一、题目

输入

10
7334774857 8461862436 540886577 5245195052 9194400521 5412986878 6694133363 1186771950 1405713915 7115286932 

输出

-29430338967

二、思考

构造差分数组：C[i] = A[i+1] - A[i]

由题目条件可知：当A[i] = A[i+1] + A[i-1] - A[i]时：

C[i](现) = A[i+1] - A[i](现) = A[i] - A[i-1]

C[i-1](现)= A[i](现) - A[i-1] = A[i+1] - A[i]

得出结论：做一次题目中的操作，例如，修改A[i]就会交换差分数组中C[i-1]和C[i]两个的顺序而数组和为A[0]+A[1]+....+A[n-1] = A[0] + (A[0] + C[0](现)) + (A[0] + C[0](现) + C[1](现)).......所以差分数组越前面的项被加的次数越多，要使数组的和最小，差分数组就要按小到大排列

 三、代码

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLong();
        }
        long[] c = new long[n - 1];
        for (int i = 0; i < n - 1; i++)
            c[i] = a[i + 1] - a[i];
        Arrays.sort(c);
        long sum = a[0];
        for (int i = 1; i < n; i++) {
            a[i] = a[i - 1] + c[i - 1];
            sum += a[i];
        }
        System.out.println(sum);
    }
}