7.7程序示例例7.4例7-5:求三個(gè)正整數(shù)的最小公倍數(shù).方法:(1)編寫一個(gè)求兩個(gè)數(shù)的最小公倍數(shù)的函數(shù)lcm(2)首先調(diào)用lcm函數(shù),求出任意兩個(gè)數(shù)的最小公倍數(shù),然后再次調(diào)用lcm函數(shù),求出前兩個(gè)數(shù)的最小公倍數(shù)與第三個(gè)數(shù)的最小公倍數(shù).求兩個(gè)數(shù)的最小公倍數(shù)算法:tem=mdowhiletemmodn<>0tem=tem+mloopprintm;"和";n;"的最小公倍數(shù)是";tem由循環(huán)編寫出函數(shù)過(guò)程PrivateFunctionlcm(ByValmAsInteger,ByValnAsInteger)Dimtem%tem=mDoWhiletemModn<>0tem=tem+mLooplcm=temEndFunction求最小公倍數(shù)函數(shù)函數(shù)名=表達(dá)式,給函數(shù)賦值PrivateSubCommand1_Click()DimaAsInteger,bAsInteger,lAsLong,cAsIntegera=Text1b=Text2c=Text3l=lcm(lcm(a,b),c)Text4=lEndSub函數(shù)嵌套調(diào)用:求三個(gè)數(shù)的最小公倍數(shù)調(diào)用程序PrivateSubCommand1_Click()Dimnumber(10)AsInteger,iAsIntegerRandomizeFori=1To10number(i)=Int(Rnd*100)+1Text1=Text1&Str(number(i))NextiCallbubble_sort(number)

Fori=1To10Text2=Text2&Str(number(i))NextiEndSub調(diào)用過(guò)程生成數(shù)組輸出例7-8冒泡法排序。(要求程序中必須包含一個(gè)通用過(guò)程)PrivateSubBubble_Sort(Sort()AsInteger)DimIAsInteger,TemAsIntegerFori=1toubound(sort)-1forj=1toubound(sort)-iIfSort(j)>Sort(j+1)ThenTem=Sort(j)Sort(j)=Sort(j+1)Sort(j+1)=TemEndIfNextjNextiEndSub書上的寫法：PrivateSubBubble_Sort(Sort()AsInteger)DimIAsInteger,TemAsIntegerDimUbAsInteger,SwitchAsBooleanUb=UBound(Sort)

Switch=TrueDoWhileSwitchSwitch=FalseUb=Ub-1ForI=1ToUbIfSort(I)>Sort(I+1)ThenSwitch=TrueTem=Sort(I)Sort(I)=Sort(I+1)Sort(I+1)=TemEndIfNextILoopEndSub冒泡法排序DimnAsInteger,numAsLong

PrivateSubCommand1_Click()DimchAsString,iAsIntegerDimchar(15)AsStringDimbin()AsStringFori=0To9char(i)=Str(i)NextiFori=0To5char(10+i)=Chr$(Asc("A")+i)NextiPrintReDimbin(1)Calltrans(bin,char)Fori=UBound(bin)To1Step-1ch=ch+bin(i)NextiText3.Text=ch

EndSub定義窗體級(jí)變量在窗體的所有過(guò)程內(nèi)都有效把char數(shù)組定義為16個(gè)元素對(duì)應(yīng)余數(shù)0—9,A--E輸出轉(zhuǎn)換后的結(jié)果,倒取余[例7-9]把一個(gè)任意十進(jìn)制正整數(shù)轉(zhuǎn)換成N進(jìn)制數(shù)(N＜=16)調(diào)用PrivateSubTrans(Vary()AsString,St()AsString)DimRAsInteger,KAsIntegerK=0DoUntilNum=0R=NumModNK=K+1ReDimPreserveVary(K)Vary(K)=St(R)Num=Num/NLoopEndSub進(jìn)制轉(zhuǎn)換的通用過(guò)程:除N倒取余，直到商為0[例7-11]編寫一個(gè)遞歸函數(shù)，求兩個(gè)整數(shù)的最大公約數(shù)。采用循環(huán)來(lái)求兩數(shù)最大公約數(shù)PrivateSubCommand1_Click()Dimm,n,rAsLongm=Val(Text1.Text)n=Val(Text2.Text)Ifm<1Orn<1ThenText3.Text="數(shù)據(jù)錯(cuò)誤！"ElseDor=mModnm=nn=rLoopUntilr=0Text3.Text=CStr(m)EndIfEndSub用遞歸思想編寫求最大公約數(shù)函數(shù)PrivateFunctionGcd(ByValXAsLong,ByValYAsLong)DimRAsLongR=XModYIfR=0ThenGcd=YElseX=YY=RGcd=Gcd(X,Y)EndIfEndFunction遞歸邊界條件DimmAsLong,nAsLongPrivateSubCommand1_Click()DimgcdvalueAsLongIfm<>0Andn<>0Thengcdvalue=gcd(m,n)Text3.Text=Str(gcdvalue)EndIfEndSubm,n是模塊級(jí)變量函數(shù)調(diào)用語(yǔ)句PrivateSubText1_Change()m=Val(Text1.Text)EndSubPrivateSubText2_Change()n=Val(Text2.Text)EndSub在Text1_Change事件中給m,n賦值[例7-12]編寫程序，驗(yàn)證大于5的奇數(shù)可以表示成三個(gè)素?cái)?shù)的和。

1.判斷1個(gè)數(shù)是否是素?cái)?shù)的通用函數(shù)(掌握)PrivateFunctionprime(nAsInteger)AsBooleanDimiAsIntegerprime=FalseFori=2ToSqr(n)IfnModi=0ThenExitFunctionNextiprime=TrueEndFunction函數(shù)值結(jié)果為邏輯型在函數(shù)體內(nèi)用函數(shù)名＝表達(dá)式（或值）給函數(shù)1個(gè)函數(shù)值提前結(jié)束函數(shù)過(guò)程3、求出N以內(nèi)的素?cái)?shù)的通用過(guò)程編寫PrivateSubprime(A()AsInteger,NAsInteger)DimiAsInteger,IdxAsIntegerDimJAsIntegerFori=2ToNForJ=2ToSqr(i)IfiModJ=0ThenExitForNextJIfJ>Sqr(i)ThenIdx=Idx+1ReDimPreserveA(Idx)A(Idx)=iEndIfNextIEndSub數(shù)組做形參對(duì)2－n以內(nèi)的數(shù)判斷,如果是素?cái)?shù),則把它放到數(shù)組a中

思路：先求出該奇數(shù)以內(nèi)的素?cái)?shù)，然后用窮舉法判斷該奇數(shù)是否可以表達(dá)成3個(gè)素?cái)?shù)之和。（注：這3個(gè)素?cái)?shù)可以出現(xiàn)重復(fù)的，比如7=2+2+3）例7-12代碼OptionExplicitOptionBase1PrivateSubCommand1_Click()DimP()AsInteger,NAsInteger,LAsIntegerDimIAsInteger,JAsInteger,kAsInteger,FlgAsBooleanN=InputBox(“輸入一個(gè)大于5的奇數(shù)！")callprime(P,n)L=UBound(P)ForI=1ToLForJ=1ToLFork=1ToLIfP(I)+P(J)+P(k)=NThenPrintN;"=";P(I);"+";P(J);"+";P(k)Flg=TrueExitForEndIfNextkIfFlgThenExitForNextJIfFlgThenExitForNextIEndSub注意此兩句代碼順序不能顛倒百元買百雞問(wèn)題P是素?cái)?shù)數(shù)組，P(I)等都是數(shù)組元素。例7-13:親密對(duì)數(shù)PrivateSubCommand1_Click()DimiAsInteger,sum1AsInteger,sum2AsIntegerFori=1To5000Callsum_factors(i,sum1)Callsum_factors(sum1,sum2)Ifi=sum2Andi<>sum1ThenText1=Text1&i&","&sum1&vbCrLfEndIfnextiEndSubPrivateSubsum_factors(ByValnAsInteger,sumAsInteger)DimiAsIntegersum=0Fori=1Ton-1IfnModi=0Thensum=sum+iEndIfNextiEndSub親密對(duì)數(shù)的定義且排除了因子和等于本身的數(shù)據(jù)[例7-7]利用級(jí)數(shù)