18個(gè)VB經(jīng)典例題Word版

1、傳播優(yōu)秀word版文檔 ，希望對(duì)您有幫助，可雙擊去除！1、隨機(jī)產(chǎn)生三個(gè)100300之間的整數(shù)，判斷這三個(gè)整數(shù)是否能構(gòu)成三角形，如果可以，求三角形的面積。dim x as integer, y as integer, z as integerdim c as single, s as singlex = int(rnd * 201 + 100)y = int(rnd * 201 + 100)z = int(rnd * 201 + 100)if x + y > z and x + z > y and y + z > x then c = (x + y + z) / 2 s = s

2、qr(c * (c - x) * (c - y) * (c - z) print "三角形面積是：" & s else print "構(gòu)不成三角形"end if2、隨機(jī)產(chǎn)生0100之間的60名學(xué)生的數(shù)學(xué)分?jǐn)?shù)，分別統(tǒng)計(jì)分?jǐn)?shù)在、之間的學(xué)生人數(shù)。dim a%, b%, c%, d%, e%, f%, g%, h%, i%, j%dim x as singlefor m = 1 to 60 '產(chǎn)生60名學(xué)生的分?jǐn)?shù) x = rnd * 101 '產(chǎn)生0100的分?jǐn)?shù) print x; if x >= 90 then a = a + 1

3、elseif x >= 80 then b = b + 1 elseif x >= 70 then c = c + 1 elseif x >= 60 then d = d + 1 elseif x >= 50 then e = e + 1 elseif x >= 40 then f = f + 1 elseif x >= 30 then g = g + 1 elseif x >= 20 then h = h + 1 elseif x >= 10 then i = i + 1 else j = j + 1 end ifnext mprint a,

4、b, c, d, e, f, g, h, i, j3、我國(guó)有13億人口，按照人口年增長(zhǎng)0.8%計(jì)算，多少年以后我國(guó)人口超過(guò)26億。dim x as double '人數(shù)一定要定義成雙精度的dim n as integerx = 1300000000n = 0do while x <= 2600000000# x = x * 1.008 n = n + 1loopprint n; "年以后，我國(guó)人口將超過(guò)26億"4、編寫一個(gè)程序求一元二次方程的根，要求對(duì)輸入系數(shù)的合法性進(jìn)行驗(yàn)證，并規(guī)范輸出結(jié)果（保留兩位小數(shù)）。（上機(jī)指導(dǎo)p44）dim a!, b!, c!, d

5、!, x1!, x2!a = text1.textb = text2.textc = text3.textd = b * b - 4 * a * cif a = 0 then msgbox "a不能為0" '判斷是否能構(gòu)成一元二次方程 elseif d >= 0 then x1 = (-b + sqr(d) / (2 * a) '求方程的兩個(gè)根 x2 = (b + sqr(d) / (2 * a) x1 = format(x1, ".#") '對(duì)根規(guī)范化輸出 x2 = format(x2, ".#")en

6、d ifprint x1, x25、某次歌手大獎(jiǎng)賽，共有10名選手，有10名評(píng)委打分。要求評(píng)委給10位選手打分，去掉一個(gè)最高分，去掉一個(gè)最低分，求出該選手的平均分。dim i as integerdim j as integerdim max as single, min as single, sum as single, aver as singlefor i = 1 to 10 '十個(gè)選手的成績(jī) max = 0 min = 100 sum = 0 for j = 1 to 10 '十個(gè)評(píng)委的分?jǐn)?shù) x = rnd * 101 '產(chǎn)生0100的分?jǐn)?shù) if x >

7、max then max = x if x < min then min = x sum = sum + x next j aver = (sum - max - min) / 8 print aver;next i6、求1到100以內(nèi)的素?cái)?shù)。dim i as integer, j as integer, prime as booleanfor i = 1 to 100 prime = true for j = 2 to sqr(i) '判斷一個(gè)數(shù)是否是素?cái)?shù) if i mod j = 0 then prime = false end if next j if prime then

8、 '如果prime = true則i是素?cái)?shù) print i & "是素?cái)?shù)" else print i & "不是素?cái)?shù)" '如果prime = false則i不是素?cái)?shù) end ifnext i7、求出100以內(nèi)的所有勾股數(shù)（勾股數(shù)為a2+b2=c2,c為自然數(shù)，且a<>b）dim a as integer, b as integer, c as integerfor a = 1 to 100 for b = 1 to 100 for c = 1 to 100 if c * c = a * a + b * b an

9、d a <> b then print a; b; c; print end if next c next bnext a8、把輸入的字符串逆序輸出。dim str as string, strre as stringdim i as integer str = text1.text '原字符串 for i = 1 to len(str) strre = mid(str, i, 1) & strre '字符串逆序 next i text2.text = strre '逆序后的字符串9、隨機(jī)產(chǎn)生0100之間的60名學(xué)生的數(shù)學(xué)分?jǐn)?shù)，分別統(tǒng)計(jì)分?jǐn)?shù)在、之間的

10、學(xué)生人數(shù)。（用數(shù)組實(shí)現(xiàn)）dim i as integer, s(1 to 10) as integer, mark(1 to 60) as integer,for i = 1 to 60 mark(i) =int( rnd * 101) '隨機(jī)產(chǎn)生0100之間的分?jǐn)?shù) print mark(i); '輸出60個(gè)分?jǐn)?shù) select case mark(i) case is < 10 s(1) = s(1) + 1 case is < 20 s(2) = s(2) + 1 case is < 30 s(3) = s(3) + 1 case is < 40 s(

11、4) = s(4) + 1 case is < 50 s(5) = s(5) + 1 case is < 60 s(6) = s(6) + 1 case is < 70 s(7) = s(7) + 1 case is < 80 s(8) = s(8) + 1 case is < 90 s(9) = s(9) + 1 case is < 100 s(10) = s(10) + 1 end selectnext iprint '換行for i = 1 to 10 print s(i); '輸出各個(gè)分?jǐn)?shù)段的人數(shù)next i10、隨機(jī)產(chǎn)生10個(gè)同學(xué)的

12、成績(jī)隨機(jī)整數(shù)存入數(shù)組，求出數(shù)組中的最高分，最低分。dim i as integer, mark(10) as integer, min as integer, max as integerfor i = 1 to 10 mark(i) = int(rnd * 41 + 60) '隨機(jī)產(chǎn)生60100之間的分?jǐn)?shù) print mark(i); '輸出分?jǐn)?shù)next imin = mark(1)max = mark(1)for i = 2 to 10 if mark(i) > max then max = mark(i) '找出最大數(shù) if mark(i) < min

13、 then min = mark(i) '找出最小數(shù)next iprint '換行print "min=" min; "max=" max11、在上一題的基礎(chǔ)上，采用動(dòng)態(tài)數(shù)組隨機(jī)生成n個(gè)同學(xué)的成績(jī)，求平均分，并統(tǒng)計(jì)高于平均分的人數(shù)。dim mark() as integer, i as integer, n as integer, aver as singlen = inputbox("請(qǐng)輸入學(xué)生人數(shù)")redim mark(1 to n) '重新定義數(shù)組aver = 0for i = 1 to nmark(i

14、) = int(rnd * 41 + 60)aver = aver + mark(i)next iredim preserve mark(1 to n + 2) '數(shù)組保留以前的數(shù)制不變mark(n + 1) = aver / nmark(2) = 0for i = 1 to nif mark(i) > mark(n + 1) then mark(n + 2) = mark(n + 2) + 1print mark(i); '輸出學(xué)生成績(jī)next iprint mark(n + 1); mark(n + 2) '輸出平均分與高于平均分的人數(shù)12、隨機(jī)生成包含10個(gè)

15、數(shù)組元素的有序數(shù)組，然后第一個(gè)與第六個(gè)進(jìn)行交換，第二個(gè)與第七個(gè)進(jìn)行交換。，并把原數(shù)組以及交換后的數(shù)組分別在窗體上顯示出來(lái)。dim s(1 to 10) as integerfor i = 1 to 10 '產(chǎn)生10個(gè)數(shù)s(i) = int(rnd * 101)print s(i); '輸出交換前的數(shù)值next iprintfor i = 1 to 5 '實(shí)現(xiàn)交換t = s(i)s(i) = s(5 + i)s(5+ i) = tnext ifor i = 1 to 10print s(i); '輸出交換后的數(shù)值next i13、已知數(shù)組a=array(1,4,8

16、,5,10)，b(6)，通過(guò)數(shù)組a給數(shù)組b賦值；用選擇法按照升序?qū)?shù)組b排序，對(duì)排序后的數(shù)組插入元素6，使b數(shù)組有序；刪除元素5，并使數(shù)組元素個(gè)數(shù)減1。option base 1private sub command1_click()dim b(1 to 6) as integer, a as varianta = array(1, 4, 8, 5, 10)for i = 1 to 5 '用數(shù)組a給數(shù)組b賦值b(i) = a(i)print b(i); '輸出數(shù)組b的前5個(gè)元素next iprint b(6); '輸出數(shù)組b最后一個(gè)元素printfor i = 1 to

17、 5 '用選擇法對(duì)數(shù)組b升序排序min = i for j = i + 1 to 6 if b(j) < b(min) then min = j next jt = b(i): b(i) = b(min): b(min) = tnext ifor i = 1 to 6 '輸出數(shù)組b的所有元素print b(i);next iprint'下面是對(duì)有序數(shù)組b插入元素6for j = 2 to 6 if 6 < b(j) and 6 > b(j - 1) then '找出插入位置 for i = 1 to j - 1 '插入位置以前的元素向前

18、移動(dòng)一個(gè)位置 b(i) = b(i + 1) next i b(j - 1) = 6 '在準(zhǔn)確的位置插入6 end ifnext jfor i = 1 to 6 '輸出插入6以后的數(shù)組b的所有元素 print b(i);next iprintdim c() as integer '定義數(shù)組c是一個(gè)動(dòng)態(tài)數(shù)組redim c(1 to 6) as integerfor i = 1 to 6 c(i) = b(i)next i'下面是刪除元素5for i = 1 to 6if c(i) = 5 then '如果數(shù)組c里面有5就進(jìn)行刪除 for j = i to

19、5 '把元素5后面的元素向前移動(dòng)一個(gè)位置 c(j) = c(j + 1) next j end if next iredim preserve c(5) '只保留數(shù)組c里面前面的5個(gè)元素，達(dá)到了刪除目的for i = 1 to 5print c(i); '輸出刪除5以后的數(shù)組元素next iend sub14、編寫一個(gè)無(wú)參數(shù)sub過(guò)程triangle，在窗體上輸出用“*”組成的如下圖所示的三角形圖案。'被調(diào)過(guò)程public sub triangle() '子過(guò)程print " *"print " * * *"pri

20、nt " * * * * *"print " * * * * * * * "print "* * * * * * * * *"end sub'主調(diào)過(guò)程private sub command1_click()call triangle '調(diào)用子過(guò)程end sub15、把上面的triangle過(guò)程改造成帶參數(shù)的過(guò)程，使之能輸出任意行三角形圖案，并嘗試用兩種方法調(diào)用。'被調(diào)過(guò)程public sub triangle(n as integer)for i = 1 to nprint tab(n - i); string

21、(2 * i - 1, "*");next iend sub'主調(diào)過(guò)程private sub command1_click()dim a as integera = val(inputbox("請(qǐng)輸入a")call triangle(a)end sub16、編寫一個(gè)函數(shù)prime，用于判斷一個(gè)自然數(shù)是否為素?cái)?shù)。調(diào)用此函數(shù)輸出3100之間的所有的素?cái)?shù)。（提示：函數(shù)prime的類型使用布爾型）'被調(diào)函數(shù)public function prime(m as integer, n as integer) as booleandim i as in

22、teger, j as integer'判斷m-n之間的所有素?cái)?shù)for i = m to nprime = truefor j = 2 to sqr(i) '判斷一個(gè)數(shù)是否是素?cái)?shù) if i mod j = 0 then prime = false end if next j if prime then '如果prime = true則i是素?cái)?shù) print i & "是素?cái)?shù)" else print i & "不是素?cái)?shù)" '如果prime = false則i不是素?cái)?shù) end ifnext iend functi

23、on'主調(diào)過(guò)程private sub command1_click()dim a as integer, b as integerdim c as booleana = val(inputbox("請(qǐng)輸入a")b = val(inputbox("請(qǐng)輸入b")c = prime(a, b)end sub17、編寫函數(shù)，判斷某數(shù)是否是“水仙花數(shù)”。并用此函數(shù)輸出100999之間的所有的水仙花數(shù)、所謂的“水仙花數(shù)”是指一個(gè)3位數(shù)，其各位數(shù)字立方和等于該數(shù)本身。'主調(diào)函數(shù)private sub command1_click()dim a as integer, b as integerdim c as booleana = val(inputbox("請(qǐng)輸入a")b = val(inputbox("請(qǐng)輸入b")c = shuixianhua(a, b)end sub'被調(diào)函數(shù)pub