,VB计算π的值
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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