joj 1062 Computer Versus Mankind 非递归最大公约数最小公倍数

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本文介绍了一个简单的程序，用于计算两个正整数的最大公约数(GCD)和最小公倍数(LCM)。通过非递归的方式实现了GCD的计算，并利用GCD计算LCM。输入为多组数据，每组包含两个正整数，输出每组数据的GCD和LCM。

1062: Computer Versus Mankind

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	2437	1119	Standard

Mike is a fan on programming. He's sure that computer can do everything faster than mankind does if a kind of task is repeated certain amount of times. Now he is proving this conclusion by calculate the gcd and lcm of two positive integers a and b, where a,b>1.

The gcd of two positive integers a and b, where a,b>1, is the greatest common dividor of them. The lcm of two positive integers a and b, where a,b>1, is the least common multiple of them. For example, the gcd of 6 and 8 is 2 while the lcm of them is 24.

Input Specification

The input consists of multilines, each line consists of two integers a and b(0<a,b<2³¹), separated by a single space. a=b=0 marks the end of input, which you shouldn't process.

Output Specification

For each pair a and b in the input, you should output the gcd and lcm on a single line, separated by a single space too.

Sample Input

6 8
5 2
0 0

Sample Output

2 24
1 10

Problem Source: 1st JOJ Cup Online VContest Warmup Problem

/*
	水题
 最大公约数，最小公倍数
 非递归版 gcd
*/

#include <stdio.h>
int m,n;
int gcd(int a,int b)
{
	int tmp;
	if(a < b)
	{
		tmp = a;
		a = b;
		b = tmp;
	}

	while(a % b != 0)
	{
		tmp = b;
		b = a % b;
		a = tmp;
	}
	return b;
}
int lcm(int a,int b)
{
	return a * b / gcd(a,b);
}
 int main()
 {
	while(scanf("%d %d",&m,&n),m != 0 && n != 0)
	{
		if(m == 0 || n == 0)
			break;
		printf("%d %d\n",gcd(m,n),lcm(m,n));
	}
	return 0;
 }