写两个函数,求最大公约数和最小公倍数

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;}

#include#include#include/*利用辗转相除法求最大公约数*/intgcd(intn,intm){intr;if(n

你的c语言写的好乱,我帮你整理如下：#include "stdio.h"int gcd(int a,int b){  &

#includeintgcd(intm,intn)//最大公约数{intt;if(m

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

include"stdio.h"intgongyue(inta,intb);intgongbei(inta,intb);voidmain(){inta,b,temp,gy,gb;clrscr();pr

#includevoidmain(){inta,b;printf("请输入两个数:");scanf("%d%d",&a,&b);intgys(inta,intb);intgbs(inta,intb);

#includevoidmain(){inthcf(int,int);intlcd(int,int,int);intu,v,h,l;scanf("%d,%d",&u,&v);h=hcf(u,v);pr

include"stdio.h"intgongyue(inta,intb);intgongbei(inta,intb);voidmain(){inta,b,temp,gy,gb;clrscr();pr

#includeintmain(){inta,b,imax,imin,tend;intmax(int,int);printf("pleaseinserttwonumbers:");scanf("%d%

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

#includeintgcd(int,int);voidmain(){inta=0,b=0;intmax=0,min=0;scanf("%d%d",&a,&b);max=gcd(a,b);min=a*

#includeintcal(intm,intn){intret=0;ret=m%n;returnret;}intmain(intargc,char**argv){intm,n,max,min

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#include"stdio.h"#include"conio.h"main(){inta,b,num1,num2,temp;printf("pleaseinputtwonumbers:\n");sc

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

#include//求a和b最大公约数：intyue(inta,intb){intk=1;intt=a>b?b:a;//a大取b,否则取afor(inti=1;i

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

functioncommonDivisor(x,y){if(isNaN(x)||isNaN(y))return"非法输入数据";varresult=[];varmax=Math.max(x,y);va