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最新推荐文章于 2024-06-28 14:18:34 发布

原创最新推荐文章于 2024-06-28 14:18:34 发布 · 513 阅读

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本文介绍了一种关于如何将不同大小的物品最优地装入固定数量的盒子中的算法问题。问题的核心在于确定能够容纳所有物品的最小盒子尺寸，同时每个盒子最多只能装两个物品。文章通过具体的示例和代码解释了不同的解决策略。

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 s_i. In fact, he keeps his cowbells sorted by size, so s_i - 1 ≤ s_i 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, ..., s_n (1 ≤ s₁ ≤ s₂ ≤ ... ≤ s_n ≤ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes s_i 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

Input

4 3
2 3 5 9

Output

Input

3 2
3 5 7

Output

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：给出 n 个球的容量，每个球的容量已经按非降序排列好，要求放入 k 个盒子，每个盒子最多放 2 个球，求放完 k 个盒子后，这 k 个盒子中球占据最多容量的盒子的容量。（1 <= n && n <= 2*k）

思路：按 n 的取值分 3 大类讨论。

1：当 n<=k 时直接输出第 n 个球的容量；

2：当 k < n && n < 2*k 时，先将后 k 个球放入 k 个盒子，再将剩余的（n-k）个球按容量较小的加到后 k 个容量较大的，详见代码；

/* 例如实例

5 3
1 2 3 4 5

（先将后 3 个球存入 3 个盒子，再将容量为 2 的球放入容量为 3 的球的盒子中，将容量为1 的球放入容量为 4 的球的盒子中，结果盒子容量为 5 5 5）

其它依次类推！

3：当 k == 2*k 时，每次取对称的头尾两个相加放入同个盒子，每两个球 1 个盒子，刚好分配完，每次取最大值，不要误以为第一个和最后一个和是最大的。例如（1 2 2 2）

！！！最坑是我以为一定要放满 k 个盒子，所以球数最少得是 k ，没想到 WA ,感觉题目说得并不清楚，所以还是得考虑全，不能凭主观意愿。！！！

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
int s[1000010];
using namespace std;
int main()
{
#ifdef OFFLINE
	freopen("t.txt", "r", stdin);
#endif
	int i, j, k, n, ans, x, y;
	while(~scanf("%d %d", &n, &k)){
		for(i=1;i<=n;i++)
			scanf("%d", &s[i]);//球容量数组
		ans=0;//初始化
		if(n<=k)  ans=s[n];//记得加上 < 号
		else if(k<n&&n<2*k){
			x=n-k, y=0;//固定 n-k 存入 x
			while(n-k){
				y++;
				ans=max(s[n-k]+s[x+y], ans);
				k++;
			}
			ans=max(ans, s[n]);
		}
		else if(n==2*k){
			i=1, j=n;
			while(i<j){//对称取球
				ans=max(s[i]+s[j], ans);//取最大值
				i++, j--;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}