Tenka 1 Computer Contest C-Align

C - Align

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Time limit : 2sec / Memory limit : 1024MB

Score : 400 points

Problem Statement

You are given N integers; the i-th of them is Ai. 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≤Ai≤109
• All values in input are integers.

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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.

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Sample Input 1

Copy

5
6
8
1
2
3

Sample Output 1

Copy

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.

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Sample Input 2

Copy

6
3
1
4
1
5
9

Sample Output 2

Copy

25

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Sample Input 3

Copy

3
5
5
1

Sample Output 3

Copy

8

题解：分奇数和偶数讨论。当为奇数时，一定是中间的两个数在左右边（两种情况）使得结果最大。偶数同理，更好枚举，只有一种情况。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
#define ll long long
//#define local

using namespace std;

const int MOD = 1e9+;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = 1e5+;

int main() {
#ifdef local
    if(freopen("/Users/Andrew/Desktop/data.txt", "r", stdin) == NULL) printf("can't open this file!\n");
#endif

    int n;
    int a[maxn];
    ll mx = ;
    ll sum[maxn];
    scanf("%d", &n);
    for (int i = ; i < n; ++i) {
        scanf("%d", a+i);
    }
    sort(a, a+n);
    for (int i = ; i < n; ++i) {
        if (!i) sum[i] = a[i];
        if (i) sum[i] = sum[i-]+a[i];
    }
    if (n&) {
        if (n == ) {
            mx = max(abs(*a[]-a[]-a[]), abs(*a[]-a[]-a[])); //当n=3时，特殊讨论一下
        } else {
            //如果是选择靠左边的两个数作为两个端点，那么他们一定是小于它们相邻的数的。
            //同理，如果是选择靠左边的两个数作为两个端点，那么他们一定是大于它们相邻的数的。
            //将需要算两次的数*2
            mx = max(abs(*(sum[n-]-sum[n/])-*(sum[n/-])-(a[n/]+a[n/-])), abs(*(sum[n-]-sum[n/+])-*(sum[n/-])+(a[n/]+a[n/+])));
        }
    } else {
        mx = abs(*(sum[n-]-sum[n/])-*(sum[n/-])+abs(a[n/]-a[n/-]));
    }
    printf("%lld\n", mx);
#ifdef local
    fclose(stdin);
#endif
    return ;
}