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I was advised to arrange for insurance______ I needed medical treatment.A.so thatB.in case
I was advised to arrange for insurance______ I needed medical treatment.
A.so that
B.in case
C.although
D.nevertheless
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I was advised to arrange for insurance______ I needed medical treatment.
A.so that
B.in case
C.although
D.nevertheless
I was advised to arrange for insurance ______ I needed medical treatment.
A.nevertheless
B.although
C.in case
D.so that
I advised them to withdraw ______.
A.so as not get not involved
B.so as not to get involved
C.so as to not get involved
D.so not as to get involved
A.nevertheless
B.although
C.in case
D.so that
I can prescribe some pills which will relieve the pain, but you'd be well advised not to overwork yourself.(英译中)
I _______her to give up taking the medicine but she refused.
A. advised
B. suggested
C. persuaded
D. hoped
阅读程序: Option Base 1 Dim arr()As Integer Private Sub Form_Click() Dim i AsInteger,j As Integer ReDim arr(3,2) For i=1 To 3 Forj=1 To 2 arr(i,j)=i*2+j Next j Next i ReDim Preserve arr(3,4) For j=3 To 4 arr(3,j)=j+9 Nextj Print arr(3,2)+arr(3,4) End Sub 程序运行后,单击窗体,输出结果为
A.21
B.13
C.8
D.25
以下函数用于求出一个2×4矩阵中的最大元素值。 max_value(arr) int arr[][4]; { int i,j,max; max=arr[O][0]; for(i=0;i<2;i++) for(j=0; 【 】;j++) if(【 】>max)max= 【 】; return(max); }
用指针作函数参数,编程序求一维数组中的最大和最小的元素值。
#define N 10
main()
{ void maxmin(int arr[],int *pt1,int *pt2,int n);
int array[N]={10,7,19,29,4,0,7,35,-16,21},*p1,*p2,a,b;
p1=&a; p2=&b;
maxmin(array,p1,p2,N);
printf("max=%d,min=%d",a,b);
}
void maxmin(int arr[],int *pt1,int *pt2,int n)
{ int i;
*pt1=*pt2=arr[0];
for(i=1;i<N;I++)
{ if(arr[i]>*pt1) (9) ;
if(arr[i]<*pt2) (10) ;
}
}
0;for(inti=0;i
A.获取数组的最大索引值
B.判断数组中是否存在重复元素
C.获取数组中元素的个数
D.获取数组中的最大元素