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1、第1章1选择题 (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选择题 (1)a b以数字开头;c是vb关键字;d中包含了“-”(减)号 (2)b a是一个字符串变量名;c什么也不是;d是一个变量名 (3)a b是整型;c是字符型;d是双精度型 (4)b(5)d(6)a (7)c 这是一个典型的四舍五入算法,根据这个可以自己写出一个四舍五入函数。 (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尽管也是3456,但
2、它是一个数值,而不是字符串,所以c是错误的 (24)d这里要注意的是表达式分为两个部分,后一部分的最小值为-1,前一部分的最大值为5,所以整个表达式的最小值就为-5,而后一部分的最大值为3,所以整个表达式的最大值为15 (25)b(26)c(27)c(28)a(29)a(30)c(31)c(32)a(33)a2判断题 (1)错;(2)错;(3)错;(4)对;(5)错;(6)错;(7)错;(8)对;(9)对;(10)(错, str函数转换正数时前面会有一个符号位表现为空格,正确的应该是7。)(11)错。3按要求写表达式 (1)转换表达式a、(3*a*a+4*b*b*b)/(a-b) (这里的乘方
3、也可以用乘方号)b、6*sin(x+y*y)/(140/(3+a)(这里的3+a可以写成分子的一部分)c、exp(x)+log(10)/sqr(x+y+1)(2)写布尔表达式a、x>y and f<2b、x<z and y>=z or y<z and x>=z 也可写成 (x-z)*(y-z)<=0 and x<>yc、a*b>0d、a*b=0 and a<>b(3)利用函数写表达式a、chr(int(rnd * (asc("l") - asc("c") + 1) + asc(&qu
4、ot;c")也可以写成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) 也可以写成 (x mod 10)*10+x10 e、round(x,2) or format(x,”#.00”) f、mid(s,5,6) 这道题很多人容易写成mid(“s”,5,6),这就严重错误了。第03章1选择题 (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编程题(1) 鸡兔同笼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 = "计算结果:鸡有" & x & "只,兔有" & y & "只"end sub(2)计算纸币张数private sub command1_click() dim x%, y%, z%, a100%, a50%, a20%, a10%, a5%, a1% x = val(inputbox("请输入支付金额") 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 & "元需要支付的钱币总张数为:" & a100 & "+" & a50 & "+" & a20 & "+" _& a10 & "+" & a5 & "+" & a1 & "
8、;=" & z & "张"end sub另解,利用数组:option base 1private sub command1_click() dim x%, z%, a(), b%(5), result$ a = array(100, 50, 20, 10, 5) x = val(inputbox("请输入支付金额") result = x & "元需要支付的钱币总张数为:" 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 & "张" print resultend sub(3)电话号码升位。private sub command1_click() dim x$ x = inputbox("请输入一个带区号的原电话号码,格式为:*-*") msgbox left(x, 4) & "8" & right(x, 7)end sub第04章1
10、选择题 (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判断题(1)对;(2)对;(3)对;(4)对;(5)错(应该是整型);(6)错;(7)对3根据程序写运行结果(略)4编程题(1) private sub command1_click()dim x%x = inputbox("请输入数值")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函数中小数位用"#"的时候当不够三位时不足的位不补0。'如果用round()函数,如果计算结果为整数时,则没有小数位。end sub(2)输入3个数,输出最大最小数private sub command1_click
12、() dim a%, b%, c% form1.fontsize = 20 a = inputbox("请输入第一个数") b = inputbox("请输入第二个数") c = inputbox("请输入第三个数") print a, b, c '先按照升序或降序排序然后再进行取最大最小数 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 "最小数是" a, "最大数是" cend subprivate sub command2_click() '定义变量max用于保存最大数,min保存最小数 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) '假设a是最大数,因此将a的值赋给max max = a 用max与b进行比较,若b大于m
14、ax,则将b赋给max,让max保存前两个数的最大数 if max < b then max = b end if '用max与c比较,若c大于max则将c赋给max,让max保存三个数中的最大数 if max < c then max = c end if '假设a是最小数,因此将a的值赋给min max = a '用min与b进行比较,若b小于min,则将b赋给min,让min保存前两个数的最小数 if min > b then min = b end if '用min与c比较,若c小于min则将c赋给min,让min保存三个数中的最小数
15、if min < c then min = c end if '显示结果 label2.caption = "最大数为:" & max & ",最小数为:" & minend sub(3)判断闰年private sub command1_click() dim x%, y% x = val(inputbox("请输入年份") if (x / 4 = x 4 and x / 100 <> x 100) or x / 400 = x 400 then print x & "
16、;是闰年" else print x & "非闰年" end ifend sub(4)判断能否被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 "您输入的数能被5整除!" else picture1.print "您输入的数不能被5整除!" end ifend sub(5)部门征税private sub co
17、mmand1_click() dim tax as single, income as single income = val(text1.text) if income <= 0 then '若输入的值小于0则提示错误,并退出该过程,不进行税额的计算 label2.caption = "输入有误." 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 = "你应缴纳的数款为:" & tax & "元"end sub(6)输出100-200之间能被3整除的数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)输出1000到1100之间的质数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)猜数字private sub command1_click() randomize timer '给一个随机种子,让每次产生的随机数都不同 x = int(101 * rnd) label1.caption = "你还有8次机会" for i = 8 to 1 step -1 y = val(inputbox("请猜一个100以内的整数", "猜数字") if y > x then label1.caption = "大了点,你还有" & i &
21、"次机会" elseif y < x then label1.caption = "小了点,你还有" & i & "次机会" else label1.caption = "恭喜你猜对了!" exit for end if next if i = 0 then label1.caption = "很遗憾,你用了8次机会也没猜正确!" & vbcrlf & "正确值为:" & x end ifend sub(9)100到999之间回文
22、数private sub command1_click() dim n as integer '记录第几个回文数 for x = 100 to 999 g = x mod 10 '获取x的各位数 b = x 100 '获取x的百位数 '若个位数和百位数相等则x就是回文数 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的近似值为:" & 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的近似值为:" & 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,重新输入" text1.text = "" text1.setfocus else s = 0 n = 0 do while s < m n = n + 1
25、 s = s + n 3 loop label1.caption = "满足不等式的最大n值是:" & n - 1 end ifend sub方法二'n的取值范围1m(1/3),按要求我们从m(1/3)(最大的数)开始找符合条件的第一个数就是了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为:" & n exit for end if nextend sub(12)'方法一private sub command1_click() picture1.cls '清除以前输出的内容 for i = 1 to 6 picture1.print spc(15 - i); '在左侧打印n个空格来填充或将spc换为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 '清除以前输出的内容 for i = 1 to 6 picture1.print tab(15 - i); string(2 * i, "0") '用string函数产生n个0 nextend sub(13)十进制转二进制 转十六进制private sub command1_click() n = val(inputbox("输入要转换的十进制整数") m = n x = "" do while n <> 0 a = n mod 2 n = n 2 x = a &
28、; x loop msgbox m & " 转换二进制数是:" & xend subprivate sub command2_click() n = val(inputbox("输入要转换的十进制整数") 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 & " 转换十六进制数是:" & xend sub(14)解灯谜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() '由题意可知abcd的取值范围是10009999之间,因此通过循环来找出符合条件的abcd for abcd = 1
31、000 to 9999 abc = abcd 10 '获取abc的值 c = abc mod 10 d = abcd mod 10 '分别获取abcd的十位数c和个位数d cdc = c * 100 + d * 10 + c '计算cdc的值'如果符合abcd-cdc=abc则输出 if abcd - cdc = abc then label1.caption = "abcd的值为:" & abcd exit for end if nextend sub(15)猴子吃桃private sub command1_click() '
32、;验证 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() '穷举发 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:假设第n天的桃子为x,前一天(第n-1天)的桃子为y '则x=y-(y/2+1)=>x=y/2-1;y=2*(x+1) '现在第十天的桃子是已知的,我们反推到第一天即可 '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、选择题 (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编程题(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)矩阵找最大元素的行列private sub command1_click() '先声明一个动态数组a,因为其大小不能确定 dim a() as integer n = val(text1.text) m = val(text2.text) '指定数组的大小 redim a(1 to n, 1 to m) as integer picture1.cls '用随机数给数组赋值,并打印到窗体 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 '假设第一个元素的值最大,用r和c分别存放最大数所在的行和列 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 = "矩阵中的最大值为:" & a(r, c) & vbcrlf label3.caption =
37、label3.caption & "位置:" & r & "行," & c & "列"end sub(3)交换数组元素值private sub command1_click() dim a$(), x, i%, n% a = split(inputbox("请输入n个数据,数据之间用逗号分隔!"), ",") '注意这里的inpubox函数的用法,输出数据时,数据之间应该用英文状态的逗号分隔 print "对换前数据序列为:"
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 "对换后数据序列为:" for each x in a print x; " " next xend sub(4)private sub command1_click() dim x$(), a%(4 to 9), b(), i%, n%, y, k% b = array("无
39、效数据", "小于60分", "6069", "7079", "8089", "90100") x = split(text1, ",") n = ubound(x) for each y in x if y < 0 or y > 100 then y = 40 elseif y < 60 then y = 50 elseif y = 100 then y = 90 end if k = y 10 a(k) = a(k) + 1 next y fo
40、r i = 4 to 9 picture1.print b(i - 4), a(i) next iend sub(5)根据身份证前17位算第18位option base 0private sub command1_click() dim a%(16), b$(), w(), i%, x$, sum% b = split("1,0,x,9,8,7,6,5,4,3,2", ",") w = array(7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2) x = inputbox("请输入您的身份
41、证前17位") for i = 0 to 16 a(i) = mid(x, i + 1, 1) sum = sum + a(i) * w(i) next i print "您的身份证号证为:" & x & b(sum mod 11)end sub(6)输出斐波拉契数列option base 1 private sub command1_click() dim a&(30), i% a(1) = 1: a(2) = 1 for i = 3 to 30 a(i) = a(i - 1) + a(i - 2) next i for i = 1 to
42、 30 print tab(i - 1) mod 5) * 10 + 1); a(i); next iend sub(7)随进产生15个不同大写字母option base 1private sub command1_click() dim a$(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)将一个数插入数组option base 1dim a() '声明窗体及变量是为了能够多次插入数据sub parray(x() '定义打印数组的子程序 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 '保证插入位置在有效范围内 k = inputbox("请输入待插
44、入的位置(" & 1 & "" & n + 1 & ")") loop until k >= 1 and k <= n + 1 redim preserve a(n + 1) '扩大数组 x = inputbox("请输入待插入的数据") 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) '给数组赋初值end sub(9)option base 1dim a() '声明窗体及变量是为了能够多次插入数据sub parray(x() '定义打印数组的子程序 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 "删除前:": call par
46、ray(a) x = inputbox("请输入待删除的数") 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 "删除后:": call parray(a) else print "查无此数!" end ifend subprivat
47、e sub form_load() a = array(1, 2, 3, 4, 5, 6, 7, 8, 9) '给数组赋初值end sub(10)求矩阵四周元素之和option base 1sub parray(x%() '定义打印数组的子程序 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 "四周元素之和为:" & send
49、sub(11)运动会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 "编号", "成绩" 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补充一:打印魔方阵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 '再放其它数 p = x: q = y '保留前一个数的位置 x = x - 1 '找下一个数的行位置 if x < 1 then x = n '如果行小于1,让行为n y = y + 1 '找下一个数的列位置 if y > n then y =
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