,VB计算π的值

PrivateSubCommand1_Click()DimsumAsDoubleDimiAsLongDimnAsDoublen=((1/(10^-5))+1)/2'正好等于10^-5时候n的值

FunctionS(rAsSingle)AsSingle'Function过程Constpi=3.1415926535898S=pi*r^2EndFunctionPrivateSubForm_Clic

OptionExplicitPrivateSubCommand1_Click()Label1.Visible=TrueDimrAs__single___Dim____s___AsDouble_cons

for循环默认有个setp1,i会每次自动增1,循环体里还有个i=i+1,所以,第一次sum从0变为1,然后i为了3,符合条件进入第二轮,sum=1+3=4,接下来由于i变为5,不满足再问：循环结束了

*优先级最高,然后到/,其后到\相当于(abs(-5)*5)\(5/5)=(5*5)\1=25

一个text控件PrivateSubForm_Load()DimRAsDoubleDimSAsDoubleR=Val(Text1.Text)S=3.14*(R^2)MsgBoxStr(S)EndSu

阶乘n!=n*(n-1)(n-2)(n-2)···3*2*1

PrivateSubCommand1_Click()DimnAsInteger,iAsIntegerDimsumAsDoublen=Val(InputBox("PleaseEnterN:"))sum=

PublicFunctions(n)Dimi,j,t,resFori=1Tont=1Forj=1Toit=t*jNextres=res+tNexts=resEndFunction

elseif连起来就行了PrivateSubCommand1_Click()DimaAsSingle,bAsSingle,cAsSingle,dAsSinglea=Val(Text1.Text)b=V

你好!PrivateSubForm_Click()DimiAsIntegerDimsAsIntegerFori=1To100s=s+iNextiPrintsEndSu

PrivateSubCommand2_Click()DimIAsLong,SAsLongForI=1To100IfIMod2=0ThenS=S-IElseS=S+IEndIfNextPrintSEnd

具体分析如下:设r=Int(Rnd*4+0.5)则Rnd=Int((r-0.5)/4)∵0≤Rnd

PrivateSubForm_Click()Dimd1AsIntegerDimd2AsSingleDimd3AsIntegerDimd4AsStringDimd5AsIntegerDimd6AsInt

DimsignAsInteger,aAsLong,piAsDoublesign=-1Fori=1To100000a=2*i-1sign=-1*signpi=pi+(1/a)*signNextpi=pi

4+5\6*7/8Mod9＝4+5\42/8Mod9说明：先计算6*7＝42＝4+5\（5.25）Mod9说明：再计算42/8＝5.25＝4+5\5Mod9说明：下一步计算5\5.25,在计算之前先把

DimcAsObjectSetc=CreateObject("MSScriptControl.ScriptControl")c.Language="vbscript"MsgBoxc.Eval("2+4

dimaaslongdimbaslongdimcaslongdimdassinglex1,x2也是dim(定义变量一般都是用dim)d=b^2-4acif再问：constaAslongbAslongc

Functionfactorial(nAsInteger)AsLongDimiAsIntegerDimretAsLongret=1Fori=1Tonret=ret*iNextifactorial=re

意思是10的-2次幂你也可以理解成小数点往坐移两位