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
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}.
题意是给n个牛铃和k个箱子，每个箱子可以装1~2个牛铃，每个牛铃大小不同，所有箱子都是s容量，求最小的s可以装所有的牛铃。
贪心，读入的数据已经排好，将k个牛铃放进k个箱子，每个箱子一个，那么还有n-k个牛铃必须放到已经装有1个的箱子里，那么就有n-k个箱子是装两个的。我们让只装一个牛铃的箱子尽量装大的，要装两个的箱子，装最小的（n-k）*2个里面最大的和最小的，第二大和第二小……
代码语言：javascript复制#include<stdio.h>
#include<algorithm>
using namespace std;
long long n,k,s[100005],maxs;
int main(){
scanf("%lld%lld",&n,&k);
for(long long i=0;i<n;i )
scanf("%lld",&s[i]);
maxs=s[n-1];
for(long long i=0;i<n-k;i )
maxs=max(maxs,s[i] s[(n-k)*2-i-1]);
printf("%lldn",maxs);
return 0;
}
