Codeforces 872B：Maximum of Maximums of Minimums（思维）

B. Maximum of Maximums of Minimums

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.

 题意

给出一个有n个整数的数组 a₁, a₂, ..., a_n 和一个整数k。你被要求把这个数组分成k 个非空的子段。 然后从每个k 个子段拿出最小值，再从这些最小值中拿出最大值。求这个最大值最大能为多少？

 思路

一共可以分三种情况：

1. 当k=1的时候，这个最大值一定是数组中的最小值
2. 当k=2的时候，可以将数组分成两部分（废话），然后找到数组的第一个数字和最后一个数字中最大的那个就可以了
3. 当k≥3的时候，可以将数组分割，让数组中最大的那个数单独放一组（这个是一定可以实现的），然后输出最大的数

代码

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
int a[maxn];
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("/home/wzy/in.txt", "r", stdin);
        freopen("/home/wzy/out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,k;
    cin>>n>>k;
    int maxx;
    int place=0;
    int minn;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        if(!i)
        {
            maxx=a[i];
            place=0;
            minn=a[i];
        }
        else if(maxx<a[i])
            place=i,maxx=a[i];
        minn=min(minn,a[i]);
    }   
    if(k==1)
        cout<<minn<<endl;
    else if(k==2)
        cout<<max(a[0],a[n-1])<<endl;
    else
        cout<<maxx<<endl;
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}