CF div2 334 B

B. More Cowbell

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.

Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 ≤ i ≤ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 ≤ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.

Input

The first line of the input contains two space-separated integers n and k (1 ≤ n ≤ 2·k ≤ 100 000), denoting the number of cowbells and the number of boxes, respectively.

The next line contains n space-separated integers s1, s2, ..., sn (1 ≤ s1 ≤ s2 ≤ ... ≤ sn ≤ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.

Output

Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.

Sample test(s)

input

2 1
2 5

output

7

input

4 3
2 3 5 9

output

9

input

3 2
3 5 7

output

8

Note

In the first sample, Kevin must pack his two cowbells into the same box.

In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.

In the third sample, the optimal solution is {3, 5} and {7}.

给你N个按体积升序排列的球，有K个盒子，每个盒子里面最多装2个球。求使得让N个球按照要求装入K个盒子时，盒子容量最小的值。

贪心，首先当k>=N时 答案就是体积最大球，当k<N时，有的盒子需要装入两个球，有的盒子需要装一个。那么能得到需要装一个球的有2K-N个，那么剩下的球需要两两装入一个盒子，这样的球有n-(2k-n)=2n-2k;

然后贪心的去合并两个球 最小的球和最大球合并，一直循环下去，找到一个最大值就是满足条件的答案。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <vector>
#include <stack>
#include <cmath>

using namespace std;

const int M = 10005;
const int maxn = 1000000;
typedef long long LL;

vector<int>G[maxn];
queue<int>Q;
stack<int>st;


LL a[maxn];

int main()
{

  int n,k;
  cin>>n>>k;
  LL res = 0;
  for(int i=0;i<n;i++){
    cin>>a[i];
    res = max(res,a[i]);
  }
  for(int i=0;i<n-k;i++){
    res = max(res,a[i]+a[2*(n-k)-i-1]);
  }
  cout << res;


    return 0;
}