# hdu 2669 Romantic (扩展欧几里得模板题)

Posted by 111qqz on Wednesday, October 12, 2016

## TOC

x,y的通解为：**(x+kgx , y-kgy ） 其中：gx= b/gcd(a,b),gy = a/gcd(a,b),k为任意整数 **

``````/* ***********************************************
Author :111qqz
Created Time :Wed 12 Oct 2016 07:30:20 PM CST
File Name :code/hdu/2669.cpp
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
int a,b;
int exgcd( int a,int b,int &x,int &y)
{
if (b==0)
{
x = 1;
y = 0;
return a;
}
int ret = exgcd(b,a%b,x,y);
int tmp = x;
x = y;
y = tmp - a/b*y;
return ret;
}
int main()
{
#ifndef  ONLINE_JUDGE
freopen("code/in.txt","r",stdin);
#endif
while (~scanf("%d%d",&a,&b))
{
int x,y;
if (exgcd(a,b,x,y)==1)
{
while (x<0)
{
x += b;
y -= a;
}
printf("%d %d\n",x,y);
}
else
{
puts("sorry");
}
}
#ifndef ONLINE_JUDGE
fclose(stdin);
#endif
return 0;
}
``````

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