扩展欧几里得

//递归写法
int exgcd(int a, int b, int &x, int &y)
{
     if (b == 0) {
         x=1; y=0;
         return a;
     } else {    
         int ret = exgcd(b, a%b, x, y);
         int t = x;
         x = y;
         y = t-a/b*x;
         return ret;
     }
}

// 有不为0的整数a,b 则定有 gcd(a,b) = ax+by 存在，其中x, y为整数
// 常用于求解不定方程， 模的逆元，  同余方程

// 返回值是最大公约数

//非递归写法。

int exgcd2(int m, int n, int &x, int &y)
{
    int x1, y1, x0, y0;
    x0 = 1; y0 = 0;
    x1 = 0; y1 = 1;
    int r = (m % n + n) % n;
    int q = (m - r) / n;
    x = 0; y = 1;
    while (r) {
        x = x0 - q * x1; y = y0 - q * y1; x0 = x1; y0 = y1;
        x1 = x; y1 = y;
        m = n; n = r; r = m % n;
        q = (m - r) / n;
    }
    return n;
}

//欧几里得求公约数

int gcd(int a, int b)
{
    return b==0 ? a : gcd(b, a%b);
} 

int gcd(int a, int b)
{
    int r = a%b;
    while (r) {
        a = b;
        b = r;
        r = a%b;
    }
    return b;
}

// lcm(a,b) = a*b/gcd(a,b)