Codeforces 604B：More Cowbell

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 12 5

output

7

input

4 32 3 5 9

output

9

input

3 23 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}.

题意是一堆东西，放入一堆箱子中。给出了东西的数量和箱子的数量，并且给出了东西从小到大的体积。然后已经说明了最多把两个东西放在一个箱子中。箱子能放入东西的条件是箱子的体积大于等于放入东西的总的体积。

考虑不同种情况，贪心。

代码：

#pragma warning(disable:4996)  #include <iostream>  #include <algorithm>  #include <cmath>  #include <vector>  #include <string>  #include <cstring>#include <queue>#include <map>using namespace std;typedef long long ll;int n, k;int val[100005];int main(){    //freopen("i.txt", "r", stdin);    //freopen("o.txt", "w", stdout);    int i, j;    int cn, yn;    int res = 0;    scanf("%d%d", &n, &k);    for (i = 1; i <= n; i++)    {        scanf("%d", &val[i]);    }    cn = n / k;    yn = n%k;if(n<=k)//如果东西的数目比箱子数少{     cout << val[n] << endl;//取最大值}    else if (cn == 1&&yn==0)//如果东西的数目和箱子数相等    {        cout << val[n] << endl;//取最大值    }    else if (cn == 1 && yn)//有一部分的东西要两个放在一起    {        res = val[n];        for (i = 1; i <= yn; i++)        {            res = max(res, val[i] + val[2 * yn - i + 1]);        }        cout << res << endl;    }    else//所有东西都要两个放在一起    {        res = val[n];        for (i = 1; i <= n/2; i++)        {            res = max(res, val[i] + val[n - i + 1]);        }        cout << res << endl;    }    //system("pause");    return 0;}