Romantic 欧几里得的扩展

Problem Description

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

思路：扩展欧几里得算法，对于两个数字最大公因子是否为1进行判断，之后再分别讨论。

#include <iostream>
#include<cstdio>
using namespace std;
int x,y;
int gcd(int a,int b)
{   int t,g;
if(b==0)
{
x=1;
y=0;
return a;
}
g=gcd(b,a%b);
t=x;
x=y;
y=t-a/b*y;
return g;
}
int main()
{
int a,b;
while(scanf("%d%d",&a,&b)==2)
{
int g=gcd(a,b);
if(g!=1)
printf("sorry\n");
else
{   while(x<=0)
{
x=x+b;
y=y-a;
}               //假设d=gcd(a,b). 那么x=x0+b/d*t; y=y0-a/d*t;其中t为任意常整数
printf("%d %d\n",x,y);
}

}
return 0;
}