Romantic

                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                             Total Submission(s): 10350    Accepted Submission(s): 4416

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 (以上全是廢話)

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

http://acm.hdu.edu.cn/showproblem.php?pid=2669

題意：

輸入a和b， 然後讓你求X和Y滿足此表示式X*a + Y*b = 1。並且輸出X最小正整數的一組，Y可以為負數。如果找到一組X，Y，則輸出sorry

思路：

利用擴充套件歐幾里德求出一個X，Y。然後找X為最小正整數的一組。

X = X0 + kb

Y = y0 - ka;

程式碼：

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

long long gcd(long long a, long long b, long long& x, long long& y)
{
	if(a == 0 && b== 0)	
		return -1;
	if(b == 0){
		x = 1;
		y = 0;
		return a;
	}
	long long d = gcd(b, a%b, y, x);
	y -= a/b*x;
	return d;
}
int main()
{
	long long a, b; 
	while(scanf("%lld %lld", &a, &b)!=EOF) {
	  	long long x, y;
	  	long long d = gcd(a, b, x, y);
	  	if(d != 1)
			cout << "sorry" << endl;
	  	else {

	  		while(x <= 0) {
	  			x += b;
	  			y -= a;
			}
	  		cout << x<< " " << y << endl;
		  }
			
	}
		
	return 0;
 }