Codeforces Round #440 B. Maximum of Maximums of Minimums_definitions of subsegment and array splitting are -优快云博客

本文探讨了一种算法问题：给定一个整数数组和一个整数k，如何将数组分割为k个非空子段，并计算每个子段的最小值，最后找到这些最小值中的最大值。文章通过具体例子解释了最优解的策略。

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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 ksequences (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.

将数组分为k个串，将每串里最小的数组合成一个新串，找新串里的最大值

如果 k=1 min(a)

如果 k=2 max(a[1],a[n])

如果k>2. max(a)

原因很好理解

#include <iostream>
#include <cstdio>
using namespace std;
int a[100005];
int main() {
    int n,k;
    int mmax=-1e9-7,mmin=1e9+7;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        mmax=max(mmax, a[i]);
        mmin=min(mmin, a[i]);
    }
    if(k>=3)
        printf("%d\n",mmax);
    else if(k==2)
        printf("%d\n",max(a[1], a[n]));
    else
        printf("%d\n",mmin);
    // insert code here...
    //std::cout << "Hello, World!\n";
    return 0;
}