下列程序段不能正确显示1！、2！、3！、4！的结果的是 。 A: For i = 1 To 4 For j = 1 To i n = 1 n = n * j Next j MsgBox( n)Next i B: Dim i%, j%, n%For i = 1 To 4 n = 1 For j = 1 To i n = n * j Next j MsgBox( n)Next i C: Dim j%, n%n = 1For j = 1 To 4 n = n * j MsgBox( n)Next j D: Dim j%, n%n = 1j = 1Do While j<=4 n=n*j MsgBox(n) j=j+1Loop E: Dim j%, n%n = 1j = 1Do n=n*j MsgBox(n) j=j+1Loop While j<=4

下面____程序段不能正确显示1!、2!、3!、4!的值。 A: For I=1 To 4 n=1 For j=1 to I n=n*j Next j Print n Next I B: For I=1 To 4 For j=1 to I n=1 n=n*j Next j Print n Next I C: n=1 For j=1 to 4 n=n*j Print n Next j D: n=1:j=1 Do While j<=4 n=n*j Print n j=j+1Loop

下列程序段能且只能显示1！、2！、3！、4！的值的是（ ）。 A: For i = 1 To 4 n = 1 For j = 1 To i n = n * j Next j MsgBox(n)Next i B: For i = 1 To 4 For j = 1 To i n = 1 n = n * j Next j MsgBox(n)Next i C: n = 1For i = 1 To 4 For j = 1 To 4 n = n * j MsgBox(n) Next jNext i D: n = 1j = 1Do While j <= 4 n = n * j j = j + 1LoopMsgBox(n)

下列程序段能分别正确显示1!、2!、3!、4!的值的是______。 A: For i = 1 To 4 n = 1 For j = 1 To i n = n * j MsgBox(n) Next jNext i B: For i = 1 To 4 For j = 1 To i n = 1 n = n * j Next j MsgBox(n)Next i C: n = 1For i = 1 To 4 For j = 1 To 4 n = n * j MsgBox(n) Next jNext i D: n = 1j = 1Do While j &lt;= 4 n = n * j MsgBox(n) j = j + 1Loop

我们读“尽管”这个词语时，“尽”在语流音变中应当读作 。 A: jīn B: jín C: jǐn D: jìn

“浸”字的读音，正确的是（ ） A: jīn B: qīn C: jìn D: Qǐn

下列程序输出结果是：（） int i,j,m,n; m=n=0; for(i=0;i<2;i++) for(j=0;j<2;j++) if(j>=i)m=1;n++; printf("n=%d ",n);

二维数组a有m行n列，则在a[i]&#91;j&#93;之前的元素个数为______。选择一项：[/i] A: i * n + j B: i * n + j + 1 C: i * n + j - 1 D: j * n + i

二维数组a有m行n列，则在a[i]&#91;j&#93;之前的元素个数为______。选择一项：[/i] A: i * n + j + 1 B: j * n + i C: i * n + j D: i * n + j – 1

在C语言中，如要给一个较大的[img=54x22]17de83d2fb35c85.png[/img]的数组赋值，以下方式性能较好的是：（） A: for ( int i = 0; i <; N; i++ ) for ( int j = N - 1; j >;= 0; j-- ) mat[i][j] = val; B: for ( int i = 0; i <; N; i++ ) for ( int j = 0; j <; N; j++ ) mat[j][i] = val; C: for ( int i = N - 1; i >;= 0; i-- ) for ( int j = N - 1; j >;= 0; j-- ) mat[j][i] = val; D: for ( int i = N - 1; i >;= 0; i-- ) for ( int j = N - 1; j >;= 0; j-- ) mat[j][i] = val;