传送门
思路：
- 若a 与 b不互质，那么令 e / f = 1，则可求得c =(a + b) / gcd(a，b) 且d = b / gcd(a，b) e = 1，f = 1。
- 可令df = b，且让d与f互质，直接用扩展欧几里得来做就行了。
- 详细代码思路也可见大佬博客。
代码实现：
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
{
1, 0}, {
-1, 0}, {
0, 1}, {
0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e6 + 5;int t, a, b;
int c, d, e, f;ll exgcd(ll a,ll b,ll &x1,ll &y1){
if(!b){
x1 = 1; y1 = 0;return a;}int g = exgcd(b,a%b,x1,y1);int tmp = x1; x1 = y1;y1 = tmp-(a/b)*y1;return g;
}vector<int> vt;
int factor[N], vis[N];void divide(int n){
me(vis);for(int i = 2; i < n; i ++){
if(!vis[i]){
vt.push_back(i);factor[i] = i;}for(int j = 0; j < vt.size() && i * vt[j] < n; j ++){
vis[i * vt[j]] = 1;factor[i * vt[j]] = vt[j];if(i % vt[j] == 0) break;}}
}signed main()
{
IOS;divide(N);cin >> t;while(t --){
cin >> a >> b;if(b == 1){
cout << "-1 -1 -1 -1" << endl;continue;}int g = __gcd(a, b);if(g != 1){
cout << (a + b) / g << " " << b / g << " 1 1" <<endl;continue;}int fc = factor[b];d = 1; f = b;while(f % fc == 0) d *= fc, f /= fc;if(f == 1){
cout << "-1 -1 -1 -1" << endl;continue;}exgcd(d, f, c, e);c = -c;while(c <= 0 || e <= 0) c += f, e += d;c *= a, e *= a;cout << e << " " << d << " " << c << " " << f << endl;}return 0;
}