next=s;
C.p.next=s;p.next=s.next;
D.p.next=s.next;p.next=s;
第5题:
在p所指结点后插入s所指结点的正确操作是:
A.s->next =p+1; p->next=s;
B.(*p).next=s; (*s).next=(*p).next;
C.s->next=p->next; p->next=s->next;
D.s->next=p->next; p->next=s;
第6题:
A.Completed,Deposited,Released,Shipped
B.Completed,Released,Shipped,Deposited
C.Deposited,Released,Shipped,Completed
D.Released,Shipped,Completed,Deposited
E.Released,Shipped,Deposited,Completed
第7题:
在单链表结点p之后插入结点s,正确的操作是()。
A.p.next=s;s.next=p.next;
B.s.next=p.next;p.next=s;
C.p.next=s;p.next=s.next;
D.p.next=s.next;p.next=s;
第8题:
在循环双链表的p节点之后插As节点的操作是______。
A.p→next=s; p→next→prior=s; s→prior=p; s→next=p→next;
B.s→next=p; s→next=p→next; p→next=s; p→next→prior=s;
C.p→next=s; s→prior=p; p→next→prior=s; s→next=p→next;
D.s→prior=p; s→next=p→next; p→next→prior=s; p→next=s;
第9题:
4、以下的描述中,必然是对Mealy型状态机的描述的是?
A.always @(*) case (state) S0: begin out = 0; if (in) next_state = S1; else next_state = S2; end ……#B.always @(*) case (state) S0: begin if (in) next_state = S1; else next_state = S0; end ……#C.always @(*) case (state) S0: begin if (in) begin next_state = S1; out=1 end else next_state = S0; end ……#D.以上答案均不正确第10题:
以下的描述中,必然是对Mealy型状态机的描述的是?
A.always @(*) case (state) S0: begin out = 0; if (in) next_state = S1; else next_state = S2; end ……#B.always @(*) case (state) S0: begin if (in) next_state = S1; else next_state = S0; end ……#C.always @(*) case (state) S0: begin if (in) begin next_state = S1; out=1 end else next_state = S0; end ……#D.以上答案均不正确