Codeforces Round 981 (Div. 3) C. Sakurako‘s Field Trip

给你一个数组，你可以交换对称的两个数，让最后数组里连续的数最小

u1s1，tag里还有dp，双指针什么的，个人感觉完全不用

从左遍历到二分之一处，每次判断换和不换谁更优就可以

import java.util.Scanner;

public class CFR981C {
    public static void main(String[] args) {
        int t;
        Scanner sc = new Scanner(System.in);
        t = sc.nextInt();
        while (t-- > 0){
            int n = sc.nextInt();
            int[] nums = new int[n];
            for (int i = 0; i < n; i++){
                nums[i] = sc.nextInt();
            }
            int[] continuous = new int[n];
            continuous[0] = 0;
            for (int i = 1; i < n/2; i++){
                int none = 0,change = 0;
                if(nums[i] == nums[i-1]){
                    none++;
                }
                if(nums[n - i -1] == nums[n - i]){
                    none++;
                }
                if(nums[i] == nums[n - i - 1 + 1]){
                    change ++;
                }
                if(nums[n - i - 1] == nums[i - 1]){
                    change++;
                }
                if(change < none){
                    int te = nums[i];
                    nums[i] = nums[n - i - 1];
                    nums[n - i - 1] = te;
                    continuous[i] += change;
                }
                continuous[i] = Math.min(continuous[i - 1] + none, continuous[i-1] + change);
            }
            int ans = 0;
            for (int i = 1; i < n; i++){
                if(nums[i] == nums[i-1])
                    ans++;
            }
            System.out.println(ans);
        }
    }
}