VB程序設(shè)計(jì)學(xué)習(xí)與實(shí)驗(yàn)指導(dǎo)答案

1、清華大學(xué)出版社：蔣銀珍 沈瑋 吳瑾 編著visual basic 程序設(shè)計(jì)學(xué)習(xí)與實(shí)驗(yàn)指導(dǎo)答案上篇 學(xué)習(xí)指導(dǎo)第1章 visual basic 概述一.選擇題123456789abddbbcac101112131415161718aaccdcabd二填空題1. 事件2. 查看對(duì)象3. 工程4. 視圖 屬性窗口5. 窗體編輯器第2章 面向?qū)ο蟪绦蛟O(shè)計(jì)初步一.選擇題1234567acadcda二填空題1. 控件2. 事件3. 類 屬性4. 屬性5. 略6. name第3章 visual basic 語(yǔ)言基礎(chǔ)一.選擇題123456789aaccaccad101112131415161718dccadda

2、cb192021222324252627bdbdabccb2829db二填空題1. (a*sin(30/180*3.14)+c*(log(b)/log(10)/(abs(d)+1)*(e+f)+12. mod3. x=int(x) and y=int(y) and x*y>04. 4.y mod 4=0 and y mod 100<>0 or y mod 400=05. x=int(99-10+1)*rnd+10)6. 0 “”7. 空格8. 6，845.309.（1）8（2）1（3）abcd（4）出錯(cuò)，類型不匹配（5）false（6）2008-6-29（7）i likevb

3、（8）4（9）4 4（10）3 4（11）10 -10（12）-1（13）aa（14）3（15）0（16）1（17）c（18）5（19）8（20）34.54（21）中國(guó)（22）上海（23）visual basic（24）visual basic10.（1）x<-100 or x>100（2）a=int(a) and b =int(b) or a<0 and b<0（3）x<=y or x>=z（4）a>0 and a<>int(a)（5）a=0 xor b=0（6）len(s)<10（7）ucase(s)=s（8）left(s,1)=

4、”a” and right(s,1)=”a”（9）instr(s,”)<>0（10）trim(s)=s（11）date=#8/8/2008#（12）weekday(date)=2第4章 程序控制結(jié)構(gòu)與算法基礎(chǔ)一.簡(jiǎn)單填空1. randomize2. n=int(99-10+1)*rnd+10)3. m=(int(100-10+1)*rnd+10)/104. exit for5. exit do二讀程序?qū)懡Y(jié)果1. 150 122. 22 23 -13. 5 3 04. 20 40 60三程序填空1. len(s) mid(s,i,1) c>=”a” and c<=”z”

5、c>=”a” and c<=”z” c>=”0” and c<=”9” n3=n3+12. 0 1 n-1 n mod i=0 sum+i sum=n3. 1000 cstr(n) cstr(n*n) right(s2,len(s1)=s14. 100 2002 n-1exit forcount+1count mod 5=0 print5. sum=1 a*x2/(2*n-1)*(2*n) sum+a abs(a)<=0.0000001第5章 數(shù)組的應(yīng)用一.選擇題123456789bccacadbc10111213abda二.讀程序?qū)懡Y(jié)果1. 120 362. 1

6、0 31 73. 26 14 26三程序填空1.a(i,j);picture1.printa(i.1)1a(i,j)j“第” & i & “行最大數(shù)是” & max data & “，第” & maxj & “列”2.text1.text & a(i);kpreserve b(k)b(k)=a(i)3.sqr(n - m) = int(sqr(n - m)redim preserve b(k)na(m) & "和" & b(m)4.int(100-50+1)*rnd+50)1b(i)+1第6章 過(guò)程一判

7、斷題123456789ttffttttt101112131415161718tfttftttt1920tt二.選擇題12345678acaaadba四讀程序?qū)懡Y(jié)果1. x=2,y=2 m=22. 9 4,5,9 273. 50 264. 5 64,55. 7,14 11,22 15,306. 29 29 37. 12 25下篇 實(shí)驗(yàn)教程實(shí)驗(yàn)2 visual basic 變量、函數(shù)與表達(dá)式的使用實(shí)驗(yàn)2-1 算術(shù)運(yùn)算符的使2-202007-10-153 2.857143 2.85714285714286實(shí)驗(yàn)2-2 字符運(yùn)算符的使用20071020071020172007-

8、10-10102007-10-20出錯(cuò)，類型不匹配實(shí)驗(yàn)2-3 關(guān)系運(yùn)算符的使用falsefalsefalsefalsefalsefalsefalsetruefalsetrue實(shí)驗(yàn)2-4 邏輯運(yùn)算符的使用truetruefalsetruefalsefalsetruetruefalse5-59實(shí)驗(yàn)2-5 常用函數(shù)的使用3.14 3.147.389056098930651 -1 .9999999982051034-1 1 0.70554755825 251250 0a a65 9748 688 -98 -89 -9aaa 4.5 aaa4.516 5visua 6.0sua basic basic

9、6.0abcde efgbasic basic basicvisual basic 6.0 basic8 8aaa dd2011-3-152011-3-15 14:35:5215320113001.731.731.7317.32e-010173%實(shí)驗(yàn)3 算法基礎(chǔ)及程序控制結(jié)構(gòu)實(shí)驗(yàn)3-1 三個(gè)數(shù)的交換a=cc=bb=tbc實(shí)驗(yàn)3-2 求周長(zhǎng)和面積const pi=3.14159262*pi*rpi*r*rpicture1.cls format(l, ".00") format(m, ".00")實(shí)驗(yàn)3-3 求分段函數(shù)的值x=val(text1.text)y

10、=1+sin(x)y=log(x)text1.text = ""text2.text = ""實(shí)驗(yàn)3-4 求一元二次方程的根val(text2.text)val(text3.text)b 2 - 4 * a * c"x1=" & (-b + sqr(delt) / (2 * a)"x2=" & (-b - sqr(delt) / (2 * a)text2.text = ""text3.text = ""end實(shí)驗(yàn)3-5 判斷素?cái)?shù)n mod k = 0k=nn

11、mod i=0k=2實(shí)驗(yàn)3-6 求最大公約數(shù)和最小公倍數(shù)m<nl=ml=nm mod k = 0 and n mod k = 0text3.text = kexit form>nl=ml=nk mod m = 0 and k mod n = 0text4.text = kexit for實(shí)驗(yàn)3-7 判斷升序數(shù)、降序數(shù)cstr(n)len(s) 1mid(s, k, 1) >= mid(s, k + 1, 1)k = len(s)n = val(text1.text)s = cstr(n)for k = 1 to len(s) - 1if mid(s, k, 1) <=

12、mid(s, k + 1, 1) then exit fornext kk = len(s)實(shí)驗(yàn)3-8 判斷回文數(shù)1len(s)2-1mid(s, k, 1) <> mid(s, len(s) - k + 1, 1)k = len(s) 2mid(s, k, 1) + s1s = s1實(shí)驗(yàn)3-9 求級(jí)數(shù)的值val(text1.text)01x ns * nt / s <= 10 -6text2.text = y實(shí)驗(yàn)3-10 查找數(shù)字串并求和text1.text = ""text2.text = ""list1.cleartext1.te

13、xt0false0while mid(s, i, 1) >= "0" and mid(s, i, 1) <= "9"flag = truek = k * 10 + mid(s, i, 1)i = i + 1flag = truesum + ktext2.text = sum實(shí)驗(yàn)3-11 隨機(jī)產(chǎn)生20個(gè)奇數(shù)int(99 - 10 + 1) * rnd) + 10a mod 2 = 1n mod 5 = 0picture1.cls實(shí)驗(yàn)3-12 加密解密chr(asc("a") + (asc(c) - asc("a&q

14、uot;) + 3) mod 26)c = chr(asc("a") + (asc(c) - asc("a") + 3) mod 26)c & s2text2.textlen(s2)mid(s2, i, 1)c = chr(asc("a") + (asc(c) - asc("a") + 23) mod 26)c = chr(asc("a") + (asc(c) - asc("a") + 23) mod 26)c & s1s1text1.text = "

15、;"text2.text = ""實(shí)驗(yàn)3-13 統(tǒng)計(jì)單詞個(gè)數(shù)text1.text0mid(s, i, 1) <> " " and i <= len(s)i = i + 1mid(s, start, i - start)ntext1.text = ""list1.clear實(shí)驗(yàn)4 數(shù)組的使用實(shí)驗(yàn)4-1 一維數(shù)組的產(chǎn)生及輸出i = 1 to 20int(9 - 0 + 1) * rnd + 0)i = 1 to 20print a(i);printi = 1 to 20picture1.print a(i);i

16、f i mod 5 = 0 then picture1.printi = 1 to 20text1.text & space(2) & a(i)if i mod 5 = 0 then text1.text = text1.text & vbcrlfv in aif i mod 5 = 0 then picture2.print實(shí)驗(yàn)4-2 選手得分int(50 - 10 + 1) * rnd + 10) / 10i = 1 to 6sum = sum + score(i)max = score(1)min = score(1)score(i) > max then

17、max = score(i)score(i) < min then min = score(i)(sum - max - min) / 4實(shí)驗(yàn)4-3 產(chǎn)生10個(gè)互不相同的整數(shù)int(999 - 10 + 1) * rnd + 10)t mod 10 = 5then exit forn= n + 1a(i) = tfor i = 1 to 10print a(i);if i mod 5 = 0 then printnext i實(shí)驗(yàn)4-4 統(tǒng)計(jì)數(shù)字出現(xiàn)的次數(shù)isnumeric(s) = falseexit sublen(s)c >= "0" and c <=

18、"9"a(c) = a(c) + 1a(k) <> 0m = m + 1實(shí)驗(yàn)4-5 排序1 to 10int(99 - 10 + 1) * rnd + 10)text1.text & a(n) & " "1 to 9i + 1 to 10a(i) > a(j)t = a(j)a(j) = a(i)a(i) = t1 to 910 - ia(j) > a(j + 1)t = a(j + 1)a(j + 1) = a(j)a(j) = t1 to 10text3.text = text3.text & a(i)

19、 & " "10t = a(i)0ji - 1 k + 1 1 to 10text4.text = text4.text & a(i) & " "實(shí)驗(yàn)4-6 二維數(shù)組的產(chǎn)生及輸出1 to 41 to 5int(9 - 0 + 1) * rnd + 0)1 to 41 to 5print a(i, j);print1 to 41 to 5picture1.print a(i, j);picture1.print1 to 41 to 5text1.text = text1.text & a(i, j) & "

20、 "text1.text = text1.text & vbcrlf實(shí)驗(yàn)4-7 查找最大和最小元素及其位置1 to 31 to 4a(i, j) = int(99 - 10 + 1) * rnd + 10)picture1.print a(i, j);picture1.printa(1, 1)， 1， 1a(1, 1)， 1， 1a(i, j)ija(i, j) < mina(i, j)ij"最大元素" & "a(" & maxi & "," & maxj & "

21、;)=" & max"最小元素" & "a(" & mini & "," & minj & ")=" & min實(shí)驗(yàn)4-8 矩陣轉(zhuǎn)置dim a() as integer"請(qǐng)輸入n的值"redim a(n, n) as integerfor i = 1 to nfor j = 1 to na(i, j) = int(99 - 10 + 1) * rnd + 10)text1.text = text1.text & a(i,

22、j) & " "next jtext1.text = text1.text & vbcrlfnext it = a(i, j)a(i, j) = a(j, i)a(j, i) = tfor i = 1 to nfor j = 1 to ntext2.text = text2.text & a(i, j) & " "next jtext2.text = text2.text & vbcrlfnext i實(shí)驗(yàn)5 過(guò)程實(shí)驗(yàn)5-1 孿生素?cái)?shù)和降序素?cái)?shù)dim i as integerfor i = 2 to n - 1if

23、n mod i = 0 then exit functionnext iprime = truedim a as integerdim b as integera = n 10b = n mod 10if a > b thendecnumber = trueelsedecnumber = falseend if10 to 100prime(i) = true and prime(i + 2) = true list2.additem i & "和" & i + 2decnumber(i) = true and prime(i) = true list3

24、.additem i實(shí)驗(yàn)5-2 求多項(xiàng)式和k = 1a = f(x, k)s + ak = k + 1single as singledim i as integer, a as singlea = (x + 1) / xfor i = 2 to ka = a * (x + i) / (2 * i - 1) * x)next ia實(shí)驗(yàn)5-3 進(jìn)制轉(zhuǎn)換asc(c) - asc("a") + 10left(s, k - 1)mid(s, k + 1)change(mid(s2, i, 1) * n (0 - i)t1 + t2tran(8, s)tran(16, s)not (c >= "0" and c <= "7" or c >= "a" and c <= "f" or c = ".")實(shí)驗(yàn)5-4 armstrong數(shù)i as long, a as long, b as longcombo1.text實(shí)驗(yàn)5-5option base 1dim a(5) as integerdim b(5) as integerb(i) = i * i + 1text2.text = text2.text & b(i) &