HDU 2669 Romantic (扩展欧几里得)

【题目链接】：click here~~

【题目大意】：

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.

求满足方程X*a + Y*b = 1.的解

【思路】：

通过扩展欧几里得算法求得x，y之后，要求所有的x和y的解的话，得求得通项公式：(x+k*gx , y-k*gy) gx= b/gcd(a,b),gy = a/gcd(a,b)互素,k为任意整数，即为答案

代码：

/* 
* Problem: HDU No.2669
* Running time: 15MS 
* Complier: C++ 
* Author: javaherongwei
* Create Time: 20:25 2015/9/2 星期三
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a,b,c;
LL ext_gcd(LL &a,LL &b,LL n,LL m)
{
    if(m==0)
    {
        a=1;
        b=0;
        return n;
    }
    LL d=ext_gcd(b,a,m,n%m);
    b-=n/m*a;
    return d;
}
int main()
{
    while(scanf("%lld %lld",&a,&b)!=EOF)
    {
        LL x,y,g;
        g=ext_gcd(x,y,a,b);
        if(g==1){
            while(x<0)
            {
                x+=b/1;
                y-=a/1;
            }
            printf("%lld %lld\n",x,y);
        }
        else puts("sorry");
    } return 0;
}