设ax1+by1=gcd(a,b), bx2+(a%b)y2=gcd(b,a%b);
由gcd(a,b)=gcd(b,a%b),可得:
ax1+by1=bx2+(a%b)y2;
即:ax1+by1=bx2+(a-(a/b)*b)y2
          =ay2+bx2-(a/b)*by2;
即:ax1+by1=ay2 + b(x2-(a/b)*y2)
根据恒等定理,对应项相等，得:x1=y2; y1=x2-(a/b)*y2;
这样我们就得到了:x1，y1的值基于x2，y2，所以我们可以通过递归求解。

C++ 代码

/* 
 * @Author: lzyws739307453 
 * @Language: C++ 
 */
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5;
//写法一
void Exgcd_1(int a, int b, int &x, int &y) {
    if (!b)
        x = 1, y = 0;
    else {
        Exgcd_1(b, a % b, x, y);
        int t = x;
        x = y;
        y = t - a / b * y;
    }
}
//写法二
void Exgcd_2(int a, int b, int &x, int &y) {
    if (!b)
        x = 1, y = 0;
    else Exgcd_2(b, a % b, y, x), y -= a / b * x;
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int a, b, x, y;
        scanf("%d%d", &a, &b);
        Exgcd_2(a, b, x, y);
        printf("%d %d\n", x, y);
    }
    return 0;
}