Euclid problem

Euclid problem

时间: 1ms        内存:64M

描述:

From Euclid, it is known that for any positive integers A and B there exist such integers X and Y that AX + BY = D, where D is the greatest common divisor of A and B. The problem is to find the corresponding X, Y, and D for a given A and B.

输入:

The input will consist of a set of lines with the integer numbers A and B, separated with space ( A, B < 1, 000, 000, 001).

输出:

For each input line the output line should consist of three integers X, Y, and D, separated with space. If there are several such X and Y, you should output that pair for which X<=Y and | X| + | Y| is minimal.

示例输入:

4 6
17 17

示例输出:

-1 1 2
0 1 17

提示:

参考答案(内存最优[752]):

#include"stdio.h"
#include"math.h"
long gcd(long p,long q,long *x,long *y)
{
	long x1,y1;
	long g;
	if(q>p)
		return gcd(q,p,y,x);
	if(q==0)
	{
		*x=1;*y=0;
		return p;
	}
	g=gcd(q,p%q,&x1,&y1);
	*x=y1;
	*y=(x1-floor(p/q)*y1);
	return g;
}
int main()
{
	long a,b,x,y,c;
	while(scanf("%ld%ld",&a,&b)!=EOF)
	{
		c=gcd(a,b,&x,&y);
		printf("%ld %ld %ld\n",x,y,c);
	}
	return 0;
}

参考答案(时间最优[0]):

#include"stdio.h"
#include"math.h"
long gcd(long p,long q,long *x,long *y)
{
	long x1,y1;
	long g;
	if(q>p)
		return gcd(q,p,y,x);
	if(q==0)
	{
		*x=1;*y=0;
		return p;
	}
	g=gcd(q,p%q,&x1,&y1);
	*x=y1;
	*y=(x1-floor(p/q)*y1);
	return g;
}
int main()
{
	long a,b,x,y,c;
	while(scanf("%ld%ld",&a,&b)!=EOF)
	{
		c=gcd(a,b,&x,&y);
		printf("%ld %ld %ld\n",x,y,c);
	}
	return 0;
}

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