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

B. Maximum of Maximums of Minimumstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou 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.InputThe 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).OutputPrint 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.Examplesinput

5 2
1 2 3 4 5

output

5

input

5 1
-4 -5 -3 -2 -1

output

-5

NoteA 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.
这题一开始没理解好题意就仓促开撸，以为不用连续取 于是排了下序，n==1或者k==1就最小值 其余都是最大值  结果居然Accept了 然后又被hack了 重新读了下题意发现不能这么做 应该多分出一种k==2的情况 暴力枚举一发 k>=3的时候肯定可以把最大点独立出来。

#include <bits/stdc++.h>

#define ll long long
using namespace std;

const int N = 1e5;

int a[N+10],n,k,premi[N+10],aftermi[N+10];

int main()
{
//freopen("F:\\rush.txt","r",stdin);
scanf("%d%d",&n,&k);
for (int i = 1;i <= n;i++)
scanf("%d",&a[i]);
if (k==1)
{
int ans = a[1];
for (int i = 2;i <= n;i++)
ans = min(ans,a[i]);
printf("%d\n",ans);
}else if (k==2)
{
premi[1] = a[1];
for (int i = 2;i <= n;i++)
premi[i] = min(premi[i-1],a[i]);
aftermi
= a
;
for (int i = n-1;i >= 1;i--)
aftermi[i] = min(aftermi[i+1],a[i]);
int ans = max(premi[1],aftermi[2]);
for (int i = 2;i <= n-1;i++)
ans = max(ans,max(premi[i],aftermi[i+1]));
printf("%d\n",ans);
}else {
int ans = a[1];
for (int i = 1;i <= n;i++)
ans = max(ans,a[i]);
printf("%d\n",ans);
}
return 0;
}