C语言解答

#include <stdio.h>
int gcd(int m, int n){
    return (n>0) ? gcd(n, m%n) : m;
}
int lcm(int m, int n){
    return m * n / gcd(m,n);
}
int main(){
    int n,a,b,c;
    scanf("%d",&n);
    while(n--){
        scanf("%d %d",&a, &b);
        printf("%d %d\n", gcd(a, b), lcm(a, b));
    }
    return 0;
}

C++解答

#include<iostream>
using namespace std;
int gcd(int a,int b){
	if(b==0)
		return a;
	else
		return gcd(b,a%b);
}
int main(){
	int n;
	cin>>n;
	while(n--){
		int a,b,max,min;
		cin>>a>>b;
		min=gcd(a,b);
		max=a*b/min;
		cout<<min<<" "<<max<<endl;

	}
}

Java解答

import java.util.Scanner;
public class Main
{
	public static void main(String[] args)
	{
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		while(n-->0)
		{
			int i=scan.nextInt();
			int j=scan.nextInt();
			int x=i,y=j;
			if(i>=j){
				int a=i%j;
				while(a!=0){
					i=j;
					j=a;
					a=i%j;
				}
				System.out.print(j+" ");
			}else{
				int t;
				t=j;
				j=i;
				i=t;
				int a=i%j;
				while(a!=0){
					i=j;
					j=a;
					a=i%j;
				}
				System.out.print(j+" ");
			}
			System.out.println((x*y)/j);
		}
	}
}