vb課后習(xí)題答案李淑琴主編西安電子科技大學(xué)出版社

1、第1章1選擇題 （1）b（2）c（3）c（4）a（5）a（6）d or ab（7）d（8）a（9）c（10）d（11）a第2章1選擇題 （1）a b以數(shù)字開頭；c是vb關(guān)鍵字；d中包含了“-”（減）號 （2）b a是一個(gè)字符串變量名；c什么也不是；d是一個(gè)變量名 （3）a b是整型；c是字符型；d是雙精度型 （4）b（5）d（6）a （7）c 這是一個(gè)典型的四舍五入算法，根據(jù)這個(gè)可以自己寫出一個(gè)四舍五入函數(shù)。 （8）b（9）c（10）c（11）d（12）b（13）a、b（14）d（15）c（16）a（17）b（18）c（19）c（20）c（21）b（22）c （23）a c盡管也是3456，但

2、它是一個(gè)數(shù)值，而不是字符串，所以c是錯(cuò)誤的 （24）d這里要注意的是表達(dá)式分為兩個(gè)部分，后一部分的最小值為-1，前一部分的最大值為5，所以整個(gè)表達(dá)式的最小值就為-5，而后一部分的最大值為3，所以整個(gè)表達(dá)式的最大值為15 （25）b（26）c（27）c（28）a（29）a（30）c（31）c（32）a（33）a2判斷題 （1）錯(cuò)；（2）錯(cuò)；（3）錯(cuò)；（4）對；（5）錯(cuò)；（6）錯(cuò)；（7）錯(cuò)；（8）對；（9）對；（10）（錯(cuò)， str函數(shù)轉(zhuǎn)換正數(shù)時(shí)前面會有一個(gè)符號位表現(xiàn)為空格，正確的應(yīng)該是7。）（11）錯(cuò)。3按要求寫表達(dá)式 （1）轉(zhuǎn)換表達(dá)式a、(3*a*a+4*b*b*b)/(a-b) （這里的乘方

3、也可以用乘方號）b、6*sin(x+y*y)/(140/(3+a)（這里的3+a可以寫成分子的一部分）c、exp(x)+log(10)/sqr(x+y+1)（2）寫布爾表達(dá)式a、x>y and f<2b、x<z and y>=z or y<z and x>=z 也可寫成 (x-z)*(y-z)<=0 and x<>yc、a*b>0d、a*b=0 and a<>b（3）利用函數(shù)寫表達(dá)式a、chr(int(rnd * (asc("l") - asc("c") + 1) + asc(&qu

4、ot;c")也可以寫成chr(int(rnd*10+67)b、int(rnd*(200-100+1)+100)c、(x mod 5) *(x mod 7)=0 d、right(x,1) & left(x,1) 也可以寫成 (x mod 10)*10+x10 e、round(x,2) or format(x,”#.00”) f、mid(s,5,6) 這道題很多人容易寫成mid(“s”,5,6)，這就嚴(yán)重錯(cuò)誤了。第03章1選擇題 （1）c（2）c（3）d（4）a（5）d（6）b（7）d（8）b（9）b（10）a（11）b（12）d（13）a（14）d（15）b（16）d（17）a

5、（18）d（19）b、a（20）b（21）c（22）d（23）a（24）c（25）c（26）a（27）c（28）c（29）d（30）b（31）c（32）b（33）c（34）a（35）c（36）b（37）c （38）c （39）d2編程題(1) 雞兔同籠private sub command1_click() dim h as integer, f as integer dim x as integer, y as integer h = val(text1.text) f = val(text2.text) x = (4 * h - f) / 2 y = (f - 2 * h) / 2 lab

6、el3.caption = "計(jì)算結(jié)果：雞有" & x & "只，兔有" & y & "只"end sub（2）計(jì)算紙幣張數(shù)private sub command1_click() dim x%, y%, z%, a100%, a50%, a20%, a10%, a5%, a1% x = val(inputbox("請輸入支付金額") y = x a100 = x 100 x = x mod 100 a50 = x 50 x = x mod 50 a20 = x 20 x = x m

7、od 20 a10 = x 10 x = x mod 10 a5 = x 5 a1 = x mod 5 z = a100 + a50 + a20 + a10 + a5 + a1 print y & "元需要支付的錢幣總張數(shù)為：" & a100 & "+" & a50 & "+" & a20 & "+" _& a10 & "+" & a5 & "+" & a1 & "

8、;=" & z & "張"end sub另解，利用數(shù)組：option base 1private sub command1_click() dim x%, z%, a(), b%(5), result$ a = array(100, 50, 20, 10, 5) x = val(inputbox("請輸入支付金額") result = x & "元需要支付的錢幣總張數(shù)為：" for i = 1 to 5 b(i) = x a(i) x = x mod a(i) z = z + b(i) result

9、= result & b(i) & "+" next i result = result & x & "=" & z & "張" print resultend sub（3）電話號碼升位。private sub command1_click() dim x$ x = inputbox("請輸入一個(gè)帶區(qū)號的原電話號碼，格式為：*-*") msgbox left(x, 4) & "8" & right(x, 7)end sub第04章1

10、選擇題 （1）c（2）a（3）c（4）b（5）a（6）a（7）d（8）c（9）c（10）b（11）c（12）元答案 （13）c（14）c （15）c（16）a（17）c（18）a (19）c （20）b（21）b （22）a（23）d（24）c（25）d （26）b（27）d（28）a（29）d2判斷題（1）對；（2）對；（3）對；（4）對；（5）錯(cuò)(應(yīng)該是整型)；（6）錯(cuò)；（7）對3根據(jù)程序?qū)戇\(yùn)行結(jié)果（略）4編程題(1) private sub command1_click()dim x%x = inputbox("請輸入數(shù)值")print tab(2); format$

11、(x * x, "#.000"); tab(12); format$(sqr(x), "#.000"); tab(22); format$(x * x * x, "#.000"); tab(32); format$(x 1 / 3, "#.000")'如果format函數(shù)中小數(shù)位用"#"的時(shí)候當(dāng)不夠三位時(shí)不足的位不補(bǔ)0。'如果用round()函數(shù)，如果計(jì)算結(jié)果為整數(shù)時(shí)，則沒有小數(shù)位。end sub(2)輸入3個(gè)數(shù)，輸出最大最小數(shù)private sub command1_click

12、() dim a%, b%, c% form1.fontsize = 20 a = inputbox("請輸入第一個(gè)數(shù)") b = inputbox("請輸入第二個(gè)數(shù)") c = inputbox("請輸入第三個(gè)數(shù)") print a, b, c '先按照升序或降序排序然后再進(jìn)行取最大最小數(shù) if a > b then t = a a = b b = t end if if a > c then t = a a = c c = t end if if b > c then t = b b = c c = t

13、end if print "最小數(shù)是" a, "最大數(shù)是" cend subprivate sub command2_click() '定義變量max用于保存最大數(shù)，min保存最小數(shù) dim max as single, min as single dim a as single, b as single, c as single a = val(text1.text) b = val(text2.text) c = val(text2.text) '假設(shè)a是最大數(shù)，因此將a的值賦給max max = a 用max與b進(jìn)行比較，若b大于m

14、ax，則將b賦給max，讓max保存前兩個(gè)數(shù)的最大數(shù) if max < b then max = b end if '用max與c比較，若c大于max則將c賦給max，讓max保存三個(gè)數(shù)中的最大數(shù) if max < c then max = c end if '假設(shè)a是最小數(shù)，因此將a的值賦給min max = a '用min與b進(jìn)行比較，若b小于min，則將b賦給min，讓min保存前兩個(gè)數(shù)的最小數(shù) if min > b then min = b end if '用min與c比較，若c小于min則將c賦給min，讓min保存三個(gè)數(shù)中的最小數(shù)

15、if min < c then min = c end if '顯示結(jié)果 label2.caption = "最大數(shù)為：" & max & ",最小數(shù)為：" & minend sub（3）判斷閏年private sub command1_click() dim x%, y% x = val(inputbox("請輸入年份") if (x / 4 = x 4 and x / 100 <> x 100) or x / 400 = x 400 then print x & "

16、;是閏年" else print x & "非閏年" end ifend sub（4）判斷能否被5整除private sub command1_click() dim a%, b%, c% form1.fontsize = 20 picture1.cls x = text1.text if x mod 5 = 0 then picture1.print "您輸入的數(shù)能被5整除！" else picture1.print "您輸入的數(shù)不能被5整除！" end ifend sub（5）部門征稅private sub co

17、mmand1_click() dim tax as single, income as single income = val(text1.text) if income <= 0 then '若輸入的值小于0則提示錯(cuò)誤，并退出該過程，不進(jìn)行稅額的計(jì)算 label2.caption = "輸入有誤." exit sub end if select case income case is <= 300 tax = 0 case is <= 500 tax = (income - 300) * 0.02 case is < 5000 tax =

18、200 * 0.02 + (income - 500) * 0.03 case else tax = 200 * 0.02 + 4500 * 0.03 + (income - 5000) * 0.04 end select label2.caption = "你應(yīng)繳納的數(shù)款為：" & tax & "元"end sub（6）輸出100-200之間能被3整除的數(shù)private sub command1_click() dim x%, n% for x = 100 to 200 if x / 3 = x 3 then print tab(n m

19、od 10) * 6 + 1); x; n = n + 1 end if next xend sub（7）輸出1000到1100之間的質(zhì)數(shù)private sub form_click() dim flag as boolean for i = 1000 to 1100 m = i: flag = true for j = 2 to m - 1 if m mod j = 0 then flag = false exit for end if next j if flag = true then n = n + 1 print tab(2 + 7 * (n - 1) mod 6); m; end

20、if next iend sub（8）猜數(shù)字private sub command1_click() randomize timer '給一個(gè)隨機(jī)種子，讓每次產(chǎn)生的隨機(jī)數(shù)都不同 x = int(101 * rnd) label1.caption = "你還有8次機(jī)會" for i = 8 to 1 step -1 y = val(inputbox("請猜一個(gè)100以內(nèi)的整數(shù)", "猜數(shù)字") if y > x then label1.caption = "大了點(diǎn)，你還有" & i &

21、"次機(jī)會" elseif y < x then label1.caption = "小了點(diǎn)，你還有" & i & "次機(jī)會" else label1.caption = "恭喜你猜對了！" exit for end if next if i = 0 then label1.caption = "很遺憾，你用了8次機(jī)會也沒猜正確！" & vbcrlf & "正確值為：" & x end ifend sub（9）100到999之間回文

22、數(shù)private sub command1_click() dim n as integer '記錄第幾個(gè)回文數(shù) for x = 100 to 999 g = x mod 10 '獲取x的各位數(shù) b = x 100 '獲取x的百位數(shù) '若個(gè)位數(shù)和百位數(shù)相等則x就是回文數(shù) if g = b then picture1.print tab(6 * (n mod 8) + 2); x; n = n + 1 end if nextend sub（10）e的近似值'方法一private sub command1_click() dim e as single, f

23、 as long, n as integer e = 1 f = 1 n = 1 do while f < 10 4 f = f * n n = n + 1 e = e + 1 / f loop label2.caption = "e的近似值為：" & eend sub'方法二private sub command2_click() dim e as single, f as long, n as integer e = 1 n = 1 do while f < 10 4 f = 1 for i = 1 to n f = f * i next n

24、 = n + 1 e = e + 1 / f loop label2.caption = "e的近似值為：" & eend sub（11）1+23+33+n3<m的最大nprivate sub command1_click() m = val(text1.text) if m < 200 then label1.caption = "m的值小于等于200，重新輸入" text1.text = "" text1.setfocus else s = 0 n = 0 do while s < m n = n + 1

25、 s = s + n 3 loop label1.caption = "滿足不等式的最大n值是：" & n - 1 end ifend sub方法二'n的取值范圍1m(1/3)，按要求我們從m(1/3)（最大的數(shù)）開始找符合條件的第一個(gè)數(shù)就是了private sub command1_click() m = val(text1.text) for n = int(m (1 / 3) to 1 step -1 s = 0 for i = 1 to n s = s + i 3 next if s < m then label2.caption = &quo

26、t;最大的n為：" & n exit for end if nextend sub（12）'方法一private sub command1_click() picture1.cls '清除以前輸出的內(nèi)容 for i = 1 to 6 picture1.print spc(15 - i); '在左側(cè)打印n個(gè)空格來填充或?qū)pc換為tab for j = 1 to 2 * i picture1.print "0" next picture1.print nextend sub'方法二private sub command2_cl

27、ick() picture1.cls '清除以前輸出的內(nèi)容 for i = 1 to 6 picture1.print tab(15 - i); string(2 * i, "0") '用string函數(shù)產(chǎn)生n個(gè)0 nextend sub（13）十進(jìn)制轉(zhuǎn)二進(jìn)制 轉(zhuǎn)十六進(jìn)制private sub command1_click() n = val(inputbox("輸入要轉(zhuǎn)換的十進(jìn)制整數(shù)") m = n x = "" do while n <> 0 a = n mod 2 n = n 2 x = a &

28、; x loop msgbox m & " 轉(zhuǎn)換二進(jìn)制數(shù)是：" & xend subprivate sub command2_click() n = val(inputbox("輸入要轉(zhuǎn)換的十進(jìn)制整數(shù)") m = n x = "" do while n <> 0 a = n mod 16 select case a case 10 b = "a" case 11 b = "b" case 12 b = "c" case 13 b = "d

29、" case 14 b = "e" case 15 b = "f" case else b = a end select n = n 16 x = b & x loop msgbox m & " 轉(zhuǎn)換十六進(jìn)制數(shù)是：" & xend sub（14）解燈謎private sub command1_click() dim a%, b%, c%, d% dim x%, y%, z% for a = 0 to 9 for b = 0 to 9 for c = 0 to 9 for d = 0 to 9 x =

30、a * 1000 + b * 100 + c * 10 + d y = c * 100 + d * 10 + c z = a * 100 + b * 10 + c if x - y = z then print "a=" a; "b=" b; "c=" c; "d=" d end if next d, c, b, aend sub方法二private sub command2_click() '由題意可知abcd的取值范圍是10009999之間，因此通過循環(huán)來找出符合條件的abcd for abcd = 1

31、000 to 9999 abc = abcd 10 '獲取abc的值 c = abc mod 10 d = abcd mod 10 '分別獲取abcd的十位數(shù)c和個(gè)位數(shù)d cdc = c * 100 + d * 10 + c '計(jì)算cdc的值'如果符合abcd-cdc=abc則輸出 if abcd - cdc = abc then label1.caption = "abcd的值為：" & abcd exit for end if nextend sub（15）猴子吃桃private sub command1_click() '

32、;驗(yàn)證 m = 1534 s = m for i = 1 to 10 print s s = s - (s / 2) - 1 next i if i = 11 and s = 1 then print send subprivate sub command2_click() '窮舉發(fā) for m = 1000000 to 1 step -1 s = m for i = 1 to 9 s = s - (s / 2) - 1 next i if i = 10 and s = 1 then print m next mend subprivate sub command3_click() &

33、#39;思路：假設(shè)第n天的桃子為x，前一天（第n-1天）的桃子為y '則x=y-(y/2+1)=>x=y/2-1;y=2*(x+1) '現(xiàn)在第十天的桃子是已知的，我們反推到第一天即可 'x = 1' for i = 9 to 1 step -1' y = 2 * (x + 1)' x = y' next' print y dim x as integer, i as integer x = 1 for i = 10 to 2 step -1 x = (x + 1) * 2 next i print xend sub第05章1

34、選擇題 （1）b（2）a（3）b（4）c（5）b（6）d（7）a（8）b（9）b（10）b（11）a（12）c（13）c（14）d（15）b （16）c （17）b （18）c2編程題（1）private sub command1_click() dim c(0 to 7) a = array(1, 3, 5, 2, 4, 18, 50, 25) b = array(5, 27, 30, 35, 60, 41, 87, 33) for i = 0 to 7 c(i) = a(i) + b(i) next print "a()", "b()", "

35、;c()" for i = 0 to 7 print a(i), b(i), c(i) nextend sub（2）矩陣找最大元素的行列private sub command1_click() '先聲明一個(gè)動態(tài)數(shù)組a，因?yàn)槠浯笮〔荒艽_定 dim a() as integer n = val(text1.text) m = val(text2.text) '指定數(shù)組的大小 redim a(1 to n, 1 to m) as integer picture1.cls '用隨機(jī)數(shù)給數(shù)組賦值，并打印到窗體 for i = 1 to n for j = 1 to m

36、a(i, j) = int(rnd * 901) picture1.print tab(5 * (j - 1) + 2); a(i, j); next next '假設(shè)第一個(gè)元素的值最大,用r和c分別存放最大數(shù)所在的行和列 r = 1: c = 1 for i = 1 to n for j = 1 to m if a(r, c) < a(i, j) then r = i c = j end if next next label3.caption = "矩陣中的最大值為：" & a(r, c) & vbcrlf label3.caption =

37、label3.caption & "位置：" & r & "行," & c & "列"end sub（3）交換數(shù)組元素值private sub command1_click() dim a$(), x, i%, n% a = split(inputbox("請輸入n個(gè)數(shù)據(jù)，數(shù)據(jù)之間用逗號分隔!"), ",") '注意這里的inpubox函數(shù)的用法，輸出數(shù)據(jù)時(shí)，數(shù)據(jù)之間應(yīng)該用英文狀態(tài)的逗號分隔 print "對換前數(shù)據(jù)序列為："

38、for each x in a print x; " " next x print n = ubound(a) for i = 0 to n 2 x = a(i): a(i) = a(n - i): a(n - i) = x next i print "對換后數(shù)據(jù)序列為：" for each x in a print x; " " next xend sub（4）private sub command1_click() dim x$(), a%(4 to 9), b(), i%, n%, y, k% b = array("無

39、效數(shù)據(jù)", "小于60分", "6069", "7079", "8089", "90100") x = split(text1, ",") n = ubound(x) for each y in x if y < 0 or y > 100 then y = 40 elseif y < 60 then y = 50 elseif y = 100 then y = 90 end if k = y 10 a(k) = a(k) + 1 next y fo

40、r i = 4 to 9 picture1.print b(i - 4), a(i) next iend sub（5）根據(jù)身份證前17位算第18位option base 0private sub command1_click() dim a%(16), b$(), w(), i%, x$, sum% b = split("1,0,x,9,8,7,6,5,4,3,2", ",") w = array(7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2) x = inputbox("請輸入您的身份

41、證前17位") for i = 0 to 16 a(i) = mid(x, i + 1, 1) sum = sum + a(i) * w(i) next i print "您的身份證號證為：" & x & b(sum mod 11)end sub（6）輸出斐波拉契數(shù)列option base 1 private sub command1_click() dim a&(30), i% a(1) = 1: a(2) = 1 for i = 3 to 30 a(i) = a(i - 1) + a(i - 2) next i for i = 1 to

42、 30 print tab(i - 1) mod 5) * 10 + 1); a(i); next iend sub（7）隨進(jìn)產(chǎn)生15個(gè)不同大寫字母option base 1private sub command1_click() dim a$(15), i%, j% for i = 1 to 15 a(i) = chr(int(rnd * 26 + 65) for j = 1 to i - 1 if a(j) = a(i) then i = i - 1 exit for end if next j next i for i = 1 to 15 print a(i); " "

43、; next iend sub（8）將一個(gè)數(shù)插入數(shù)組option base 1dim a() '聲明窗體及變量是為了能夠多次插入數(shù)據(jù)sub parray(x() '定義打印數(shù)組的子程序 dim i%, y for each y in x print y; next y printend subprivate sub command1_click() dim i%, k%, n%, x n = ubound(a) print "插入前：": call parray(a) do '保證插入位置在有效范圍內(nèi) k = inputbox("請輸入待插

44、入的位置(" & 1 & "" & n + 1 & ")") loop until k >= 1 and k <= n + 1 redim preserve a(n + 1) '擴(kuò)大數(shù)組 x = inputbox("請輸入待插入的數(shù)據(jù)") for i = n to k step -1 a(i + 1) = a(i) next i a(k) = x print "插入后：": call parray(a)end subprivate sub form_lo

45、ad() a = array(1, 2, 3, 4, 5, 6, 7, 8, 9) '給數(shù)組賦初值end sub（9）option base 1dim a() '聲明窗體及變量是為了能夠多次插入數(shù)據(jù)sub parray(x() '定義打印數(shù)組的子程序 dim i%, y for each y in x print y; next y printend subprivate sub command1_click() dim x%, i%, n%, k%, f as boolean n = ubound(a) print "刪除前：": call par

46、ray(a) x = inputbox("請輸入待刪除的數(shù)") for i = 1 to n if a(i) = x then f = true for k = i to n - 1 a(k) = a(k + 1) next k end if if f = true then redim preserve a(n - 1) exit for end if next i if f = true then print "刪除后：": call parray(a) else print "查無此數(shù)！" end ifend subprivat

47、e sub form_load() a = array(1, 2, 3, 4, 5, 6, 7, 8, 9) '給數(shù)組賦初值end sub（10）求矩陣四周元素之和option base 1sub parray(x%() '定義打印數(shù)組的子程序 dim i%, j% for i = 1 to ubound(x, 1) for j = 1 to ubound(x, 2) print tab(j * 5); x(i, j); next j next i printend subprivate sub command1_click() dim i%, j%, m%, n%, s%,

48、a%() m = inputbox("m=") n = inputbox("n=") redim a(m, n) for i = 1 to m for j = 1 to n a(i, j) = int(rnd() * 10) next j next i call parray(a) for i = 1 to m s = s + a(i, 1) + a(i, n) next i for i = 2 to n - 1 s = s + a(1, i) + a(m, i) next i print "四周元素之和為：" & send

49、sub（11）運(yùn)動會private type cj bh as string * 3 cj as singleend typeprivate sub command1_click() dim a(0 to 9) as cj, x$(), y$(), t as cj dim i%, j%, k% x = split("011,095,041,070,008,009,021,061,006,004", ",") y = split("12.4,11.1,13.4,12.1,12.4,10.4,14.4,15.1,15.4,11.4", &

50、quot;,") for i = 0 to 9 a(i).bh = x(i) a(i).cj = y(i) next i print "編號", "成績" print "=" for i = 0 to 9 print a(i).bh, a(i).cj next i for i = 0 to 8 k = i for j = i + 1 to 9 if a(k).cj > a(j).cj then k = j next j t = a(k): a(k) = a(i): a(i) = t next i print "

51、;=" for i = 0 to 9 print a(i).bh, a(i).cj next iend sub補(bǔ)充一：打印魔方陣option base 1private sub command1_click() dim a%(), i%, j%, k%, n%, x%, y%, p%, q% do n = inputbox("n=") loop while n 2 = n / 2 redim a(n, n) x = 1: y = n 2 + 1 a(x, y) = 1 '先放1 for i = 2 to n * n '再放其它數(shù) p = x: q = y '保留前一個(gè)數(shù)的位置 x = x - 1 '找下一個(gè)數(shù)的行位置 if x < 1 then x = n '如果行小于1,讓行為n y = y + 1 '找下一個(gè)數(shù)的列位置 if y > n then y =