1364B. Most socially-distanced subsequence

题目：在给定的数组中找到长度最短而且两两之间差的绝对值最大的子序列

思路：将数组想象成一座山，，越大的元素代表越高的地理位置,越小的元素代表越低的地理位置，然后要寻找出所有的“山峰” 和 “山谷”，数组头元素和尾元素肯定是为“山谷”或者“山峰”，因为相对于左边或者右边的元素，它要么更大要么更小，而另外一边没有元素。所以我们只要寻找到每个转折点

for _ in range(int(input())):n = int(input())a = [*map(int, input().split())]ans = [a[0]]i = 0for i in range(1, n-1):if (a[i-1] < a[i] > a[i+1]) or (a[i-1] > a[i] < a[i+1]):ans+=[a[i]]ans += [a[-1]]print(len(ans))print(*ans)

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1364B. Most socially-distanced subsequence

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