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【简答题】
以下程序的功能是；求出数组X中各相邻两个元素的和依次存放到a数组中,然后输出。请填空。 main() {int x[10],a[9],i； for(i=O；i<10；i++)scanf("％d",&x[i])； for(【 】；i<10；i++) a[i-1]=X[i]+【 】; for(i=O；i<9；i++)printf("％d",a[i])； printf("")； }
参考答案：
参考解析：
刷刷题刷刷变学霸
举一反三
【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
【单选题】设有声明和语句:int a[]={2,4,6,8,10},*p; p=&a[1]; 则printf("%d\n",p[1]);输出的结果为
A.
4
B.
6
C.
8
D.
10