构造出一个数列,使得它的前缀 $GCD$ 的字典序最大,数组范围只有 $1000$,所以在构造的过程中,计算当前的前缀 $GCD$,线性选出一个和之前结果最大的 $GCD(pre, a[i])$ 即可。
// Date: Tue Dec 5 23:51:39 2023
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
using namespace std;
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;
const ull Pr = 131;
#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])
#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())
#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z) \
cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif
const int N = 1010;
int t, n, a[N], b[N];
bool vis[N];
int gcd(int a, int b) { return (a % b == 0) ? b : gcd(b, a % b); }
int main(void) {
#ifdef _DEBUG
freopen("1407b.in", "r", stdin);
#endif
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> t;
while (t--) {
cin >> n;
memset(vis, false, sizeof vis);
For1(i, 1, n) { cin >> a[i]; }
sort(a + 1, a + 1 + n, greater<int>());
b[1] = a[1];
vis[1] = true;
int pre = b[1];
For1(i, 2, n) {
int g = 0, idx = -1;
For1(j, 2, n) {
if (vis[j])
continue;
int g1 = gcd(pre, a[j]);
if (g1 > g) {
g = g1;
idx = j;
}
}
vis[idx] = true;
b[i] = a[idx];
pre = gcd(pre, b[i]);
}
For1(i, 1, n) { cout << b[i] << ' '; }
cout << '\n';
}
return 0;
}