D - Romantic

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. 

InputThe input contains multiple test cases. 
Each case two nonnegative integer a,b (0<a, b<=2^31) 
Outputoutput 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
这个题目的坑就在 输出上，，x必须为正数，，y必须为负数，想想也很有道理 因为 X*a + Y*b = 1.  a,b,都是大于1 的所以x和Y要有一个小于0，一个大于0，结果才会等于1

//直接用拓展欧几里得公式求出x和y，再除以a，b的最大公约数q，如果依然是整数，则输出，否则，x+=b,y-=a,再判断
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll x,y;

ll exgcd(ll a,ll b){
    if(b==) {
        x=;
        y=;
        return a;
    }
    ll r=exgcd(b,a%b);
    ll t=y;
    y=x-(a/b)*y;
    x=t;
    return r;
}

int main(){
    ll a,b;
    while(scanf("%lld%lld",&a,&b)!=EOF)
    {
        ll r=exgcd(a,b);
        if(r!=){
            puts("sorry");
            continue ;
        }
        ll x1=x;
        ll y1=y;
        x1=(x1+b)%b;
//        while(x1<0){
//            x+=b/r;
//        }
//        while(y1>0){
//            y1-=a/r;
//        }
        y1=(y1-a)%a;

        printf("%lld %lld\n",x1,y1);
    }
    return ;
}