CodeForces 604B More Cowbell_凯文是一个一丝不苟的牛铃收藏家,知道他的第i(1≤i≤n)牛铃的大小是一个整数si.事-优快云博客

本文介绍了一个关于计算最小箱容量的问题，Kevin需要将他的牛铃收藏品装入固定数量的箱子中，每个箱子最多只能装两个牛铃。文章提供了完整的代码实现，并通过样例输入输出展示了如何解决这个问题。

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因为读错题没有思路，没能正确求解。题目中说一个箱子最多能装俩，我却没看到。

Description

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 s_i. In fact, he keeps his cowbells sorted by size, so s_i - 1 ≤ s_i 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, ..., s_n (1 ≤ s₁ ≤ s₂ ≤ ... ≤ s_n ≤ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes s_i 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 Input

Input

2 1
2 5

Output

Input

4 3
2 3 5 9

Output

Input

3 2
3 5 7

Output

Sample Output

Hint

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}.

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
using namespace std;
//#define LOCAL
#define MAX_N 100000

int BOX[MAX_N];
int main()
{
	#ifdef LOCAL
		freopen("b:\\data.in.txt", "r", stdin);
	#endif

    int n , k;
    cin >> n >> k;
    for(int i = 0; i < n; i++)
    {
        cin >> BOX[i] ;
    }
    //先给ans最大的量
    int ans = BOX[ n - 1 ];
    if( n <= k)
        cout << ans << endl ;
    else
    {
        //n > k && n <= 2*k时
        int len = n - k ;//记录前向遍历的长度
        int j = len ;
        for(int i = len-1 ; i >= 0 ; i--)
        {
            ans = max( ans , BOX[i] + BOX[j++] );
        }
        cout << ans << endl ;
    }

    return 0;
}