1、鄭州大學(xué)vb實驗報告書 參考 答案 以下內(nèi)容均來自百度文庫，所以僅供參考。不過經(jīng)親自測試 習(xí)題及實驗（一）的答案基本正確，除了實驗題的3，4小題沒有改字形字體顏色什么的。 實驗一一、選擇題ddbab,c(cd)ddc,addca二、填空題windows中斷工具箱代碼分類form2.show有語法錯誤錯誤事件結(jié)構(gòu)化文件名和結(jié)構(gòu)工程編寫代碼雙擊任意控件或右健快捷菜單選“代碼窗口”vb的安裝文件夾屬性方法fontload對象三、編程1-1private sub command1_click()print sqr(8), 8 (1 / 3)end sub1-2private sub command2_

2、click()for i = 100 to 110 print sqr(i)next iend sub1-3（基本設(shè)置在屬性窗口完成?。﹑rivate sub command3_click()print 鄭州大學(xué)在前進!end sub1-4（基本設(shè)置在屬性窗口完成?。﹑rivate sub command4_click()label1 = 藍天、白云、綠水、青山end sub實驗二一bcbcc,cbbda,cccba,cbdcc二form_click()interval,5000vbp,frmtop,left,height,width,像素pictureloadlockedfont,align

3、mentgotfocus,lostfocus屬性窗口，代碼三2-1private sub command1_click()label3 = text1 * text2end subprivate sub command2_click()label3 = text1 / text2end sub2-2private sub command1_click()picture1.visible = truepicture1 = loadpicture(c:documentsand settingsall usersdocumentsmy pictures示例圖片water lilies.jpg)end

4、 subprivate sub command2_click()picture1.visible = truepicture1 = loadpicture(c:documentsand settingsall usersdocumentsmy pictures示例圖片winter.jpg)end subprivate sub command3_click()picture1.visible = falseend sub2-3private sub command1_click()label3 = text1 * text2end sub2-4private sub command1_click

5、()text2 = text1end subprivate sub command2_click()text1 = text2 = end subprivate sub text1_change()command1.visible = truecommand2.visible = trueend sub2-5private sub command1_click()timer1.interval = 500end subprivate sub command2_click()timer1.interval = 0end subprivate sub timer1_timer()image1.le

6、ft = int(rnd * (form1.scalewidth -image1.width)image1.top = int(rnd * (form1.scaleheight -image1.height)end sub實驗三一a(bc)(bc)aa,a(bcd)dbb,b二3，4，5，6，8，10，11，12，14，15，16三abs(x+y)(3+x*y)2(-b+sqr(b2-4*a*c)/(2*a)sin(30*3.14/180)+exp(2)四int(rnd*71+20)x*y=10 and x=a and x=zleft(s,3) 或者,mid(s,1,3)五3-2private

7、 sub command1_click()dim x as singlex = val(text1)print format(x 2, #.000),format(sqr(x), #.000), format(x 3, #.000), format(x (1 / 3), #.000)end sub3-3private sub command1_click()dim x as integerx = int(rnd * 900 + 100)text1 = trim(str(x)text2 = right(text1, 1) + mid(text1, 2, 1)+ left(text1, 1)end

8、 sub3-4private sub command1_click()dim max, min as integermin = val(text1)max = val(text2)randomizeprint int(rnd * (max - min + 1) + min),int(rnd * (max - min + 1) + min), int(rnd * (max - min + 1) + min)end sub3-5private sub command1_click()text1 = datetext2 = timetext3 = weekdayname(weekday(date)e

9、nd sub實驗四一cdcda,dbbac二(1) 48.71e-01,-4.8712(2) 關(guān)系，邏輯，數(shù)值(3) 900，100，mid(a,2,1),c(4)m1,2,3print m & 月是第二季度end select三(1) x=4 y=13(2) 10四4-1private sub command1_click()dim num, a1, a2, a3 as singlenum = val(text1) * 180 / 3.14a1 = int(num)num = num - a1a2 = int(num * 60)num = num * 60 - a2a3 = format(n

10、um * 60, #.000)print a1, a2, a3end sub4-2private sub command1_click()dim y as integery = val(text1)if (y mod 400 = 0 or y mod 4 = 0 and y mod100 0) then msgbox y & 是閏年else msgbox y & 不是閏年end ifend sub4-3private sub command1_click()dim x, y, z, t as integerx = val(inputbox(input x:)y = val(inputbox(i

11、nput y:)z = val(inputbox(input z:)if x y then t = x: x = y: y = tif y z thent= y: y = z: z = t if x y then t = x: x = y: y = tend ifprint x, y, zend sub4-4private sub command1_click()dim nettime as integer, money as singlenettime = val(inputbox(每月上網(wǎng)時數(shù)：, 上網(wǎng)費用)if nettime = -1 then money = 0elseif nett

12、ime 10 then money = 25elseif nettime 50 then money = nettime * 2elseif nettime 100 then money = nettime * 1.5elseif nettime 200 then money = nettimeelse money = 200end ifif money 0 then msgbox 該月上網(wǎng)費用為 & money &元實驗五一bcbac二（1）for x=1 to 10s*(1-1/x)(2)0t1e-6(3)jt(4)511 5三5-1private sub command1_click()

13、dim i, sum as integersum = 0for i = 1 to 100sum = sum + inext iprint 1+2+.+100=; sumend sub5-2private sub command1_click()dim x, y as integerfor x = 0 to 20 for y = 0 to 33 if 5 * x + 3 * y + (100 - x - y) / 3 = 100 then print x, y, 100 - x - y end if next ynext xend sub5-3private sub command1_click

14、()dim i as integeri = 7doif i mod 2 = 1 and i mod 3 = 2 and i mod 5= 4 and i mod 6 = 5 and i mod 7 = 0 then print i exit doelsei= i + 7end ifloopend sub5-4private sub command1_click()dim x, y, z as integerfor x = 1 to 9 for y = 1 to 9 for z = 0 to 9 if 100 * x + 10 * x + z + 100 * y + 10 * z + z = 5

15、32 thenprint x, y, z end if next z next ynext xend sub5-5private sub command1_click()dim sum as single, n as integersum = 1do while sum = 0.0001 ti = (-1) (i + 1) / (2 * i - 1) pi = pi + tii= i + 1loopend sub5-9private sub command1_click()dim i, j, k as integerprint 水仙花數(shù)有：for i = 1 to 9 for j = 0 to

16、 9 for k = 0 to 9if i * 100 + j * 10 + k = i 3 + j 3+ k 3 thenprint i, j, k end if next k next jnext iend sub5-10private sub command1_click()dim i, j, k as integerk = 0for i = 999 to 100 step -1 for j = 2 to int(sqr(i) if i mod j = 0 then exit for next j if j = int(sqr(i) + 1 then k = k + 1: print i

17、 if k = 3 then exit fornext iend sub5-11private sub command1_click()dim i, n, min, max as integerdim aver as singlemin = 100max = 50for i = 1 to 20n= int(rnd * (99 - 51) + 51) print n; if i mod 5 = 0 then print if n max then max = n if n max then max = ia(i): imax = iif ia(i) min then min = ia(i): i

18、min = inext iprint max=; max,imax=; imaxprint min=; min,imin=; iminprint aver=; aver / 10end sub6-2private sub command1_click()dim ia(20), num(5) as integerdim max, min, aver as integer, imax, iminas integerfor i = 1 to 20 ia(i) = int(rnd * 101) print ia(i); select case ia(i) case is 60 num(1) = num

19、(1) + 1case60 to 69 num(2) = num(2) + 1 case 70 to 79 num(3) = num(3) + 1 case 80 to 89 num(4) = num(4) + 1 case 90 to 100 num(5) = num(5) + 1 end selectnext iprintprint 各分?jǐn)?shù)段的人數(shù)分別是(059,6069,7079,8089,90100)：for i = 1 to 5 print num(i);next iend sub6-3private sub command1_click()dim a(10) as integer,

20、 t%for i = 1 to 10a(i) = int(rnd * 51)print a(i);next iprintfor i = 1 to 9for j = i + 1 to 10 ifa(i) max then max = ia(i, j): imax = i: jmax = j next j printnext iprint max=; max,imax=; imax, jmax=; jmaxend sub實驗七一ccdab ddaaa二（1）n?。?）1到n累加和（3）n是否是素數(shù)（4）冒泡排序（5）遞歸調(diào)用實現(xiàn)斐波那契數(shù)列（6）求最大公約數(shù)和最小公倍數(shù)三7-1private su

21、b command1_click()dim n, sum as integern = val(inputbox(n=?)sum = jsh(n)print 1!+2!+.+; n;!=; sumend subpublic function jsh(n) as integerdim i, s, t as integers = 0t = 1for i = 1 to nt= t * is= s + tnext ijsh = send function7-2private sub command1_click()dim n, flag as integern = val(inputbox(n=?)fl

22、ag = 1sushu n, flagprint n=; flagend subpublic sub sushu(n, flag)dim i as integerfor i = 2 to n - 1 if n mod i = 0 then flag = 0: exit fornext iend sub7-3private sub command1_click()dim flag as integer, str as stringstr = inputbox(n=?)flag = huiwen(str)if flag = 1 then print str; 是回文數(shù) else print str

23、; 不是回文數(shù)end subpublic function huiwen(str) as integerdim lens, m, i as integerlens = len(str)for i = 1 to int(lens / 2) if mid(str, i, 1) mid(str, lens - i + 1, 1) then exit fornext iif i = int(lens / 2) + 1 then huiwen = 1else huiwen = 0end function7-4private sub command1_click()dim n, m, c1, c21, c

24、22, c23 as integern = val(inputbox(n=?)m = val(inputbox(m=?)c1 = jc1(n) / (jc1(m) * jc1(n - m)print 函數(shù)計算c1=; c1jc2 n, c21jc2 m,c22jc2 n - m, c23print 子過程計算c2=; c21 / (c22 * c23)end subpublic sub jc2(n, c)dim i as integerc = 1for i = 1 to nc= c * inext iend subpublic function jc1(n)dim i as integers

25、= 1for i = 1 to ns= s * inext ijc1 = send function7-5private sub command1_click()dim str1 as stringstr1 = inputbox(請輸入一個實數(shù)：)msgbox sum(str1)end subpublic function sum(str as string)dim lens, i, n as integerdim c as string * 1lens = len(str)n = 0for i = 1 to lensc= mid(str, i, 1) if c = 0 then n = n

26、+ val(c) end ifnext isum = nend function7-6private sub command1_click()dim str1 as stringstr1 = inputbox(請輸入一個字符串：)print str1; 中有數(shù)字字符; num(str1); 個。end subpublic function num(str as string)dim lens, i, n as integerdim c as string * 1lens = len(str)n = 0for i = 1 to lensc= mid(str, i, 1) if c = 0 the

27、n n = n + 1 end ifnext inum = nend function7-7private sub command1_click()可參考本實驗填空第六題或采用本方法，注意兩法的形參是傳地址還是傳值的區(qū)別dim m, n, y, b as integerm = val(inputbox(please input ainteger(m):)n = val(inputbox(please input ainteger(n):)print m; 和; n;gysgbs m, n, pprint 的最大公約數(shù)為; nprint 最小公倍數(shù)為; pend subpublic sub gy

28、sgbs(a, b, c)dim r as integerc = a * br = a mod bdo while r 0a= bb= rr= a mod bloopc = c / bend subvisual basic程序設(shè)計課程考試大綱（本學(xué)期19周，1-16周上課，周4學(xué)時，課程總學(xué)時64，其中：理論32+上機32）【基本要求】熟悉visual basic集成開發(fā)環(huán)境；了解visual basic中對象的概念和事件驅(qū)動程序的基本特性；掌握visual basic的基本語句；了解簡單的數(shù)據(jù)結(jié)構(gòu)和算法；掌握visual basic常用控件及其控件的三個要素；掌握模塊化程序設(shè)計的過程設(shè)計（包括函數(shù)過程、子過程）方法；能針對簡單的實際問題構(gòu)造算法，并能正確地實現(xiàn)算法與程序之間的轉(zhuǎn)換，具有使用visual basic編制和上機調(diào)試簡單程序的能力；培養(yǎng)利用計算機解決問題的意識、思路和能力，為今后進行計算機應(yīng)用奠定基礎(chǔ)。【考試范圍】1.visual basic程序設(shè)計概述q發(fā)展、特點、版本、安裝、啟動、退出qvisual basic 6.0集成開發(fā)環(huán)境（7個主要窗口）2.visual basic程序設(shè)計的相關(guān)概念q對象和類的概念q對象的建立和編輯q對象的屬性、事件和方法