codeforces Round #440 B Maximum of Maximums of Minimums【思维/找规律】

B. Maximum of Maximums of Minimums

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

【题意】：最大化数组分割为k个区间里的最小值。

【分析】：观察，发现最大化的值和k有关。

【代码】：

#include<bits/stdc++.h>

using namespace std;

int n,k;
const int maxn = 1e5+;
int a[maxn];
#define inf 0x3f3f3f3f
int main()
{
    int mm,ma;
    memset(a,,sizeof(a));
    while(cin>>n>>k)
    {
        mm=inf;
        ma=-inf;
        for(int i=;i<n;i++)
        {
            scanf("%d",&a[i]);
            mm=min(mm,a[i]);
            ma=max(ma,a[i]);
        }
        if(k==)
        {
            cout<<mm<<endl;
            return ;
        }
        if(k>)//分割大于2时，最大化策略为总把最大值划为单独，就可以最大化结果刚好为最大值
        {
            cout<<ma<<endl;
            return ;
        }
        if(k==)//分割为2个区间时，想要最大化，策略1 5 9 7 2/9 5 1 7 8可以试试，无论在哪里隔板，肯定符合两端的最大值。
        {
            cout<<max(a[],a[n-])<<endl;
        }
    }
    return ;
}