B - Maximum of Maximums of Minimums （思维题）

最新推荐文章于 2021-01-09 23:14:28 发布

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本文探讨了如何在给定的整数数组中通过分割成k个非空子段来找到最大的最小值，并提供了解决该问题的算法实现。

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 kobtained 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

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.

题意：给你两个数n，k。你要在n个数字的队列中分割k段，然后问你选出每一段的最小值，然后选这些值里面的最大值。问你这个值最小是多少。

思路：当k大于3时候时确定值

代码“”

#include<bits/stdc++.h>
using namespace std;

const int INF = 0x3f3f3f3f;
int main(){
    ios_base::sync_with_stdio(0);
    int mx = -INF,mn = INF;
    int n,k;cin>>n>>k;
    int begin,end;  //记录下左右端点
    for(int i = 0;i < n;i++){
        int x;cin>>x;
        if(i == 0) begin = x;
        if(i == n-1) end = x;
        mx = max(mx,x);
        mn = min(mn,x);
    }
    if(k >= 3) {
        cout<<mx<<endl;
    }
    else if(k == 2) {
        cout<<max(begin,end)<<endl;
    }
    else if(k == 1) {
        cout<<mn<<endl;
    }
    return 0;
}