Codeforces Round #334 (Div. 2)

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 kboxes 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}.

错因：没有仔细分析题意，一直以为是分组背包问题，，自己把自己吓到了

解答：贪心问题，先将几个最大的物体一个一个的放入背包中，然后再将小物体加入已有大物体的背包中，要倒序（贪心）

<span style="font-size:18px;">#include<iostream>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[100005];
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int n,k;
    while(~scanf("%d %d",&n,&k))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        if(n<=k)
        {
            printf("%d
",a[n]);
            continue;
        }
        int ans=a[n],cnt=n-k,j=n-k+1;
        for(int i=cnt;i>=1;i--)
            ans=max(ans,a[i]+a[j++]);//倒序加进背包，体现贪心思想
        printf("%d
",ans);
    }
    return 0;
}
</span>