写出两个函数,分别求两个整数的最大公约数和最小公倍数

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){  &

#includemain(){inta,c,b,d;scanf("%d%d",a,b);c=a+b;d=a*b;printf("%d%d",c,d);}再问：采用函数的方法再答：先输入两个数，然后执行

这个简单：#includeusingnamespacestd;intHe(intx,inty){intz;z=x+y;returnz;}intCha(intx,inty){intz;z=x-y;ret

#include"stdafx.h"#includeintsct(intm,intn){inttemp,a,b;if(m>y;g=sct(x,y);cout再问：是fun函数吗？测试用的主函数

#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

publicintmax(inta,intb){returna>b?a:b;}publicdoublemax(doublea,doubleb,doublec){doublet=a>b?a:b;retu

1925=5×5×7×11；两个商都是1925的因数且互质，而且和为16，所以这两个商分别为5和11；这两个整数分别是：1925÷5=385，1925÷11=175；答：这两个整数分别是385与175

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

#includeintmaxY(intm,intn){inti;for(i=n;i>0;i--)if(m%i==0&&n%i==0)break;return(i);}intminB(intm,intn

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

#includeclassdigital{private:intm_num;public:digital(intnum=0){this->m_num=num;}digital(){};intGetNu

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%

#include"stdio.h"intfunc(inta,intb){inttemp;while(b!=0)/*利用辗除法,直到b为0为止*/{temp=a%b;a=b;b=temp;}retu

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

按照问题给出三个重载,并在main中演示.已在私信中给出完整的例子,请注意查收!

#include<iostream.h>classdigital{private:intm_num;public:digital(intnum=0){this->m_num=num;

templateTAdd(constT&a,constT&b){returna+b;}voidmain(){//根本没必要重载inta=0,b=0;CRealRealA(1.0,),RealB(1.0

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