Align AtCoder - 4371 

Problem Statement

You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Constraints

• 2≤N≤10⁵
• 1≤A_i≤10⁹
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A1
:
AN

Output

 

Print the maximum possible sum of the absolute differences between the adjacent elements after arranging the given integers in a row in any order you like.

Sample Input 1

5
6
8
1
2
3

Sample Output 

21

When the integers are arranged as 3,8,1,6,2, the sum of the absolute differences between the adjacent elements is |3−8|+|8−1|+|1−6|+|6−2|=21. This is the maximum possible sum.

Sample Input 

6
3
1
4
1
5
9

Sample Output 

25

Sample Input 

3
5
5
1

Sample Output 

8
题意：对于原序列进行排序，找到一种排序方法，使得相邻两个数字的差值的绝对值的和最大。
解题思路：将序列从小大到达排序，取后面一般的数列 减去 前面一半的数列 可以得到一个最大差值
　　分奇数情况 和 偶数情况讨论 （叫前面一半的数列的元素为较小数， 后面一半数列的的元素为较大数）
奇数情况：假设序列的排序情况只有 大小大小大 或者 小大小大小 两种形式 较大数和较小数较差排列 
　　　　　　大小大小大情况 ： 两端都少加了一次
　　　　　　小大小大小情况 ： 两端都少减了一次
偶数情况：大小大小 和 小大小大 情况相同
注意：数据范围！！！

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<map>
#include<string>
#include<set>
#include<stack>
#include<queue>
using namespace std;
const int maxn  = 200000;
long long n,ans;
long long num[maxn];
int main(){
    scanf("%lld",&n);
    for(int i = 1; i <= n ; i++){
        scanf("%lld",&num[i]);
    }
    sort(num + 1, num+ n + 1);
    if(n&1){
        long long temp1 = 0,tem1 = 0,ans1 = 0;
        long long temp2 = 0,tem2 = 0,ans2 = 0;
        int t1 = n / 2 + 1;
        for(int i = 1 ;i <= t1 ; i++) temp1 += 2*num[i];
        for(int i = t1 + 1 ; i <= n ; i++) tem1 += 2*num[i];
        ans1 = tem1 - temp1 + num[t1] + num[t1 - 1];
        int t2 = n/2; 
        for(int i = 1 ; i <= t2 ; i++) temp2 += 2*num[i];
        for(int i = t2 + 1 ; i <= n ; i++) tem2 += 2*num[i];
        ans2 = tem2 - temp2 - num[t2 + 1] - num[t2 + 2];
        ans = max(ans1,ans2);    
    }else{
        long long t = n/2;
        long long tem = 0,temp = 0;
        for(int i = 1 ; i <= t ; i++) temp += 2*num[i];
        for(int i = t + 1 ; i <= n ; i++) tem += 2 * num[i];
        ans = tem - temp - num[t + 1] + num[t];
    }
    printf("%lld\n",ans);
    return 0;
}

AC代码

一个从很久以前就开始做的梦。