VB上机考试最全调试题及答案

1、第二部分 浙江省vb二级试题解a程序调试题一、选择结构程序调试01option explicitpublic sub getanswer() 该过程是用于计算一元二次方程的根，。 dim dalt!, a#, b#, c# a = inputbox(输入系数a) b = inputbox(输入系数b) c = inputbox(输入系数c) dalt = b * b - 4 * a * c if -1- then dalt = sqr(dalt) msgbox format(-b + dalt) / 2 / a), 0.00 + chr(13) + chr(10) + format(-b -

2、dalt) / 2 / a, 0.00) elseif -2- then msgbox format(-b / 2 / a, 0.00) + chr(13) + chr(10) + format(-b / 2 / a, 0.00) else dalt= -3- msgbox format(-b / 2 / a, 0.00) + +i + format(dalt / 2 / a, 0.00) + chr(13) + chr(10) + format(-b / 2 / a, 0.00) + -i + format(dalt / 2 / a, 0.00) end ifend sub二、单重循环程序调

3、试02option explicit 这条语句要求在该模块中，使用变量前必须先声明。public sub uppersen() 该过程是将一个英文句子的每个单词的首字母都变成大写。 从键盘上任意输入一条英文句子， 将句子中的每个单词的首字母都变成大写 例如：输入i am a good student. 要求输出i am a good student. dim oldsen as string, newsen as string dim char as string, lastchar as string dim k as integer, i as integer oldsen = input

4、box(请输入英文句子：) k = -1- 以空格作为单词的界定，空格后的字母转换为大写字母 lastchar = -2- for i = 1 to k char = -3- if lastchar = then char = ucase(char) end if newsen = newsen & char lastchar = char next i form1.print input:; oldsen form1.print output:; newsenend sub程序调试03public sub jisuan() 该过程用于计算1-(1/2)+(1/3)-.+(1/99)-(1/1

5、00)的值并打印出来。 dim i as integer * 错误1 * dim k as integer dim s as single s = 0 * 错误2 * for i = 1 to 100 k = 1 / i s = s + k * 错误3 * k = 1 / i + 1 s = s + k next i form1.print s=; send sub程序调试04option explicitpublic sub same() 该过程找出1-1000之间所有的同构数 所谓同构数是指一个数出现在它的平方数的右端 如25在25平方625的右端，则25为同构数 利用数字转字符再取出右端

6、字符的方法进行判断 dim i as long dim x1 as string, x2 as string for i = 1 to 1000 x1 = -1- 将i转字符型 x2 = -2- 将i2转字符型 if -3- then form1.print i; 是同构数 end if next iend sub程序调试05public sub armstrongnumber() 该过程是用于求出1-999之间所有的armstrong数，并输出。 所谓armstrong数是指一个数等于它每位上数字的立方和。例如：371=33+73+13，那么371就是一个armstrong数。 dim ar

7、mstrong as integer dim i as integer dim hundred as integer 百位上的数字 dim ten as integer 十位上的数字 dim one as integer 个位上的数字 for i = 1 to 999 hundred = -1- ten = int(i mod 100) 10) one = -2- if -3- then form1.print i; is armstrong number end if next iend sub程序调试06public sub sum() 该过程是用于计算并输出f=1-1/(2*3)+1/(

8、3*4)-1/(4*5)+1/(19*20)。 dim f as single dim i as integer dim sign as integer -1- f = 1 -2- f = f + sign / (i * (i + 1) -3- next i form1.print f=; fend sub程序调试07option explicitpublic sub eat() 该过程是用于计算猴子共摘了多少个桃子。 （小猴在一天摘了若干个桃子，当天吃掉了一半多一个； 第二天吃了剩下的一半多一个； 以后每天都吃尚存的一半零一个，到第7天早上要吃时只剩下一个了）， 并将结果输出来。 dim n

9、%, i%, x% x = 1 for i=6 to -1- step -2- -3- next i form1.print 共有; x; 个桃子end sub程序调试08public sub fabonia() 该过程是对以下数列进行运算： 有一个数列，它的前三个数是0，1，2， 从第四个数起，每个数都是它前面的两个数之和 求出该数列的第17个数是多少 求出该数列的第几个数起每个数都超过1e+8 dim last_one as long dim last_two as long dim this_one as long dim i as integer last_one = 1 数列的第二个

10、数 last_two = 2 数列的第三个数 i = 4 从数列的第四个数求起 do this_one = last_one + last_two last_one = -1- last_two = -2- if i = 17 then form1.print no:17=; this_one end if -3- loop while this_one 1e+8end sub程序调试09public sub findat() 该过程是用于在一个字符串变量中查找”at”，并用消息框给出查找结果的报告：没有找到或找到的个数。 dim str1 as string dim length as in

11、teger 字符串长度 dim sum as integer 查到的个数 dim i as integer str1 = inputbox(请输入一个字符串) length = -1- i = 1 sum = 0 do while i -2- if -3- = at then sum = sum + 1 end if i = i + 1 loop if -4- then msgbox 没有找到！ else msgbox 找到了 & str(sum) & 个 end ifend sub程序调试10注：此题题目和第41题相同，但算法不一样。option explicitpublic sub com

12、div() 该过程是求任意两个正整数的最大公约数 dim m as integer, n as integer 任意两个正整数 dim i as integer dim div as integer 最大公约数 输入两个正整数，要求m与n都必须大于零 do m = val(inputbox(m=) n = val(inputbox(n=) loop while -1- 先将两个整数中的较小数假设为最大公约数，再依次往下 寻找能同时除尽m和n的数即为最大公约数 -2- if n m then div = n end if do while m mod div 0 or n mod div 0 -

13、3- loop form1.print m; 和; n; 的最大公约数是：; divend sub程序调试11public sub wrap()该过程用于判断一个字符串是否“回文”。所谓“回文”是指字符串顺读与倒读都是一样的，如“潮起潮落，落潮起潮”。 dim length as integer dim str1 as string dim strleft as string dim strright as string dim k as integer str1 = inputbox(请输入任意的字符串) 输入任意字符串 *错误1* length = val(str1) k = 1 do *

14、错误2* strleft = left(str1, k) 从左边起逐个取出一个字符 *错误3* strright = right(str1, k) 从右边起逐个取出一个字符 *错误4* if strleft = strright then exit do end if k = k + 1 loop while k length / 2 then form1.print str1 & 是回文 else form1.print str1 & 不是回文 end ifend sub程序调试12option explicitpublic sub dtob() 该过程是将一个十进制正整数转换成为一个二进制

15、数。 采用连除2取余数的方法，将一个十进制数转换为二进制数。 dim dec as integer dim bin as string dim res as integer dim i as integer dec = val(inputbox(x=) 输入一个十进制数 form1.print 十进制数：; dec do res = -1- 求出除以2的余数 bin = trim(str(res) & -2- dec = -3- loop while -4- form1.print 转换为二进制数为：; binend sub程序调试13public sub guess() 该过程是猜数游戏，由

16、计算机产生一个1，100的任意整数， 输入猜数后计算机给出提示，如果5次后还没有猜中就结束游戏并公布正确答案。 dim r as integer dim x as integer dim times as integer randomize r = int(rnd * 100) + 1 产生一个在区间1,100 的任意整数 times = 1 do x = val(inputbox(输入猜数x) select case x case r form1.print 猜中了 exit do * 错误1 * case is 5 if times 5 then form1.print 猜数失败,游戏结束

17、! * 错误3 * form1.print 正确答案为 & str(x) end ifend sub程序调试14option explicitpublic sub summary()该过程是用于计算s=1+2+22+23+，直至s超过1e+16 dim s as single dim i as integer -1- i = 1 -2- s = s + 2 i -3- loop form1.print s=; send sub程序调试15option explicitpublic sub find() 该过程是用于找出被3、5、7除，余数为1的最小的5个整数，并将结果输出来。 dim coun

18、tn%, n% countn = 0 n = 1 do n=-1- if -2- then form1.print n countn = countn + 1 end if loop -3-end sub程序调试16option explicitpublic sub e() 该过程是用于计算e的值并将结果输出，要求精确到0.000000000000001， e的计算公式为：e=1+1/1!+1/2!+1/n!。 dim n, term, t n = 0: term = 1: t = 1 do n = n + 1 t= -1- term= -2- loop while t 0.000000000

19、000001 form1.print e=& -3-end sub程序调试17option explicitpublic sub qiun() 该过程是用于求出满足不等式1+2x+3x2+4x3+(n+1)xn1000的最大n值， 其中x是大于等于1的实数，其值由键盘输入。 dim x as single, s as single, n as integer, s1 as single, p as single x = val(inputbox(x=) s = 1: n = 1 p = x while -1- s1 = s s = s + (n + 1) * p p=p* -2- n= -3-

20、 wend n= -4- form1.print the maxism of n; n, s=; s1end sub三、多重循环程序调试18public sub prt() 该过程是输出由数字组成的如下所示金字塔图案 9 888 77777 6666666 555555555 44444444444 3333333333333 222222222222222 11111111111111111 dim i as integer, j as integer * 错误1 * for i = 9 to 1 form1.print space(i); for j = 1 to 19 - 2 * i *

21、错误2 * form1.print i next j form1.print next i *错误3 *end function程序调试19option explicitpublic sub pyramid() 该过程用于打印数字金字塔 1 222 33333 4444444 555555555 6666666 77777 888 9 dim i as integer dim j as integer dim start as integer 每行起始空格数 dim num as integer 每行数字个数 for i = 1 to 9 if i = -1- then start = 20

22、- i num = 2 * i - 1 else start = 10 + i num = 19 - 2 * i end if for j = 1 to -2- form1.print tab(start+j);-3- next j form1.print next iend sub程序调试20public sub prime() 该过程是输出100到200之间所有的素数，并统计素数的个数 dim i as integer dim j as integer dim k as integer dim t as integer 统计素数的个数 dim b as boolean for i = 10

23、0 to 200 b = true k= - 1 - j = int(sqr(i) do while k ave then -3- form1.print x(i); space(2); if -4- then form1.print end if next i form1.print form1.print 共有; count; 个在平均值之上.end sub程序调试22option explicitpublic sub del() 该过程从一个含有10个数组元素的数组中删除一个指定的数据，若该数据不存在，则给出提示。 先产生10个1，100的随机整数 从键盘上输入一个数，将该数从数组中删除

24、 randomize dim x as integer dim a() as integer dim i as integer, pos as integer redim a(10) as integer for i = 1 to 10 a(i) = int(rnd * 100) + 1 next i form1.print 原始数据： for i = 1 to 10 form1.print a(i); ; next i form1.print x = val(inputbox(输入删除的数x=) pos = 0 for i = 1 to 10 if x = a(i) then -1- exi

25、t for end if next i if pos 0 then for i = -2- to 9 -3- next i redim preserve a(9) 保留原数组中的9个数 form1.print 删除后数据： for i = 1 to 9 form1.print a(i); ; next i form1.print else msgbox 数组中未找到 & str(x) & ,删除不成功 end ifend sub程序调试23option explicitpublic sub counteven() 该过程是用于统计在随机产生的10个两位整数中的偶数的个数并用消息框输出。 dim

26、 a(10) as integer 存放10个随机数第一个为a(0) dim count as integer dim i as integer, n as integer randomize *错误1* for i = 1 to 10 *错误2* n = int(rnd * 90) a(i) = n form1.print a(i) if a(i) mod 2 = 0 then count = count + 1 end if next i *错误3* msgbox(偶数个数：,count)end sub程序调试24option explicitpublic sub sortchar() 该

27、过程是对字符串进行整理， 首先从键盘上输入一个任意的字符串，将该字符串的所有组成字符拆分开， 再按照字符ascii码从小到大的顺序将这些字符重新组成新的字符串。 例如输入a4fkze5，重新组合的字符串为45aefkz。 dim x as string 原始字符串 dim y as string 重新组合的字符串 dim c() as string 拆分出的字符 dim k as integer 字符串长度 dim i as integer, j as integer dim temp as string x = inputbox(输入一个字符串) k = -1- redim c(k) as

28、string 字符串拆分 for i = 1 to k c(i) = -2- next i 字符排序 for i = 1 to k - 1 for j = -3- if c(i) c(j) then temp = c(i) c(i) = c(j) c(j) = temp end if next j next i 排序后的字符组成新字符串 y = for i = 1 to k y = y & c(i) next i form1.print 原始字符串; x form1.print 重新组合的字符串; yend sub程序调试25option explicitpublic sub composit

29、or() 该过程是用于产生20个不大于1000的自然数， 并使用选择分类法按从大到小的顺序输出，相同数不输出。 const n = 20 dim a(n), i, p, j, t for i = 1 to n a(i) = -1- next i for i = 1 to n - 1 p = i for j = i to n if a(p) a(j) then -2- next j if -3- then t = a(i): a(i) = a(p): a(p) = t next i form1.print a(1) for i = 2 to n if a(i) a(i - 1) then for

30、m1.print -4- next i form1.printend sub程序调试26option explicitpublic sub prt() 该过程是输出由数字组成的如下所示杨辉三角形图案 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 const n = 8 dim a(n + 1) as byte, i as integer, j as integer dim s as string 给数组a赋初值 for i = 1 to n + 1 a(i) = 0 next i 计算并打印第一个数 a(1) = 1

31、form1.print space(2 * n + 2); 计算数字前的空格数 form1.print a(1) 计算其余行的数并打印 for i = 2 to -1- a(i) = a(i - 1) for j = -2- to 2 step -1 -3- next j form1.print space(n - i + 1) * 2); for j = 1 to i s = trim(str(-4-) form1.print space(int(4 - len(s) / 2 + 0.5) + s + space(int(4 - len(s) / 2 + 0.5); next j form1

32、.print next iend sub五、多维数组程序调试27public sub transpose() 该过程用于产生一个66的转置矩阵，将二维数组中所有行和对应列的元素进行交换。 dim a(1 to 6, 1 to 6) as integer dim i as integer, j as integer form1.print 原始数据 for i = 1 to 6 for j = 1 to 6 a(i, j) = int(rnd * 10) form1.print a(i, j); next j form1.print next i for i = 2 to 6 for j = 1

33、 to - 1 - - 2 - next j next i form1.print 转置后数据 for i = 1 to 6 for j = 1 to 6 - 3 - next j form1.print next iend subpublic sub swap(a as integer, b as integer) dim temp as integer temp = a a = b b = tempend sub程序调试28public sub printarray() 下面程序用于显示5行5列的数字方阵 使两对角线上元素均为1，其余均为2 要求显示的数字方阵两列数字之间空3格，两行之间空

34、一行 只要修改标出出错位置的下面那一条语句 dim a(5, 5) as integer dim i as integer, j as integer for i = 1 to 5 for j = 1 to 5 * 错误1 * if i j or i 6 - j then a(i, j) = 2 else a(i, j) = 1 end if * 错误2 * form1.print a(i, j) + space(3); next j * 错误3 * form1.print: form1.cls next iend sub程序调试029printarray过程是输出一个右上三角元素（含对角线）

35、为1，其余元素为0 的55矩阵public sub printarray()dim a(1 to 5, 1 to 5) as integerdim i as integer, j as integerfor i = 1 to 5 for j = 1 to 5 if - 1 - then - 2 - end if next jnext ifor i = 1 to 5 for j = 1 to 5 form1.print ; a(i, j); next j - 3 -next iend sub六、sub过程程序调试30private const n = 10private a(1 to n) as

36、 integertj过程是将一批数据中小于零的数及它们的积打印出来数据由scsj过程产生，数据的取值范围为-10 10 之间的随机整数数public sub tj() dim i as integer dim t as single t= - 1 - for i = 1 to 10 if - 2 - then t = t * a(i) end if next i form1.print t=; tend subpublic sub scsj() randomize form1.print 原始数据 dim i as integer dim j as integer for i = 1 to n

37、 随机产生0或1，为0时取负，为1时取正 j = int(rnd * 2) if - 3 - then j = -1 a(i) = j * int(rnd * (n + 1) form1.print a(i); next i form1.printend sub程序调试31dim a(30) as integerpublic sub findmax() 该过程是用于查找一批数据中的最大数，并输出最大数的值及最大数在数组中的下标值。 dim position as integer 最大数在数组中的下标 dim max as integer 最大数 dim i as integer generat

38、edata max = -1- position = -2- for i = 2 to 30 if a(i) max then max = -3- position = -4- end if next i form1.print 最大值= & str(max) form1.print 最大值的下标= & str(position)end subpublic sub generatedata() dim i as integer for i = 1 to 30 a(i) = int(500 * rnd() + 1) form1.print a(i), if i mod 5 = 0 then form1.print next iend sub程序调试32dim a(1 to 5, 1 to 4) as integerdim ave(1 to 5) as integerpublic sub maxline() 该过程用于查找一个5行4列的二维数组中行平均值最大的行，并将该行所有数据调整到第一行的位置。 dim i as integer, j as intege