CodeForces 604B More Cowbell

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
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 Input

Input

2 1
2 5

Output

7

Input

4 3
2 3 5 9

Output

9

Input

3 2
3 5 7

Output

8

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;
}