Codeforces Round #334 (Div. 2) B. 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 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}.

题意：给一个长度为n的有序的数列（从小到大）分成k组（如ai和aj一组分组和变为ai+aj），求分组后最大的数是多少

贪心

当n<=k时

每个数一组

当n>k时

尽量让前面的数两两一组

前面的有尽量最大的和最小的一组

#include
   
    
#include
    
     
#include
     
      
#include
      
       
using namespace std;

const int N = 100010;
int a[N];

int main(void)
{
    int n, k;
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d%d", &n, &k))
    {
        int i;
        for(i = 0; i < n; i++) scanf("%d", &a[i]);
        sort(a, a+n);
        if(n <= k)
        {
            printf("%d\n", a[n-1]);
        }
        else
        {
            int tmp = n - k;
            int MAX = -1;
            for(i = 0; i < tmp; i++)
            {
                MAX = max(MAX, a[i]+a[2*tmp-i-1]);
                //printf("i=%d \t 2*tmp-i-1=%d\n", i, 2*tmp-i-1);
            }

            if(n > tmp*2) MAX = max(a[n-1], MAX);
            printf("%d\n", MAX);
        }
    }
    return 0;
}