1楼：匿名用户
'其实不难，运行效果如
上，**如下
'若要判断三位数、四位数、五位数……以内的略改一下**即可'有问题可找我
privatesub***mand1_click()dimiasinteger
dima,basinteger
dimnasdouble
print"100以内的灵巧数有："
fori=1to99
a=int(i/10)'取十位
b=imod10'取个位
n=sqr(i+(b*10+a))ifint(n)=nthen'判断n是否为整数printi
endif
nexti
endsub
2楼：匿名用户
innodb_data_home_dir = /longxibendi/mysql/mysql/var/
#innodb_data_file_path = ibdata1:1g:autoextend
innodb_data_file_path = ibdata1:500m;ibdata2:2210m:autoextend #表空间
innodb_file_io_threads = 4 #io线程数
vb编写程序找灵巧数
3楼：庆年工坊
subs()
fori=10to99
a=sqr(i+(imod10)*10+i\10)
ifint(a)=athenprinti;
next
endsub
编写vb程序，找出给定范围内所有的回文数
4楼：匿名用户
privatesub***mand1_click()dimaasstring,nasinteger,iasinteger
a=text1.text
n=len(a)
fori=1ton
ifmid(a,i,1)<>mid(a,n-i+1,1)thenexitfor
nexti
ifi=n+1then
print"是回文数
"else
print"不是回文数"
endif
endsub
5楼：刘师傅的一生
回文？都不知道你在说什么，我要举报你
求vb高手:编写程序，找出10000~99999的所有回文数!
6楼：匿名用户
private sub ***mand1_click()dim i as long, j as longdim x1 as long, x2 as longdim str as string
dim flag as boolean
str = "从 10000 到 99999 之间的回文数有：" & vbcrlf
for i = 10000 to 99999flag = false
for j = 1 to len(cstr(i))if mid(cstr(i), j, 1) <> mid(cstr(i), len(cstr(i)) - j + 1, 1) then
flag = true
exit for
end if
next j
if flag = false then str = str & i & " "
next i
msgbox str
end sub
7楼：匿名用户
dim str as string
dim i as integer
dim j as integer
dim k as integer
dim l as integer
dim hwdata(899) as integerl=0for i=1 to 9
for j=0 to 9
for k=0 to 9
str=trim(str(i)) & trim(str(j)) & trim(str(k)) & trim(str(j)) & trim(str(i))
hwdata(l)=val(str)
l=l+1
next k
next j
next i
hwdata中保存得是所有得回文数据
8楼：匿名用户
把这些数的个位，十位。百位，千位，万位分别用变量保存起来，只要对比个位跟万位相等，同时十位跟千位相等的就是了。
9楼：桑嫒运昕
楼上有点投机取巧的嫌疑哦，呵呵~~~也试试下面这个啦：
dima
asinteger,bas
integer,cas
integer
fora=1
to9forb=0
to9forc=0
to9printa&
b&c&
b&anext
**ext
bnexta
编制vb程序，找出1--1000之间的全部“完备数”
10楼：匿名用户
添加一个按钮
**如下：
private sub ***mand1_click()
dim perfect as string
for i = 1 to 1000
if isperfect(i) = true then '如果i完备数
perfect = perfect & "，" & i '将所有的完备数存放在字符串里
end if
next i
msgbox "1到1000的完备数有：" & mid(perfect, 2, len(perfect) - 1) '输出完备数
end sub
function isperfect(byval n as long) as boolean '判断是否为完备数
dim i as long
dim x as long
for i = 1 to int(n / 2)
if n mod i = 0 then
x = x + i
end if
next
if x = n then
isperfect = true
else
isperfect = false
end if
end function
结果如下：
1到1000的完备数有：6，28，496
11楼：喝多了跑偏
两个函数，一个找约数，一个做for循环，然后测试。
vb编写function过程，判断数字是否是回文数。程序要求输入一系列数字，找出所有的回文数并显示在文本框中 50
12楼：网海1书生
private sub ***mand1_click()dim x as long, a as integerfor a = 1 to 5
x = val(inputbox("请输入一个整数"))if fac(x) then text1.text = text1.text & x & " "
next
end sub
function fac(x as long) as booleanfac = (x = val(strreverse(x)))end function
怎么样，我这函数只用一行语句就解决了！原理就是把一个数左右反转，如果与原来的数相等，那么就是回文数。
用vb编写程序，找出所有小于或等于100的自然数对。
13楼：匿名用户
private sub form_activate()dim i%, j%, x%, y%
for i = 1 to 100
for j = 1 to i
x = i + j
y = i - j
if y > 0 then
if sqr(x) = int(sqr(x)) and sqr(y) = int(sqr(y)) then
print i, j
end if
end if
next j
next i
end sub
vb设计一个程序：找出1～9999之间的全部的同构数
14楼：灰色
publicfunctionlength(byvalresasinteger)asinteger
length=0
dores=res/10
length=length+1
loopwhileres<>0
endfunction
publicfunctionisisomorphic(byvalnumasinteger)asboolean
pow=num^2
mult=1
fori=1tolength(num)
mult=mult*10
ifpowmodmult=numthen
isisomorphic=ture
else
isisomorphic=false
endif
next
endfunction
publicsubjudge()
fori=1to9999
ifisisomorphic(i)thendebug.printi;" ";i*i
endif
next
endsub
这个不难，复制上去的时候不知道空格有没有出来
15楼：匿名用户
dim i as long
dim l as integer
dim j as long
for i = 1 to 9999
j = i * i
l = len(cstr(i))
if right(cstr(j), l) = cstr(i) then debug.print i
next i
16楼：仁晏五淑然
#include
intmain()}
用vb编写程序(找出三个数中的最大数，通过inputbox函数获得三个数)
17楼：匿名用户
private sub ***mand1_click()dim a as single
dim b as single
dim c as single
dim max as single
a = val(inputbox("输入第1个数"))max = a
b = val(inputbox("输入第2个数"))if b > max then max = bc = val(inputbox("输入第3个数"))if c > max then max = cmsgbox "最大数是" & max
end sub