vb产生20个随机数,奇数一列

试一下：PrivateSubCommand1_Click()Dimarr(1To20),iFori=1To20Randomizearr(i)=Int(Rnd*101)NextfcarrPrintJoi

PrivateSubCommand1_Click()Me.Timer1.Enabled=TrueEndSubPrivateSubForm_Load()Me.Label1.Caption="第一个数："

是不是这样：（fomr的autodraw属性要设定为true）PrivateSubCommand1_Click()DimiAsIntegerDimRndNUmAsInteger'定义随机数DimN1A

PrivateSubCommand1_Click()Dima(10To99)AsIntegerFori=1To1000x=Int(90*Rnd())+10‘产生10~99的随机数a(x)=a(x)+1

PrivateSubCommand1_Click()DimxxAsIntegerDimNAsLong,FlagsAsBooleanDimiiAsIntegerDim素数个数AsIntegerDim素数

dima,max,min,sum,averagesum=0max=20min=50fori=1to20a=(Rnd*50)+50sum=sum+aifamaxthenmax=aprintanextpr

PrivateSubForm_click()Dimn(1To20)AsIntegerDimi,j,ave,tempRandomizeFori=1To20n(i)=Int((51*Rnd)+50)Pri

Dima(1To20)AsIntegerPrivateSubCommand1_Click()ClsScale(0,0)-(100,100)Foru=1To20a(u)=Int(Rnd*50)+50b=

PrivateSubCommand1_Click()RandomizeTimerFori=1To100x=Int(Rnd()*90+10)IfxMod2=0Thenm=m+xElsen=n+xEndI

PrivateSubCommand1_Click()Dima(1)AsIntegerFori=0To1a(i)=Int(Rnd*10+1)Ifa(0)a(1)ThenPrinta(i);NextiEn

OptionExplicitPrivateSubCommand1_Click()DimAAsString,iAsInteger,DDAsSingleRandomizeDoDD=Int(Rnd*1000

PrivateFunctionAverageSub()AsDoubleDimintTotalAsIntegerDimiAsIntergerintTotal=0Randomize()Fori=0To9i

可以先产生一个随机数来选择范围然后根据这个范围来选择随机数范围再产生一个随机数.比如OptionExplicitPrivateSubCommand1_Click()Dimx,yRandomizex=I

Dimsum,max,minSubaaa(ByRefsum,ByRefmax,ByRefmin)min=100'给最小值先给一个比较大的值Fori=1To100s=Int(Rnd(1)*(67-0))

窗体加一按钮,一个文本框PrivateSubCommand1_Click()RandomizeFori=1To100Text1.Text=Int(Rnd*90)+10NextEndSu

'窗体放一个按钮Command1PrivateSubCommand1_Click()Randomize'初始化随机数生成器DimJiShuHangAsString'奇数行DimOuShuHangAsS

PrivateSubCommand1_Click()RandomizeFori=1To20a=Int(Rnd()*21+30)Printa;IfiMod4=0ThenPrintNextEndSub这里

产生2~99之间的1次1个随机数randomizeInt((99-2)*Rnd+1)产生1~n之间的1次20个随机数RandomizeDimtemAsInteger,jAsInteger,iAsInt

sum=val(label1.caption)+val(label2.caption)

PrivateSubForm_Click()Dima(30)AsIntegerDimiAsIntegerRandomizeFori=1To30a(i)=Int(100*Rnd)NextiFori=1T