C - 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.

Example

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 = nli = 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

 

水题,关键是搞清楚题意,还有当k==2时,为什么是max(a[0], a[n-1]),为什么a[0],和a[n-1]一定是序列里的最小值????

 

 

C - Maximum of Maximums of Minimums(数学)_编程C - Maximum of Maximums of Minimums(数学)_i++_02
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

int main()
{
    int n,k;
    int i,j;
    int ans;
    int mmax,mmin;
    int a[100010];
    mmax = -1e9-7;
    mmin = 1e9+7;
    scanf("%d %d",&n, &k);
    for(i = 0; i < n; i++)
    {
        scanf("%d",&a[i]);
    }
    if(k == 1)
    {
        for(i = 0; i < n; i++)
        {
            mmin = min(a[i], mmin);
        }
        printf("%d\n",mmin);
    }
    else if(k == 2)
    {
        ans = max(a[0], a[n-1]);        
        printf("%d\n",ans);
    }
    else if(k >= 3)
    {
        for(i = 0; i < n; i++)
        {
            mmax = max(a[i], mmax);
        }
        printf("%d\n",mmax);
    }     
    return 0;
}
View Code

 

永远渴望,大智若愚(stay hungry, stay foolish)