codeforces626E.Simple Skewness（三分）

Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list of n (not
necessarily distinct) integers. Find the non-empty subset (with repetition) with the maximum simple skewness.

The mean of a collection is the average of its elements. The median of a collection is its middle element when all of its elements are sorted, or the average of its two middle elements if it has even size.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200
000) — the number of elements in the list.

The second line contains n integers xi (0 ≤ xi ≤ 1 000 000) —
the ith element of the list.

Output

In the first line, print a single integer k — the size of the subset.

In the second line, print k integers — the elements of the subset in any order.

If there are multiple optimal subsets, print any.

Examples

input

4
1 2 3 12

output

3
1 2 12

input

4
1 1 2 2

output

3
1 1 2

input

2
1 2

output

2

1 2

题意：给你n个数，让你找到一个非空子集合，使得这个子集合的平均数和中位数的差最大。

思路：首先，这个产生最大值的子集合内含的数的个数一定是奇数（平均数不等于中位数），因为如果个数是偶数，那么我们可以减去中间较打的一个数，那么平均数减去中位数的值就会变大。我们可以枚举每一个数为中位数，然后三分找到最大的平均数。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 200050
ll a[maxn];
ll n,m;
ll sum[maxn];

int main()
{
    ll i,j;
    while(scanf("%lld",&n)!=EOF)
    {
        for(i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
        }
        sort(a+1,a+1+n);
        sum[0]=0;
        for(i=1;i<=n;i++){
            sum[i]=sum[i-1]+a[i];
        }
        if(n==1 || n==2){
            printf("1\n");
            printf("%I64d\n",a[1]);continue;
        }
        ll l,r,d,m1,m2,t1,t2;
        ll mid=1,len=0,zong=0;
        ll zong1,mid1,len1;
        for(i=2;i<=n-1;i++){
            l=1;r=min(i-1,n-i);
            for(j=1;j<=100;j++){
                m1=(2*l+r)/3;
                m2=(l+2*r+2)/3; //向上取整
                /*
                d=(l+r)/2;
                m1=d;
                m2=(d+r)/2;
                */
                t1=sum[i]-sum[i-m1-1]+sum[n]-sum[n-m1];
                t2=sum[i]-sum[i-m2-1]+sum[n]-sum[n-m2];
                if(t1*(2*m2+1)<t2*(2*m1+1)){
                    l=m1+1;
                }
                else r=m2-1;
            }
            zong1=sum[i]-sum[i-l-1]+sum[n]-sum[n-l]-(2*l+1)*a[i];
            len1=l;
            mid1=i;
            if(zong1*(2*len+1)>zong*(2*len1+1) ){
                zong=zong1;
                len=len1;
                mid=mid1;
            }
        }
        printf("%I64d\n",len*2+1);
        int flag=1;
        for(i=mid-len;i<=mid;i++){
            if(flag){
                flag=0;printf("%I64d",a[i]);
            }
            else{
                printf(" %I64d",a[i]);
            }
        }
        for(i=n-len+1;i<=n;i++){
            printf(" %I64d",a[i]);
        }
        printf("\n");
    }
    return 0;
}