湖南省计算机二级考试VB编程题目

1、l一个数如果刚好与它的因子之和相等，则称完数。求5到100间的完数之和，结果：34sum = 0for n = 5 to 100 s = 0 for i = 1 to n - 1 if n mod i = 0 then s = s + i next i if s = n then sum = sum + nnext nprint sumlijk+kji=1333,其中i,j,k的范围是0到9，求有几组（i,j,k)满足条件，结果：6c = 0for i = 0 to 9for j = 0 to 9for k = 0 to 9if i * 100 + j * 10 + k + k * 100 +

2、 j * 10 + i = 1333 then c = c + 1next k, j, iprint cl宴会上共有1225次握手，每一位参加的人对其他的人都有同样的礼节，那么与会人士有多少，结果：50人数握手数1 02 13 34 6n个人的握手数为：1+2+3+。+n-1=1225，解得n=50l司机开车时里程表读数为一回文数12321公里，开了2小时后里程表又是一个最小回文数，求司机的开车速度。结果：50下一个最小回文数肯定是12421，（12421-12321）/2=50n = 12321do n = n + 1 a = n 10000 b = n 1000 mod 10 c = (n

3、 mod 100) 10 d = n mod 10loop while a <> d or b <> cprint (n - 12321) / 2l好啊好+ 真的好- 真的好啊如果是数字，求该四位数。结果：1098for h = 1 to 9for a = 0 to 9for z = 1 to 9for d = 0 to 9if h * 100 + a * 10 + h + z * 100 + d * 10 + h = z * 1000 + d * 100 + h * 10 + a thenprint h; a; hprint z; d; hprint z; d; h;

4、 aend ifnext d, z, a, hl求500以内含500能被5或9整除的所有自然数的倒数之和?按四舍五入的方式精确到小数点后第二位1.48private sub command1_click()s = 0for i = 1 to 500if i mod 5 = 0 or i mod 9 = 0 then s = s + 1 / inext iprint round(s, 2)end subl求s=1/2+2/3+3/5+5/8的前30项的和（注：该级数从第二项开始，其分子是前一项的分母，'其分母是前一项的分子与分母的和）要求：按四舍五入的方式精确到小数后第二位结果：18.4

5、6private sub command1_click()fz = 1: fm = 2: s = fz / fmfor i = 2 to 30 t = fz: fz = fm: fm = t + fz: s = s + fz / fmnextprint round(s, 2)end subl当m的值为50时，计算：t=1-1/(2*2)-1/(3*3)-1/(m*m) 四舍五入精确到小数点第四位 结果：0.3749private sub command1_click()t = 1for n = 2 to 50t = t - 1 / (n * n)nextprint round(t, 4)end

6、 subl求1 . 50之间所有整数能构成直角三角形的三边的组数。例如；3*3+4*4=5*5.它构成直角三角形，所以3、4、5为一组，但视4、3、5与3、4、5为同一组。结果：20private sub command1_click()n = 0for a = 1 to 50for b = 1 to 50for c = 1 to 50if a 2 + b 2 = c 2 and b > a and c > b then n = n + 1: print a, b, cnext c, b, aprint nend subl求100.900之间相差为12的素数对的对数,(难题)，结果

7、：7private sub command1_click()dim a(800)num = 0for i = 100 to 900if isp(i) then a(num) = i: num = num + 1end ifnextfor i = 1 to num if a(i) - a(i - 1) = 12 then n = n + 1nextprint nend subpublic function isp(n)isp = truefor i = 2 to sqr(n) if n mod i = 0 then isp = falsenextend functionl求200,300之间的有

8、奇数个不同因子的最大整数，（在计算因子个数时，包括1和该数本身）结果：289private sub command1_click()for n = 300 to 200 step -1 c = 0 for i = 1 to n if n mod i = 0 then c = c + 1 next i if c mod 2 = 1 then print nnext nend subl求（200,300）有奇数个不同因子的最小整数。（在计算因子个数时，包括1和该数本身），结果：225 for n = 200 to 300 c = 0 for i = 1 to n if n mod i = 0 th

9、en c = c + 1 next i if c mod 2 = 1 then print nnext nl求1000,2000范围内由小到大第100个索数，结果： 1721private sub command1_click()n = 0for i = 1000 to 2000 if isp(i) then n = n + 1 if n = 100 then print i: exit for end ifnext iend subpublic function isp(n)isp = truefor i = 2 to sqr(n) if n mod i = 0 then isp = fal

10、senextend functionl求具有abcd=(ab+cd)2性质的四位数的个数。结果：3n = 0for i = 1000 to 9999 ab = i 100 cd = i mod 100 if i = (ab + cd) 2 then n = n + 1nextprint nl求四位奇数中，各位数之积（积不为0)是60的倍数的数之和。结果：3456254sum = 0for n = 1001 to 9999 step 2 a = n 1000 b = n 100 mod 10 c = n 10 mod 10 d = n mod 10 e = a * b * c * d if e

11、<> 0 and e mod 60 = 0 then sum = sum + nnext nprint sumls=1+1/(1+2)+1/(1+2+3)+1/(1+2+3+n)，当n的值为50时，求s得值，结果：1.96078sum = 0s = 0for n = 1 to 50 s = s + n sum = sum + 1 / snextprint suml300,800范围内同时满足以下两个条件的十进制数，（1）其个位数与十位数字之和除以10所得的余数是百位数字；（2）该数是素数，求满足上述条件的最大的三位十进制数。结果：761private sub command1_cl

12、ick()for n = 800 to 300 step -1 b = n 100 s = n 10 mod 10 g = n mod 10 if (g + s) mod 10 = b and isp(n) then print nnextend subpublic function isp(n)isp = truefor i = 2 to sqr(n) if n mod i = 0 then isp = falsenextend functionl斐波那契数列的前二项是1 、1 其后每一项都是前面两项之和，求：10000000以内最大的斐波那契数？9227465a = 1: b = 1: c

13、 = a + bdoa = b: b = c: c = a + bloop while c < 10000000print b注意输出的是b而不是cl一个数如果恰好等于它的所有真因子之和，这个数就称为完数。求（1，100）之间的最大完数？结果：28，（100以内只有6和28两个完数）for n = 100 to 1 step -1 s = 0 for i = 1 to n - 1 if n mod i = 0 then s = s + i next if s = n then print nnextl一个数如果恰好等于它的所有真因子之和，这个数就成为完数。例如：6的真因子为1 2 3 而

14、6=1+2+3 ，因此，6是“完数”。求(8100,8200)之间的完数？结果：8128for n = 8200 to 8100 step -1 s = 0 for i = 1 to n - 1 if n mod i = 0 then s = s + i next if s = n then print nnextl已知：f(0)=f(1)=1 f(2)=0 f(n)=f(n-1)-2f(n-2)+f(n-3),(n>2)求f(0)到f(50)的所有51个值中的最大值？程序设计题：结果：598325dim f(50)f(0) = 1: f(1) = 1: f(2) = 0: max =

15、f(0)for n = 3 to 50 f(n) = f(n - 1) - 2 * f(n - 2) + f(n - 3) if f(n) > max then max = f(n)nextprint maxl已知：a1=1 a2=1/(1+a1) a3=1/(1+a2） a4=1/(1+a3).求a50(按四舍五入的方式精确到小数点后第三位）观察：an=1/(1+an-1)，结果：0.618dim a(50) as doublea(1) = 1for n = 2 to 50 a(n) = 1 / (1 + a(n - 1)nextprint a(50)l已知fibonacci数列：1,

16、1,2,3,5,8.它可由下面公式表达：f(1)=1 if n=1 f（2）=1 if n=2 f(n)=f(n-1)+f(n-2) if n>2 试求f(45)值 提示：最好使用递推法求解（用递推法要用好长的时间还可能得不到解，下面是递推法）结果：1134903170private sub command1_click()print f(45)end subpublic function f(n)if n = 1 or n = 2 then f = 1else f = f(n - 1) + f(n - 2)end ifend function最好用下面方法：dim x(45) as l

17、ongx(1) = 1: x(2) = 1for i = 3 to 45 x(i) = x(i - 1) + x(i - 2)nextprint x(45),l有一个三位数满足下列条件（1）此三位数的三位数字各不相同（2）此三位数等于它的各位数字的立方和。试求所有这样的三位数之和(即水仙花数之和)，结果：1301sum = 0for b = 1 to 9 for s = 0 to 9 for g = 0 to 9 n = b * 100 + s * 10 + g if b <> s and s <> g and b <> g and n = b 3 + s

18、3 + g 3 then sum = sum + n end ifnext g, s, bprint suml已知s1=2 s2=2+4 s3=2+4+6 s4=2+4+6+8 s5=2+4+6+8+10.求s=s1+s2+s3+s4+s5+.+s20的值，(仔细观察得出：sn=sn-1+2*n)，结果：3080sum = 0s = 0for n= 1 to 20 s = s + 2 * n sum = sum + snextprint sum 一l爱因斯坦上楼梯，一次跨2级剩1级，一次跨4级剩3级，一次跨5级剩正好（其腿也太长），求楼梯最少有多少级？同题：韩信点兵，2人一组剩1人，4人一组剩

19、3人，5人一组正好，求这队士兵最少多少人？private sub form_click()s = 1do while s mod 2 <> 1 or s mod 4 <> 3 or s mod 5 <> 0 s = s + 1loopprint send sub二l当n=50时，求下列级数和：s=1/（1*2）+1/（2*3）+1/（n*(n+1)按四舍五入的方式精确到小数点后第四位。结果：0.9804private sub form_click()s = 0for n = 1 to 50 s = s + 1 / (n * (n + 1)next nprin

20、t round(s, 4)end sub三l计算y=1+2/3+3/5+4/7+n/(2*n-1)的值，n=50,要求：按四舍五入的方式精确到小数点后第二位。结果：26.47private sub form_click()y = 0for n = 1 to 50 y = y + n / (2 * n - 1)next nprint round(y, 2)end sub四l水仙花数和。结果是：1301private sub form_click()y = 0for b = 1 to 9 for s = 0 to 9 for g = 0 to 9 n = 100 * b + 10 * s + g

21、if n = b 3 + s 3 + g 3 then y = y + nnext g, s, bprint yend sub五l把一张一元钞票，换成一分、二分和五分硬币，每种至少11枚，有多少种方案？结果：13（同题：百马驮百瓦、百钱买百鸡。）private sub form_click()s = 0for a = 11 to 100 for b = 11 to 50 for c = 11 to 20 if a + b * 2 + c * 5 = 100 then s = s + 1next c, b, aprint send subvb期末范围题总汇 1 求随机10个整数的最大值、最小值、

22、平均值以及和； dim a(1 to 10) private sub command1_click() randomize picture1.print "产生的随机数为：" for i = 1 to 10 a(i) = int(rnd * 99 + 1) picture1.print a(i); next i picture1.print end sub private sub command2_click() dim max, min, ave max = a(1) min = a(1) ave = 0.1 * a(1) for i = 2 to 10 if a(i)

23、> max then max = a(i) if a(i) < min then min = a(i) ave = ave + 0.1 * a(i) next i picture1.print "最大数为：" max picture1.print "最小数为：" min picture1.print "平均数为：" ave end sub 2.求水仙花数 private sub form_click() dim a, b, c as integer 'a(个)b(十)c(百) for a = 0 to 9 for

24、b = 0 to 9 for c = 1 to 9 if a 3 + b 3 + c 3 = a + 10 * b + 100 * c then msgbox 100 * c + 10 * b + a end if next c next b next a end sub 3.百元买百鸡问题； option explicit const a = 5, b = 3, c = 1 private sub form_activate() dim i as integer, j as integer, k as integer, n as integer for i = 1 to 100 for j

25、= 1 to 100 for k = 1 to 100 if i * a + j * b + k * c = 100 then n = n + 1 list1.additem " 公鸡：" & i & " 母鸡：" & j & " 小鸡：" & k end if next: next: next msgbox "共有组合：" & n end sub 求1000以内的所有完数 一个按钮里调用的过程 private sub command1_click() outputw

26、annumber 1000 end sub 4.'求完数的过程 private sub form_click() dim i, j as integer for i = 4 to 100 temp = 1 for j = 2 to i / 2 if i mod j = 0 then temp = temp + j end if next j if temp = i then print i end if next i end sub 5.求各位数字之和 input a:'输入任意数 do b=a mod 10:'取a的末位数 sum=sum+b:'求和 a=a

27、10:'去掉末位数 loop until a=0 print "sum="sum:'输出 end （dim n as string input n for i = 1 to len(n) sum=sum+val(mid(n,i,1) next i print sum） 6.求最小公倍数 private sub form_load() form1.autoredraw = true dim n1%, m1%, m%, n%, r% n1 = inputbox("输入n1") m1 = inputbox("输入m1") i

28、f m1 > n1 then '为了求最小公倍数，增加m,n变量 m = m1: n = n1 else m = n1: n = m1 end if do r = m mod n if r = 0 then exit do m = n n = r loop print n1; "," m1; "的最大公约数为" n print "最小公倍数=", m1 * n1 / n end sub 7.求逆序数（感觉题目类型太多） 8. 级数有限项求和问题（题目类型太多） 9. 求质因子问题 private sub command1

29、_click() dim n as integer, i as integer n = val(inputbox("请输入2的整数:") i = 2 do if n mod i = 0 then print i; n = n i else i = i 1 end if loop while n 1 end sub 10. 字符统计 option base1 option explicit private sub command1_click() dim i as integer，a(26) as integer，n as integer dim s as string*1,

30、strl as string strl=text1 n=len(strl) for i=1 to n s=mid(strl,i,1) if ucase(s)>=”a” and ucase(s)<=”z” then a(asc(ucase(s)-64)+1 end if next i for i=1 to 26 list1.additem chr(64+i) & “:” & a(i) netx i end sub private sub command_click() end end sub 第二大题 1. 判定素数过程 function isprime(num as long) as boolean if num < 2 then isprime = false: exit function dim i as long for i = 2 to sqr(num) if (num mod i) = 0 then isprime = false exit function end if next i isprime = true end function private sub command1_click() dim i as long for i = 1 to 1000 if isprime(i) then pr