CF1027C Minimum Value Rectangle【贪心/公式化简】

https://www.luogu.org/problemnew/show/CF1027C

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define mod(x) ((x)%M)
#define pi acos(-1.0)
#define rep(i,x,n) for(int i=(x); i<(n); i++)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const int maxn = 1e4+10; //1e5不行！
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[2]={-1,1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int t,n;
int vis[maxn];
int b[maxn];
int x;
inline int read() //要读入挂！
{
    char x;
    while((x = getchar())<'0' || x>'9');
    int u = x-'0';
    while((x = getchar())>='0' && x<='9') u = (u<<3)+(u<<1)+x-'0';
    return u;
}
/*
1345 4409 1345 4409 8664 8664
4409 4409 8664 8664=4225(v)
1345 1345 4409 4409=3064(x)
*/
int main()
{
    t=read();
    while(t--)
    {
        int f=1;
        ms(vis,0);
        ms(b,0);
        n=read();
        int m=0;
        for(int i=0;i<n;i++)
        {
            x=read();
            vis[x]++;
            if(vis[x]==4 && f)
            {
                printf("%d %d %d %d\n",x,x,x,x);
                f=0;
            }
            if(vis[x]==2 && f)
                b[m++]=x;
        }
        if(f)
        {
            sort(b,b+m);
            int l,r;
            double Min=INF;
            for(int i=0;i<m-1;i++)
            {
                double tmp = (double)b[i]/b[i+1]+(double)b[i+1]/b[i];
                if(tmp<Min)
                {
                    Min = tmp;
                    l=b[i];
                    r=b[i+1];
                }
            }
            printf("%d %d %d %d\n",l,l,r,r);
        }
    }
    return 0;
}
/*
【题意】
3
4
7 2 2 7
8
2 8 1 4 8 2 1 5
5
5 5 5 5 5
【类型】
【分析】
【时间复杂度&&优化】
【trick】
【数据】
*/