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下面算法的功能（栈和队列的元素类型均为int）。 void algo(Queue &Q) { Stack S; int d; InitStack(S); while(!QueueEmpty(Q)) { DeQueue(Q, d); Push(S, d); } while(!StackEmpty(S)) { Pop(S, d); EnQueue(Q, d); } } 算法的功能：利用“栈”作辅助，将“队列”中的数据元素进行逆置
A.
正确
B.
错误
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参考答案：
举一反三
【单选题】在n个结点的线性表的数组实现中,算法的时间复杂度是O(1)的操作是 。
A.
访问第i（1<=i<=n）个结点和求第i个结点的直接前驱（1
B.
在第i（1<=i<=n）个结点后插入一个新结点< span="">
C.
删除第i（1<=i<=n）个结点< span="">
D.
以上都不对
【单选题】有如下事件过程：Private Sub Form_Load() Show Dim Arr() As Variant Arr = Array(1, 3, 5, 7, 9, 11, 13, 15) Call converse(Arr) For i = 0 To 7 Print arr(i); Next iEnd Sub 以上程序的功能是，通过调用Converse过程，将数组Arr中的元素逆序存放并输出...
A.
Private Sub converse(dim a() as Variant ) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
B.
Private Sub converse(dim a() as Variant) Dim T Dim J As Integer J = (LBound(A) + UBound(A)) / 2 For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
C.
Private Sub converse(a() As Variant) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
D.
Private Sub converse(a() As Variant) Dim T Dim J As Integer J = Int((LBound(A) + UBound(A)) / 2) For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
相关题目：
【单选题】在n个结点的线性表的数组实现中,算法的时间复杂度是O(1)的操作是 。
A.
访问第i（1<=i<=n）个结点和求第i个结点的直接前驱（1
B.
在第i（1<=i<=n）个结点后插入一个新结点< span="">
C.
删除第i（1<=i<=n）个结点< span="">
D.
以上都不对
【单选题】有如下事件过程：Private Sub Form_Load() Show Dim Arr() As Variant Arr = Array(1, 3, 5, 7, 9, 11, 13, 15) Call converse(Arr) For i = 0 To 7 Print arr(i); Next iEnd Sub 以上程序的功能是，通过调用Converse过程，将数组Arr中的元素逆序存放并输出...
A.
Private Sub converse(dim a() as Variant ) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
B.
Private Sub converse(dim a() as Variant) Dim T Dim J As Integer J = (LBound(A) + UBound(A)) / 2 For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
C.
Private Sub converse(a() As Variant) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
D.
Private Sub converse(a() As Variant) Dim T Dim J As Integer J = Int((LBound(A) + UBound(A)) / 2) For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
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