数论--扩展欧几里得exgcd

算法思想

我们想求得一组(x,y)使得

(ax+by = gcd(a,b))

根据

(gcd(a,b) = gcd(b,amod b))

如果我们现在有(x',y') 使得

(bx'+(amod b)y' = gcd(b,amod b))

那么

$ax+by = bx'+( a-lfloorfrac a b floor b)y'$

移项之后

(ax+by = ay'+b(x'-lfloorfrac a b floor y'))

我们可以得到一组特解 $x = y',y = x' - lfloorfrac a b floor y'$

递归求解,当(b = 0)时候,(x = 1,y = 0);

模板

int exgcd(int a,int b,int &x,int &y)
{
    if(b==0)
    {
        x = 1; y = 0;return a;
    }
    int x1,y1;
    int ans = exgcd(b,a%b,x1,y1);
    x = y1;
    y = x1-(a/b)*y1;
    return ans;
}

例题

https://www.luogu.com.cn/problem/P1082

参考博客

https://www.zybuluo.com/samzhang/note/541890