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1、第五章VB控制结构算法总结1、求最大公约数P67，例5-3PrivateSubCommand1_Click()DimnAsLong,mAsLong,rAsLongn=Int(Val(Text1.Text))m=Int(Val(Text2.Text))If(m<1Orn<1)ThenText3.Text="请输入合法数据"ElseDor=mModnm=nn=rLoopWhile(r<>0)Text3.Text=CStr(m)EndIfEndSub2、求N！。P70例5-4PrivateSubCommand1_C
2、lick()DimNAsInteger,iAsInteger,ResultAsLongResult=1'累乘前Result的初始值为1N=CInt(Text1.Text)Fori=1ToNResult=Result*iNextiLabel2.Caption=Str(N)+"!="+Str(Result)EndSub3、判断素数P72例5-5PrivateSubCommand1_Click()DimiAsInteger,jAsLongFori=2To30Forj=2ToSqr(i)IfiModj=0ThenEx
3、itForNextjIfj>Sqr(i)ThenPrintStr(i);EndIfNextiEndSub4、最小公倍数P78一阅读程序题（1）5、求一个数的因子及因子和，并判断完数。P75例5-8PrivateSubCommand1_Click()DimiAsInteger,jAsIntegerDimsumAsInteger,sAsStringText1.Text=""Fori=1To1000sum=0s=i&"="Forj=1Toi-1IfiModj=0Thensum=sum+js=s&j&"+"EndIf
4、NextIfi=sumThenText1.Text=Text1.Text&Left(s,Len(s)-1)&vbCrLfEndIfNextEndSub6、级数求和P75例5-9PrivateSubCommand1_Click()DimxAsDouble,tAsDouble,EpsAsDouble,SAsDouble,iAsIntegeri=1x=Val(Text1.Text)Eps=Val(Text2.Text)t=1S=1Dot=-t*x^2/((2*i-1)*(2*i))S=S+ti=i+1LoopUnt
5、ilAbs(t)<=EpsLabel3.Caption="cos("&x&")的值是"&SEndSub7、反序数P77例5-10PrivateSubCommand1_Click()DimaAsLong,iAsIntegerDimsAsStringa=InputBox("输入一个正整数","逆向输出")s=""Fori=1ToLen(CStr(a))s=Mid(CStr(a),i,1)&","&sNextiPrintsMsgBox"输出的结果为："&Left(s,Len(s)-1),vbDefaultButto
6、n1,"逆向输出"EndSub或PrivateSubCommand1_Click()DimaAsLongDimsAsStringa=InputBox("输入一个正整数","逆向输出")s=""Dok=aMod10s=s&Str(k)&","a=a10LoopWhile(a<>0)MsgBox"输出的结果为："&Left(s,Len(s)-1),vbDefaultButton1,"逆向输出"EndSub8、回文数。找出所有五位的回文数。PrivateSubForm_Click()DimiAsLong,jAsL
7、ongDimsAsString,s1AsStringFori=10000To99999s=CStr(i)s1=""Forj=1ToLen(s)s1=Mid(s,j,1)&s1NextjIfs=s1ThenList1.AddItemsNextiEndSub9、十进制转二进制PrivateSubCommand1_Click()DimxAsInteger,RAsInteger,SAsStringx=Val(Text1)DoR=xMod2x=x2S=CStr(R)&SLoopUntilx=0Text2=SEndSu
8、b10、十进制转十六进制PrivateSubCommand1_Click()DimxAsInteger,RAsInteger,SAsStringx=Val(Text1)DoR=xMod16x=x16IfR<=9ThenS=CStr(R)&SElseSelectCaseRCase10S="A"&SCase11S="B"&SCase12S="C"&SCase13S="D"&SCase14S="E"&S
