Codeforces Round #440 (Div. 2）B. Maximum of Maximums of Minimums

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.

题目意思：

给你两个数字n和k   把这n个数字分成k段 让这k段中的k个最小值中的最大值最大

思路：

当k==1的时候我们直接输出最小值就好了

当k>2的时候我们可以单独把最大值隔开 然后能找到的最大值就一定是原序列的最大值本身

当k==2的时候  我用了一个前缀最小值数组和一个后缀最小值数组来维护前n项中的最小值和后n项中的最小值

本来这个操作需要一个for来遍历的 但是后来发现这个array[0]  and array[n - 1] 一定是前缀最小值数组和后缀最小值数组中最大的值

所以只需要比较这两个数字的大小输出大的就好~~

比赛的时候我这道题莫名TL4次 就一个for循环 我至今不知道为什么~~~

#include <cstdio>
#include <cstring>
#include <algorithm>
const int inf = 0x7fffffff;
const int N = 100005;
using namespace std;
int array
;
int main()
{
int n , k;
scanf("%d%d",&n,&k);
int maxx = -inf;
int minn = inf;
int max_id = 0;
for (int i = 0 ; i < n ; i++) {
scanf("%d",&array[i]);
if(array[i] > maxx) {
maxx = array[i];
max_id = i;
}
minn = min(minn , array[i]);
}
if(k == 1) {
printf("%d\n",minn);
} else if (k > 2) {
printf("%d\n",maxx);
}
else {
printf("%d\n",max(array[n-1],array[0]));
}
}