传送门
题面：
C. 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 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：
给你n个木棍，让你构成一个矩形，使得矩形的周长的平方除以面积的值最小。
题目分析：
很简单的一个题目，但是最近脑子瓦特了，打的时候傻傻做不粗来。
首先我们设长和宽分别为a和b，则有
很显然要使得最小，我们只需要使得
最小即可。而对于一个矩形而言，什么时候这个值是最小呢，显然最优的策略是选择的大小接近的合法的边长。（而倘若一种边可以构成正方形，显然最优）。
因此我们只需要先筛选出合法的边长，并从大到小枚举相邻的矩形，并不断更新最小值即可。
代码：
#include <bits/stdc++.h>
#define maxn 1000005
using namespace std;
int a[maxn];
const double eps=1e-8;
int sgn(double x){
if(fabs(x)<eps) return 0;
if(x<eps) return -1;
else return 1;
}
map<int,int>mp;
vector<int>vec;
int main()
{
int t;
scanf("%d",&t);
while(t--){
mp.clear();
vec.clear();
int n;
scanf("%d",&n);
memset(a,0,sizeof(int)*n);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
mp[a[i]]++;
}
bool flag=false;
sort(a,a+n);
int sz=unique(a,a+n)-a;
int ansa=0,ansb=0;
double res=1e100;
for(int i=0;i<sz;i++){
if(mp[a[i]]>=2) vec.push_back(a[i]);//筛选出合法的边长
if(mp[a[i]]>=4){//倘若能够构成正方形，证明已经最优
printf("%d %d %d %d\n",a[i],a[i],a[i],a[i]);
flag=true;
break;
}
}
if(!flag){
sz=vec.size();
for(int i=1;i<sz;i++){//枚举剩下的合法的边长，更新最小值
double tmp=(double)vec[i]/vec[i-1]+(double)vec[i-1]/vec[i];
if(res>tmp){
res=tmp;
ansa=vec[i],ansb=vec[i-1];
}
}
printf("%d %d %d %d\n",ansa,ansa,ansb,ansb);
}
}
return 0;
}