Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)C. Laboratory Work

Anya and Kirill are doing a physics laboratory work. In one of the tasks they have to measure some value n times, and then compute the average value to lower the error.

Kirill has already made his measurements, and has got the following integer values: x1, x2, ..., xn. It is important that the values are close to each other, namely, the difference between the maximum value and the minimum value is at most 2.

Anya does not want to make the measurements, however, she can't just copy the values from Kirill's work, because the error of each measurement is a random value, and this coincidence will be noted by the teacher. Anya wants to write such integer values y1, y2, ..., ynin her work, that the following conditions are met:

• the average value of x1, x2, ..., xn is equal to the average value of y1, y2, ..., yn;
• all Anya's measurements are in the same bounds as all Kirill's measurements, that is, the maximum value among Anya's values is not greater than the maximum value among Kirill's values, and the minimum value among Anya's values is not less than the minimum value among Kirill's values;
• the number of equal measurements in Anya's work and Kirill's work is as small as possible among options with the previous conditions met. Formally, the teacher goes through all Anya's values one by one, if there is equal value in Kirill's work and it is not strike off yet, he strikes off this Anya's value and one of equal values in Kirill's work. The number of equal measurements is then the total number of strike off values in Anya's work.

Help Anya to write such a set of measurements that the conditions above are met.

Input

The first line contains a single integer n (1 ≤ n ≤ 100 000) — the numeber of measurements made by Kirill.

The second line contains a sequence of integers x1, x2, ..., xn ( - 100 000 ≤ xi ≤ 100 000) — the measurements made by Kirill. It is guaranteed that the difference between the maximum and minimum values among values x1, x2, ..., xn does not exceed 2.

Output

In the first line print the minimum possible number of equal measurements.

In the second line print n integers y1, y2, ..., yn — the values Anya should write. You can print the integers in arbitrary order. Keep in mind that the minimum value among Anya's values should be not less that the minimum among Kirill's values, and the maximum among Anya's values should be not greater than the maximum among Kirill's values.

If there are multiple answers, print any of them.

Examples

input

6
-1 1 1 0 0 -1

output

2
0 0 0 0 0 0 

input

3
100 100 101

output

3
101 100 100 

input

7
-10 -9 -10 -8 -10 -9 -9

output

5
-10 -10 -9 -9 -9 -9 -9 

题意：给出a串，保证每个数字是整数且所有数间差值最大不超过2，输出b串使其：1.两串总和相等 2.b串数取值范围和a串数相同 3.两串相同数尽量少
找规律题，分类讨论

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define eps 0.00000001
#define pn printf("\n")
using namespace std;
typedef long long ll;

const int maxn = 1e5+7;
int n, a[maxn];
int MIN = INF, MAX = -INF, tot = 0;
int k = 0, fl = 0;
vector <int> vi;
map <int,int> mp;

int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",a+i);
        mp[a[i]] ++;
        tot += a[i];
        MIN = min(MIN, a[i]);
        MAX = max(MAX, a[i]);
    }

    if(MAX - MIN <= 1)
    {
        printf("%d\n", n);
        for(int i=0;i<n;i++)
        {
            if(i) printf(" ");
            printf("%d", a[i]);
        }pn;
    }
    else
    {
        if(!mp.count(MAX-1))
        {
            int mt, res;
            if(mp[MIN] < mp[MAX])
            {
                mt = mp[MIN];
                res = MAX;
            }
            else
            {
                mt = mp[MAX];
                res = MIN;
            }
            int mid = 2 * mt, fl = 0;
            printf("%d\n" ,n - mid);
            for(int i=0;i<mid;i++)
            {
                if(!fl) fl = 1;
                else printf(" ");
                printf("%d", MIN  +1);
            }
            for(int i=mid;i<n;i++)
            {
                if(!fl) fl = 1;
                else printf(" ");
                printf("%d", res);
            }
        }
        else // 1 2 2 2 2 3
        {
            if(min(mp[MIN],mp[MAX])*2 > mp[MIN+1]/2*2)
            {
                int mt, res;
                if(mp[MIN] < mp[MAX])
                {
                    mt = mp[MIN];
                    res = MAX;
                }
                else
                {
                    mt = mp[MAX];
                    res = MIN;
                }
                int mid = 2 * mt + mp[MIN+1], fl = 0;
                printf("%d\n" ,n -2 * mt);
                for(int i=0;i<mid;i++)
                {
                    if(!fl) fl = 1;
                    else printf(" ");
                    printf("%d", MIN  +1);
                }
                for(int i=mid;i<n;i++)
                {
                    if(!fl) fl = 1;
                    else printf(" ");
                    printf("%d", res);
                }

            }
            else
            {
                int mt = mp[MIN+1]/2;
                int ans = (mp[MIN+1]&1) + mp[MIN] + mp[MAX];
                printf("%d\n",ans);
                int fl = 0, cn = mp[MIN] + mt, cx = mp[MAX] + mt;
                if(mp[MIN+1]&1)
                {
                    printf("%d",MIN + 1); fl = 1;
                }
                for(int i=0;i<cn;i++)
                {
                    if(!fl) fl = 1;
                    else printf(" ");
                    printf("%d", MIN);
                }
                for(int i=0;i<cx;i++)
                {
                    if(!fl) fl = 1;
                    else printf(" ");
                    printf("%d", MAX);
                }
            }
        }
        pn;
    }
    
}

 

转载于:https://www.cnblogs.com/HazelNut/p/8621344.html