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

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 int a[maxn];
12 int main(int argc, char const *argv[])
13 {
14     #ifndef ONLINE_JUDGE
15         freopen("/home/wzy/in.txt", "r", stdin);
16         freopen("/home/wzy/out.txt", "w", stdout);
17         srand((unsigned int)time(NULL));
18     #endif
19     ios::sync_with_stdio(false);
20     cin.tie(0);
21     int n,k;
22     cin>>n>>k;
23     int maxx;
24     int place=0;
25     int minn;
26     for(int i=0;i<n;i++)
27     {
28         cin>>a[i];
29         if(!i)
30         {
31             maxx=a[i];
32             place=0;
33             minn=a[i];
34         }
35         else if(maxx<a[i])
36             place=i,maxx=a[i];
37         minn=min(minn,a[i]);
38     }
39     if(k==1)
40         cout<<minn<<endl;
41     else if(k==2)
42         cout<<max(a[0],a[n-1])<<endl;
43     else
44         cout<<maxx<<endl;
45     #ifndef ONLINE_JUDGE
46         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
47     #endif
48     return 0;
49 }