大学程序设计基础-Visual_Basic教材课后习题答案

1、 教材课后习题答案 第1章 一、选择题1234567bdbcbbd第2章 一、选择题12345678910bcbccbcaab111213141516abbcaa二、填空题1. rem或 _ ：2. “” # #3. 5x2-3x-2sina/34. 3214565. 300三、操作题1、程序段：private sub command1_click()dim x as integer, y as integerdim s as long, c as longx = text1.texty = text2.texts = x * yc = (x + y) * 2label3.caption =

2、长方形的面积为 + str(s)label4.caption = 长方形的周长为 + str(c)end sub3、程序段：private sub command1_click()text1.text = 第一end subprivate sub command2_click()text1.text = 第二end sub4、程序代码：private sub text1_change()text1.maxlength = 10end sub5、程序代码：private sub text1_click()text1.selstart = 0text1.sellength = len(text1.

3、text)end sub第3章 一、选择题1234567dadbadb二、填空题1. false2. b2-4*a*c else三、操作题1从键盘输入三个值，判断它们能否构成三角形的三个边。如果能构成一个三角形，则计算三角形的面积。 dim a as integer, b as long, c as long, s as single, p as singlea = val(inputbox(请输入一个值：)b = val(inputbox(请输入一个值：)c = val(inputbox(请输入一个值：)if (a + b c and a + c b and b + c a) thenp =

4、 (a + b + c) / 2s = sqr(p * (p - a) * (p - b) * (p - c)print a & , & b & , & c & 能构成一个三角形,它的面积为: & send if2编写程序，任意输入一个整数，判定该整数奇偶性。 dim n as integern = val(inputbox(请输入一个整数：)if n mod 2 = 0 then print n & 是偶数else print n & 是奇数end if3求一元二次方程的a*x2+b*x+c=0的根。 dim a as integer, b as integer, c as integer,

5、 delta as single, x1 as single, x2 as singlea = val(inputbox(请输入一个整数：)b = val(inputbox(请输入一个整数：)c = val(inputbox(请输入一个整数：)delta = b 2 - 4 * a * cif delta = 0 then x1 = -b / (2 * a) print 一元二次方程有两个相等的实根x1=x2= & x1else if delta 0 then x1 = (-b + sqr(delta) / (2 * a) x2 = (-b - sqr(delta) / (2 * a) pri

6、nt 一元二次方程有两个不等的实根x1= & x1 & ,x2= & x2 else x1 = (-b) / (2 * a) x2 = sqr(-delta) / (2 * a) print 一元二次方程有两个不等的虚根x1= & x1 & + & x2 & i,x2= & x1 & - & x2 & i end ifend if4. 任意输入三个数，输出最大数。dim a as integer, b as integer, c as integer, max as integera = val(inputbox(请输入一个整数：)b = val(inputbox(请输入一个整数：)c = v

7、al(inputbox(请输入一个整数：)max = aif max b then max = bend ifif max c then max = cend ifprint 最大数是 & max5任意输入三个数，由大到小排序输出。dim a as integer, b as integer, c as integer, temp as integera = val(inputbox(请输入一个整数：)b = val(inputbox(请输入一个整数：)c = val(inputbox(请输入一个整数：)if a c then print a, b, celse if a c then pri

8、nt a, c, b else print c, a, b end ifend if第4章 一、选择题12345678910abacdbadbb111213141516dcdcbb二、填空题1. 92. 1 2 33. 74. 100+100；x mod 5=05. 66. 57. * 2 5 15 30 75 16 32 80三、 编程题1键盘输入几个不同的数，将它们从大到小排序。 dim i%, j%, k%, s(10) as integerfor i = 1 to 10 s(i) = int(rnd() * 100)nextfor i = 1 to 9 for j = i + 1 to

9、 10 if s(i) s(j) then k = s(i) s(i) = s(j) s(j) = k end if nextnextfor i = 1 to 10 print s(i);next2设计程序，求出s=11+（1+2）+（1+2+3）+（1+2+3+4+n）的值。dim i as integer, sn as long, s as long, n as integersn = 0s = 0n = inputbox()for i = 1 to n sn = sn + i s = s + snnextprint s运行结果：若n为10，s为220。3输入x的值，求s=x-x3/3!+

10、x5/5!- 的前20项的和值。dim x as single, i as integer, s as doublex = inputbox(输入x值：)s = xfor i = 1 to 19 p = 1 for j = 1 to 2 * i + 1 p = p * j next s = s + (-1) (i) * x (2 * i + 1) / pnextprint s运行结果：当x为2时，s为0.91(精确到小数点后两位)4勾股定理中3个数的关系是：a2+b2=c2。编写程序，输出30以内满足上述关系的整数组合，例如3、4、5就是一个整数组合。dim a as integer, b a

11、s integer, c as integer, k as integerfor a = 1 to 30 for b = 1 to 30 for c = 1 to 30 if a 2 + b 2 = c 2 then k = k + 1: print a, b, c next nextnextprint k运行结果：225输出1000以内的完全数。完全数就是它的各个约数之和等于该数本身（如：6=1+2+3，6是完全数）。dim i as integer, j as integer, a as integerfor i = 2 to 1000 a = 0 for j = 1 to i - 1 i

12、f i mod j = 0 then a = a + j 判断j是否能整除i，如果能，则a增加j next j if a = i then print i 判断i是否是完全数，如果是，则输出inext i 运行结果：6284966编程求出所有的水仙花数。水仙花数就是一个三位数，其各位数字立方和等于该数。（如153=13+53+33）。dim i as integer, s as integerdim a as integer, b as integer, c as integerprint 100到999所有水仙花数：;for i = 100 to 999 a = i 100 取百位数 b =

13、 i 10 mod 10 或 b = i mod 100 10 取十位数 c = i mod 10 取个位数 s = a 3 + b 3 + c 3 水仙花数的判断依据 if s = i then print i;next i运行结果：100到999所有水仙花数：153 370 371 4077编写程序，输出100300间的所有素数。 dim i as integer, j as integer, c as integer, k as integerfor i = 100 to 300 for j = 2 to i - 1 if i mod j = 0 then exit for next i

14、f i = j then print i; k = k + 1 s = s + i if k mod 10 = 0 then print end ifnextprintprint 100300间总共有 + str(k) + 个素数， + 和值为 + str(s)运行结果：8我国现有人口为13亿。设年增长率为1，编写程序，计算多少年后增加到20亿。dim i as long, s as singledim n as single, m as singlen = inputbox(请输入增长率：)i = 0s = 13m = sdo while m 0.000001 i = i + 2 flag

15、= -flag n = 1 / i pi = pi + flag * n loopprint pi * 4运行结果：3.14159465358569第5章一、选择题12345678910dcddbaabcb1112131415dbbcb二、填空题1. 242. 73. a=30 b=704. “123456789”三、编程题1 编写一个求3个数中最大值max和最小值min的过程，然后用这个过程分别求3个数、5个数、7个数中的最大值和最小值。dim a()sub maxmin(b() 过程dim max, min, kmax = b(1)min = b(1)for k = 1 to ubound

16、(b)if max b(k) thenmin = b(k)end ifend ifnextprint 最大值=; max, 最小值=; minend subprivate sub command1_click() 主程序n = inputbox(n=)redim a(n)for i = 1 to na(i) = val(inputbox(a=)print a(i),nextprintcall maxmin(a()end sub2编写程序，求sa!+b!+c!，阶乘的计算分别用sub过程和function过程两种方法来实现。public sub fact2(m, ss)dim p, kp = 1

17、for k = 1 to mp = p * knextss = ss + pend subpublic function fact1(n)dim p, kp = 1for k = 1 to np = p * knextfact1 = pend functionprivate sub command1_click()dim s, a, b, ca = val(inputbox(a=)b = val(inputbox(b=)c = val(inputbox(c=)s = fact1(a) + fact1(b) + fact1(c) 函数print sss = 0call fact2(a, ss)