编写函数gcd_lcm求两个整数的最大公约数和最小公倍数

编写主函数根据用户输入的两个整数调用函数gcd_lcm求解并输出它们的最大公约数和最小公倍数。

输出语句格式：
cout<<"Please input two integers: ";
cout<<"The GCD of them is "<<…<<endl;
cout<<"The LCM of them is "<<…<<endl;

输入输出实例1：
Pleae input two integers: 12 24
The GCD of them is 12
The LCM of them is 24

输入输出实例2：
Pleae input two integers: 12 56
The GCD of them is 4
The LCM of them is 168

参考程序模板：

#include <iostream>
using namespace std;
//only necessary output statments are given here
void gcd_lcm( int num1, int num2, int &gcd, int &lcm ) ;
int main()
{
    //to add other statements you need
    cout<<"Please input two integers: ";

    //to add other statements you need
    cout<<"The GCD of them is "<<gcd<<endl;
    cout<<"The LCM of them is "<<lcm<<endl;
    return 0;
}
void gcd_lcm( int num1, int num2, int &gcd, int &lcm ) 
{
    //to add statements to solve gcd  and lcm

}

#include <stdio.h>
int gcd(int n,int m)
{
    int i=1,x,y;  /*始终让x最大，y最小*/ 
    if(n>m) {x=n; y=m;}
    else x=m;y=n;
    while (i!=0)
    {
        i=x%y;
        if(i==0) return y;
        else {x=y;y=i;}
    }
}

int lcm(int GCD,int n,int m)
{
    int x;
    x=n*m/GCD;
    return x;
}

int main()
{
    int n,m,GCD,LCM;
    while (scanf("%d,%d",&n,&m)!=EOF&&m*n!=0)
    {
        GCD=gcd(n,m);
        LCM=lcm(GCD,n,m);
        printf("GCD=%d,LCM=%d\n",GCD,LCM);
    }
    return 0;
}