1415 - C语言-公约公倍

5 7

1 35

Solution C

#include<stdio.h>
int main()
{
	int a,b,max,min,c,d,temp,t;
	scanf("%d%d",&a,&b);
	c=a,d=b;
	for(;;)
	{
		if(a<b)
			temp=a,a=b,b=temp;
		if(a%b==0)
		{
			max=b;
			break;
		}
		else
		t=a%b,a=b,b=t;
	}
	min=(c*d)/max;
	printf("%d %d\n",max,min);
	return 0;
}

Solution C++

#include<iostream>
#include<fstream>
#include<string>
#include<cstring>
#include<cmath>
using namespace std;
int  gcd(int x,int y)
{
	int temp;
	if(x<y)  {temp=x;x=y;y=temp;}
	if(y==0) return x;
	gcd(y,x%y);
}
int go(int n,int m)
{
	return (n/gcd(m,n)*m);
}
int main()
{
	//ifstream cin("aaa.txt");
	int i,j,n,m;
	cin>>n>>m;
	cout<<gcd(n,m)<<" "<<go(n,m)<<endl;


return 0;
}