Tenka 1 computer contest C-align

C-align

Time Limit: 2sec/memory limit: 1024 MB

Score:400Points

Problem Statement

You are givenNIntegers;I-Th of them isAI. 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.

Input

Input is given from standard input in the following format:

NA1: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 copy

568123

Sample output 1 copy

21

When the integers are arranged3, 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 2 Copy

6314159

Sample output 2 Copy

25

Sample input 3 copy

3551

Sample output 3 copy

8

Question: it is divided into odd and even numbers. When the value is an odd number, the two numbers in the middle must be at the left and right sides (in both cases) to maximize the result. The same is true for even numbers. There is only one case for better enumeration.

# Include <iostream> # include <cstring> # include <algorithm> # include <cmath> # include <cstdio> # include <vector> # include <queue> # include <set> # include <map> # include <stack> # define ll long // # define localusing namespace STD; const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; const double Pi = ACOs (-1.0); const int maxn = 1e5 + 10; int main () {# ifdef local if (freopen ("/users/Andrew/desktop/data.txt", "r", St Din) = NULL) printf ("can't open this file! \ N "); # endif int N; int A [maxn]; ll MX = 0; ll sum [maxn]; scanf (" % d ", & N ); for (INT I = 0; I <n; ++ I) {scanf ("% d", A + I) ;}sort (A, A + n ); for (INT I = 0; I <n; ++ I) {If (! I) sum [I] = A [I]; if (I) sum [I] = sum [I-1] + A [I];} If (N & 1) {If (n = 3) {MX = max (ABS (2 * A [2]-A [1]-A [0]), ABS (2 * A [0]-A [1]-A [2]); // when n = 3, special discussion} else {// If two numbers on the left are selected as two endpoints, they must be smaller than their adjacent numbers. // Similarly, if two numbers on the left are selected as two endpoints, they must be greater than their adjacent numbers. // Calculate the number of times X 2 MX = max (ABS (2 * (sum [n-1]-sum [n/2]) -2 * (sum [n/2-2])-(A [n/2] + A [n/2-1]), ABS (2 * (sum [n-1]-sum [n/2 + 1])-2 * (sum [n/2-1]) + (A [n/2] + A [n/2 + 1]);} else {MX = ABS (2 * (sum [n-1]-sum [n/2])-2 * (sum [n/2-2]) + ABS (A [n/2]-A [n/2-1]);} printf ("% LLD \ n", MX ); # ifdef local fclose (stdin); # endif return 0 ;}

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Tenka 1 computer contest C-align