【Codeforces Round #440 (Div. 2) B】Maximum of Maximums of Minimums

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探讨了一个整数数组分割成k个非空子段的问题，目标是最大化这些子段中的最小值的最大值。通过预处理前后缀最小值并考虑特殊情况来解决。

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You are given an array a1, a2, ..., an 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, ..., an ( - 109 ≤ ai ≤ 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.

Examples
input
5 2
1 2 3 4 5
output
5
input
5 1
-4 -5 -3 -2 -1
output
-5

Note

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

Splitting of array a of n elements into k subsegments [l1, r1], [l2, r2], ..., [lk, rk] (l1 = 1, rk = n, li = ri - 1 + 1 for all i > 1) is k sequences (al1, ..., ar1), ..., (alk, ..., ark).

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.

这题一开始没理解好题意就仓促开撸，以为不用连续取于是排了下序，n==1或者k==1就最小值其余都是最大值结果居然Accept了然后又被hack了重新读了下题意发现不能这么做应该多分出一种k==2的情况暴力枚举一发 k>=3的时候肯定可以把最大点独立出来。

#include <bits/stdc++.h>

#define ll long long
using namespace std;

const int N = 1e5;

int a[N+10],n,k,premi[N+10],aftermi[N+10];

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    scanf("%d%d",&n,&k);
    for (int i = 1;i <= n;i++)
        scanf("%d",&a[i]);
    if (k==1)
    {
        int ans = a[1];
        for (int i = 2;i <= n;i++)
            ans = min(ans,a[i]);
        printf("%d\n",ans);
    }else if (k==2)
    {
        premi[1] = a[1];
        for (int i = 2;i <= n;i++)
            premi[i] = min(premi[i-1],a[i]);
        aftermi[n] = a[n];
        for (int i = n-1;i >= 1;i--)
            aftermi[i] = min(aftermi[i+1],a[i]);
        int ans = max(premi[1],aftermi[2]);
        for (int i = 2;i <= n-1;i++)
            ans = max(ans,max(premi[i],aftermi[i+1]));
        printf("%d\n",ans);
    }else {
        int ans = a[1];
        for (int i = 1;i <= n;i++)
            ans = max(ans,a[i]);
        printf("%d\n",ans);
    }
    return 0;
}