求任意两个整数的最大公约数和最小公倍数,用java编程

intfun(inta,intb){inti,j,m,n;if(a>b){m=a;a=b;b=m;}i=a;j=b;while((n=j%i)!=0){j=i;i=n;}returni;}

packagecom.fmzrt;/***求两个数的最大公约数和最小公倍数*@authorkele**/publicclassGongyueGongbeiShu{/***求两个数的最大公约数*@par

*用visualfoxpro语言求任意两个整数的最大公约数程序INPUT'第1个正整数：'TOmINPUT'第2个正整数：'TOnFORi=MIN(m,n)TO1STEP-1IFm%i=0.and.n

#includeintgongyue(intm,intn){intr;if(m==n)returnm;elsewhile((r=m%n)!=0){m=n;n=r;}returnn;}intgongbe

#includeintgongyue(intm,intn){intr;if(m==n)returnm;elsewhile((r=m%n)!=0){m=n;n=r;}returnn;}voidmain(

#includeintyue(inta,intb){inti,y;for(i=b;i>=1;i--)if(a%i==0&&b%i==0){y=i;break;}returny;}intbei(inta

公约数公倍数vara,b,i,p,n,m:longint;beginread(a,b);ifa>bthenp:=aelsep:=b;whilei

PrivateSubCommand1_Click()DimSAsLongS=YueOrBei(24,32)MsgBox"24和32的最大公约数是："&S,vbInformationS=YueOrBei

intmax_common_divisor(inta,intb){//最大公约数intlarge_num,small_num,r;if(a>b){large_num=a;small_num=b;}el

一般思维：#includeusingnamespacestd;intmain(){intx,y,max,min,k;inttem0,tem1;couty;max=x>y?x:y;min=x

EXCELVBA求公约数函数过程如下：PublicFunctionGYS(MAsInteger,NAsInteger)AsIntegerDimYS,YS1,MX,MNMX=WorksheetFunct

这种问题太基础了,证法很多,先举两种：法一：设(a,b)=c,则a=cm,b=cn且(m,n)=1则[a,b]=mnc所以(a,b)[a,b]=c*mnc=mnc^2=ab法二：对a,b进行标准分解a

PrivateSubForm_Load()DimXAsLong,YAsLong,KAsLongDimAAsLong,BAsLongA=6B=8X=IIf(A>=B,A,B)'x存入最大值Y=IIf(A

例子一:#includevoidmain(){voidz1(inta,intb);intx,y,temp;scanf("%d%d",&x,&y);if(x

这种方法是数学里面的辗转相除法.具体思路为：假设a=15,b=9那么r=a%b则r=15%9=6,由于余数不为0,所有9不是最大公约数现在令a=b,b=r,那么a=9,b=6,继续求r=a%b则r=9

#include"stdio.h"voidmain(){\x05intnum1,num2,temp,a;\x05printf("pleaseinputtwonumbers:\n");\x05scanf

programex1;varj,m,n,m1,n1,a,b:integer;beginreadln(m,n);ifm再问：拜托，错了再答：programex1;varj,m,n,m1,n1,a,b:i

//输入的两个数请用空格分隔,如：1218#includeintmain(){intp,r,n,m,temp;printf("请输入两个正整数n,m:");scanf("%d%d,",&n,&m);i

①#includeinthcf(inta,intb){\x09intc;\x09while(b){c=b;b=a%b;a=c;}\x09returna;}intlcd(inta,intb,intc){

完整程序如下：#includefun(intx,inty){intr;if(x>y){x=x;y=y;}r=x;x=y;y=r;r=x%y;while(r!=0){x=y;y=r;r=x%y;}ret