Maximum of Maximums of Minimums

最新推荐文章于 2025-05-20 15:48:31 发布

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最新推荐文章于 2025-05-20 15:48:31 发布

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探讨了给定数组和整数k的情况下，如何将数组分割成k个非空子段，并计算每个子段的最小值后得到的最大值。通过不同情况的分析，给出了一种高效的解决方案。

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You are given an array a₁, a₂, ..., a_n consisting of n integers, and an integer k. You have to split the array into exactly k non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the k obtained minimums. What is the maximum possible integer you can get?

Definitions of subsegment and array splitting are given in notes.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 10⁵) — the size of the array a and the number of subsegments you have to split the array to.

The second line contains n integers a₁, a₂, ..., a_n ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

Print single integer — the maximum possible integer you can get if you split the array into k non-empty subsegments and take maximum of minimums on the subsegments.

Example

Input

5 2
1 2 3 4 5

Output

Input

5 1
-4 -5 -3 -2 -1

Output

-5

Note

A subsegment [l, r] (l ≤ r) of array a is the sequence a_l, a_l + 1, ..., a_r.

Splitting of array a of n elements into k subsegments [l₁, r₁], [l₂, r₂], ..., [l_k, r_k] (l₁ = 1, r_k = n, l_i = r_i - 1 + 1 for all i > 1) is k sequences (a_l₁, ..., a_r₁), ..., (a_{l_k}, ..., a_{r_k}).

In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are min(1, 2, 3, 4) = 1 and min(5) = 5. The resulting maximum is max(1, 5) = 5. It is obvious that you can't reach greater result.

In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4, - 5, - 3, - 2, - 1). The only minimum is min( - 4, - 5, - 3, - 2, - 1) = - 5. The resulting maximum is - 5.

题解: 当k=1时，很显然，输出数组的最小值就行了；当k>2时，也很显然，只要输出数组的最大值就行了，因为分成大于等于3段的数组，我们可以将最大值单独分成一份，这样不管其他怎么分答案永远是最大值。当k=2的时候，我们可以记录一下1~i和i~n两段的最小值，然后O(n)扫一遍就好了..不过k=2的时候还有更简单的方法，就是直接输出max(a[1],a[n])。可以发现，两段中必有一段含有数列的最小值，这一段的最小值是确定的。而这两段肯定分别包含一个端点，因此另一段的最小值不会小于端点值。可以发现，这个值一定能取到。因此，输出两个端点的最大值即可。

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 1000000005 
int a[100005];
int main()
{
	int n,k;
	int maxn=-MAX,minn=MAX;
	cin>>n>>k;
	for(int i=0; i<n; i++) {
		cin>>a[i];
		if(a[i] > maxn) maxn = a[i];
		if(a[i] < minn) minn = a[i];
	}
	if(k==1) {
		cout<<minn<<endl;
	}
	else if(k==2){
	    cout<<max(a[0],a[n-1])<<endl;
	}
	else {
		cout<<maxn<<endl;
	}
	return 0;
}