CF1027C Minimum Value Rectangle 贪心 数学

Minimum Value Rectangle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have nn sticks of the given lengths.

Your task is to choose exactly four of them in such a way that they can form a rectangle. No sticks can be cut to pieces, each side of the rectangle must be formed by a single stick. No stick can be chosen multiple times. It is guaranteed that it is always possible to choose such sticks.

Let SS be the area of the rectangle and PP be the perimeter of the rectangle.

The chosen rectangle should have the value P2SP2S minimal possible. The value is taken without any rounding.

If there are multiple answers, print any of them.

Each testcase contains several lists of sticks, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer TT (T≥1T≥1) — the number of lists of sticks in the testcase.

Then 2T2T lines follow — lines (2i−1)(2i−1) and 2i2i of them describe the ii-th list. The first line of the pair contains a single integer nn (4≤n≤1064≤n≤106) — the number of sticks in the ii-th list. The second line of the pair contains nn integers a1,a2,…,ana1,a2,…,an (1≤aj≤1041≤aj≤104) — lengths of the sticks in the ii-th list.

It is guaranteed that for each list there exists a way to choose four sticks so that they form a rectangle.

The total number of sticks in all TT lists doesn't exceed 106106 in each testcase.

Output

Print TT lines. The ii-th line should contain the answer to the ii-th list of the input. That is the lengths of the four sticks you choose from theii-th list, so that they form a rectangle and the value P2SP2S of this rectangle is minimal possible. You can print these four lengths in arbitrary order.

If there are multiple answers, print any of them.

Example

input

Copy

3
4
7 2 2 7
8
2 8 1 4 8 2 1 5
5
5 5 5 5 5

output

Copy

2 7 7 2
2 2 1 1
5 5 5 5

Note

There is only one way to choose four sticks in the first list, they form a rectangle with sides 22 and 77, its area is 2⋅7=142⋅7=14, perimeter is 2(2+7)=182(2+7)=18. 18214≈23.14318214≈23.143.

The second list contains subsets of four sticks that can form rectangles with sides (1,2)(1,2), (2,8)(2,8) and (1,8)(1,8). Their values are 622=18622=18, 20216=2520216=25 and 1828=40.51828=40.5, respectively. The minimal one of them is the rectangle (1,2)(1,2).

You can choose any four of the 55 given sticks from the third list, they will form a square with side 55, which is still a rectangle with sides (5,5)(5,5).

题意：给你n根木棒，从中挑出四根木棒组成一个矩形，矩形面积为S，周长为P，求使P^2/S最小的四根木棒长度

分析：假设矩形宽为a，长为b，则P^2/S=(2*(a+b))^2/a*b=4*(a*a+2*a*b+b*b)/a*b=4*(2+a/b+b/a)

　　即求a/b+b/a的最小值

　　a/b+b/a>=2*sqrt(a/b*b/a)=2(当且仅当a/b=b/a时等式成立)

　　即当a=b时取最小，a!=b时，a，b越接近值越小

　　所以我们先求出所有可以用来做矩形的边，将这些边排序后再枚举求相邻两边a/b+b/a的最小值

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;
const ll mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll n, T, cnt, a[maxn], b[maxn];
int main() {
    ios::sync_with_stdio(0);
    cin >> T;
    while( T -- ) {
        cnt = 0;
        cin >> n;
        for( ll i = 1; i <= n; i ++ ) {
            cin >> a[i];
        }
        sort(a+1,a+n+1);
        bool flag = false;
        for( ll i = 1; i <= n; i ++ ) {
            if(!flag) {
                flag = true;
            } else {
                if( a[i] == a[i-1] ) {
                    b[++cnt] = a[i];
                    flag = false;
                }
            }
        }
        double ans = 1e9;
        ll x, y;
        for( ll i = 1; i < cnt; i ++ ) {
            if( ans>(b[i+1]*1.0)/b[i]+(b[i]*1.0)/b[i+1]) {
                ans = (b[i+1]*1.0)/b[i]+(b[i]*1.0)/b[i+1];
                x = b[i], y = b[i+1];
            }
        }
        cout << x << " " << x << " " << y << " " << y << endl;
    }
    return 0;
}