扩展欧几里得

保存推导部分：

gcd = b*x1 + (a-(a/b)*b)*y1
    = b*x1 + a*y1 – (a/b)*b*y1
    = a*y1 + b*(x1 – a/b*y1)

代码：

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

int exgcd(int a, int b, int &x, int &y)
{
    if(!b)
    {
        x = 1; y = 0;
        return a;
    }
    int ans = exgcd(b,a%b,x,y);
    int t = x;
    x = y;
    y = t-a/b*y;
    return ans;
}

int main()
{
    int a, b, x, y;
    cin >> a >> b;
    cout << exgcd(a, b, x, y) << endl;
    cout << x << " " << y << endl;
    return 0;
}