Codeforces 1027C. Minimum Value Rectangle

本文解析了C.MinimumValueRectangle问题,目标是在给定的小棍中找到四个能组成矩形的组合,使得周长的平方除以面积的值最小。通过数学推导,发现最优解出现在矩形的两组边长尽可能接近时。

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https://blog.youkuaiyun.com/qq_40727946/article/details/81841865

C. Minimum Value Rectangle

题目链接

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 the ii-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:找出四个小棍可以拼成矩形,要求(周长的平方)/(面积)最小。

归类:数论。

题解:可以根据不等式推导出当 a=b 的时候可以取到最小值,所以我们只要求 a 与 b 最接近的两个数就行。经过转化可以变为:a/b = 1 。只要求 a/b 的最大值即可。

代码:

题目很简单 耐克函数而已 输入被坑了

我特判了4个数相同的情况,没把testcase读完就直接Continue了,导致后面读入的不对,而且还能过样例lol

还有,memset太大也是会超时的lol

#include<bits/stdc++.h>
#define rep(i, j, k) for (int i=j; i<k; i++)
#define dprintf if (debug) printf
#define ll long long
using namespace std;
ll st[2], ans;
int debug = 0;
string s;
const int INF = 0x7fffffff;
const int maxn = 1e6+60;
const int maxm = 1e4+5;
int n, q, x, y, zu, flag, nowx, nowy, cnt, a[maxn], show[maxm], ans1, ans2, pro; 
int main(){
	scanf("%d", &q);
	rep(i, 0, q){
		scanf("%d", &n);
		memset(show, 0, sizeof(show));
		cnt = 0; double minn = INF, pro = 0;
		rep(i, 0, n){
			scanf("%d", &x);
			if (pro) continue;
			show[x]++;
			if (show[x] == 4){
				printf("%d %d %d %d\n", x, x, x, x);
				pro = 1;
			}
			else if (show[x] == 2){
				a[cnt++] = x;
			}
		}

		if (pro) continue;
		sort(a, a+cnt);
		rep(i, 0, cnt-1){
			if ((double) a[i+1]/a[i] < minn){
				minn = (double) a[i+1]/a[i];
				ans1 = a[i]; ans2 = a[i+1];
			}
		}
		printf("%d %d %d %d\n", ans1, ans1, ans2, ans2);
	}
}

 

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