扩展欧几里得求逆元模板

int exgcd(int a,int b,int &x,int &y)
{
    int d=a;
    if(b!=0)
    {
        d=exgcd(b,a%b,y,x);
        y-=a/b*x;
    }
    else
    {
        x=1;
        y=0;
    }
    return d;
}
int inverse(int a,int m)
{
    int x,y;
    exgcd(a,m,x,y);
    return (x%m+m)%m;
}

猜你喜欢

转载自blog.csdn.net/w1304636468/article/details/90182303