hdu 2669(拓展欧几里得)

Romantic

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

The Sky is Sprite.
The Birds is Fly in the Sky.
The Wind is Wonderful.
Blew Throw the Trees
Trees are Shaking, Leaves are Falling.
Lovers Walk passing, and so are You.
................................Write in English class by yifenfei



Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead.

 

Input

The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)

 

Output

output nonnegative integer X and integer Y, if there are more answers than the X smaller one will be choosed. If no answer put "sorry" instead.

 

Sample Input

   
   
   
   

    
    
    
    77 51
10 44
34 79
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2 -3
sorry
7 -3
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    解题思路：这道题目其实直接套模板就ok了，只是要稍微做点修改。。首先gcd(a,b)=1,其次x>=0,第一个好办，第二个条件可以这么处理：在刘汝佳的《算法竞赛入门经典》里，有一个这样的结论：设a，b，c为任意整数。若方程ax+by=c的一组整数解为(x0,y0),则它的任意整数解都可以写成(x0+kb',y0-ka'),其中a'=a/gcd(a,b),b'=b/gcd(a,b),k取任意整数。那么可以根据这个结论，如果x<0,可以把x一直加b',y一直减a'，直到x>=0为止。。
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    AC:
   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

typedef __int64 LL;
LL extend_gcd(LL a,LL b,LL &x,LL &y)
{
	if(a == 0 && b == 0) return -1; //无最大公因子
	if(b == 0)
	{
		x = 1; y = 0;
		return a;
	}
	LL d = extend_gcd(b,a%b,y,x);
	y -= (a/b)*x;
	return d;
}

int main()
{	
	LL a,b;
	while(cin>>a>>b)
	{
		LL x,y,d;
		d = extend_gcd(a,b,x,y);
		if(d == 1)
		{
			while(x < 0)
			{
				x += b/d;
				y -= a/d;
			}
			cout<<x<<' '<<y<<endl;
		}
		else
			cout<<"sorry"<<endl;
	}
	return 0;
}