国开山东00035《VisualBasic程序设计》期末综合练习程序阅读分析题复习资料答案Word文件下载.docx

1、A. 12 B. 20 C. 60 D. s=60 s=60static x as integerclsy = 1y = y + 5x = 5 + xprint x, y程序运行后，单击命令按钮command1三次后，窗体上显示的结果为（ ）。A. 15 16 B. 15 6 C. 15 15 D. 5 6 15 6n = text1.textselect case ncase 1 to 20x = 10case 2, 4, 6x = 20case is 10x = 30case 10x = 40end selecttext2.text = x程序运行后，如果在文本框text1中输入10，然后

2、单击命令按钮，则在text2中显示的内容是（ ）。A. 10 C. 30 D. 40 10static y as integerclsfor i = 0 to 2x = x + yy = y + 3next iprint x, y程序运行后，连续二次单击command1按钮后，窗体上显示的值是（ ）。A. 9 9 B. 30 12 C. 36 18 D. 63 27 36 18dim s as doubledim i as integers = 5i = 1do while i 6i = i + 2s = s + ilooptext1.text = s程序运行后，单击命令按钮，在文件框text

3、1中的输出结果是（ ）。A. 13 C. 11 D. 18 20x = 0do while x 60x = （x + 2） * （x + 3）n = n + 1text1.text = str（n）text2.text = str（x）text3.text = val（text1.text） + val（text2.text）程序运行后，单击命令按钮在文件框text3中显示的结果是（ ）。A. 72 B. 74 C. 7 D. 16 74static sum as integerdosum = sum + ii = i + 1loop while i 2j = j + n * （n + 1）p

4、rint n; j程序运行后，单击命令按钮，在窗体上显示的值是（ ）。A. 0 1 B. 3 7 C. 3 21 D. 3 13 3 7p = 0s = 0p = p + 2s = s + ploop while p 3 thenc = c + 5exit forelsek = k + 1end ifnext jprint c, k程序运行后，单击窗体，在窗体上显示的c,k的值分别是（ ）。A. 7 3 B. 7 4 C. 2 4 D. 7 5 7 4dim i as integer, j as integerdim c as integerc = 0for i = 1 to 3for j =

5、 1 to 2c = c + 3next jprint c程序运行后，单击窗体，在窗体上显示的结果是（ ）。A. 18 B. 15 C. 12 18dim i as integer, k as integer, c as integerfor i = 1 to 5if i mod 2 = 0 thenk = k + 2elseend ifprint k, c程序运行后单击窗体，form1上的输出结果是（ ）。A. 2 12 B. 4 9 C. 4 6 D. 2 9 4 9dim i as integer, n as integerfor i = 0 to 50 i = i + 3if i 10

6、 then exit for程序运行后，单击命令按钮，在文本框中显示的值是（ ）。A. 2 B. 3 C. 4 D. 5 3for i = 1 to 10k = k + 3c = c + 4A. 10 20 B. 20 15 C. 15 20 D. 10 30 15 20c = c + 6exit forA. 10 8 B. 8 10 C. 8 7 D. 8 4 8 10dim i as integer, k as integerstatic c as integerfor i = 1 to 6程序运行后，连续两次单击窗体，在窗体上显示的值是（ ）。A. 6 12 B. 6 24 C. 6 9

7、 D. 6 18 6 18b = 1do while b 40b = b * （b + 1）print b程序运行后，单击窗体，在窗体上显示的值为（ ）。A. 42 B. 39 C. 6 D. 1 42dim a（1 to 3） as stringdim j as integera（1） = 3a（2） = 6a（3） = 9c = 1for j = 1 to 3c = c + val（a（j）B. 19 C. 16 D. 17 19dim a（1 to 5） as integerdim x as integera（i） = ii = 1a（i） = a（i） + xx = x + 2i =

8、i + 2text1.text = a（3）text2.text = a（5）程序运行后，单击命令按钮，在两个文本框中显示的值是（ ）。A. 5 9 B. 5 4 C. 9 5 D. 4 5 5 9a（i） = ia（i） = a（i） * xx = x + 1text1.text = a（1）text2.text = a（2）程序运行后，单击command1后按钮，文件框text1、text2中的结果是（ ）。A. 2 0 B. 0 2 D. 2 5 0 2dim m（5）m（i） = i * iprint m（i - 1）A. 5 B. 25 C. 05 25sub change（x as

9、 integer, y as integer）dim t as integert = xx = yy = tdim a as integer, b as integera = 30: b = 40change a, bprint a, bA. 30 4030 40 B. 30 4040 30 C. 40 30D. 40 3040 30sub test（i as integer）i = i + 4x = 5 xcall test（x）程序运行后,单击窗体，在窗体上显示的值是（ ）。A. x= 5 x= 9 B. x= 9 x= 5 C. x= 5x= 9 D. 9 5 x= 5x= 9funct

10、ion f（a as integer）dim b as integerb = b + 2c = c + 3f = a + b + cend functiondim a as integera = 4print f（a）;A. 9 9 9 B. 9 12 15 C. 9 14 19 D. 9 11 13 9 12 15dim x, y as integerx = 5: y = 10nc x, yprivate sub nc（byval a%, b%）a = a + bb = aa = a * bA. 5 15 B. 15 15 C. 225 15 D. 5 10 5 15sub abcd（byval n as integer）n = n + 5nx% = 3call abcd（nx%）print nx%C. 9 D. 10 dim y as integerx = 1y = 2call aa（y + 1）z = bb（x）print x, y, zpublic sub aa（byval z as integer）y = z + xz = x + ypublic function bb（byref y as integer）y = y + 1bb = x + yA. 2 2 2 B. 2 3 2 C. 3 2 3 D. 2 3 3 2 3 2