B. More Cowbell

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/604/problem/B

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

Sample Input

2 1
2 5

Sample Output

7

HINT

题意

给你n个物品，k个盒子，每个盒子最多可以塞进去2个物品，但是塞进去的物品的权值和必须小于盒子的权值

问你盒子的权值最小可以为多少

保证n<=2*k

题解：

二分答案，check的时候，有一个贪心

最小的+最大的这样扔进去，比两个最小的这样扔进去更加优越

代码：

#include<iostream>

#include<math.h>

#include<stdio.h>

#include<algorithm>

using namespace std;

int n,k;

int a[];

int check(int x)

{

    for(int i=;i<=n;i++)

        if(a[i]>x)return ;

    int ans = ;

    int st = ,ed = n;

    while()

    {

        if(st>ed)break;

        if(a[st]+a[ed]<=x)

        {

            st++,ed--;

            ans++;

        }

        else

        {

            ed--;

            ans++;

        }

    }

    if(ans<=k)return ;

    return ;

}

int main()

{

    scanf("%d%d",&n,&k);

    for(int i=;i<=n;i++)

        scanf("%d",&a[i]);

    int l = ,r = * ;

    while(l<=r)

    {

        int mid = (l+r)/;

        if(check(mid))r = mid-;

        else l = mid+;

    }

    printf("%d\n",l);

}