Codeforces Round #334 (Div. 2) B. More Cowbell 二分

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

 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

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[100005];
int check(int x)
{
    for(int i=1;i<=n;i++)
        if(a[i]>x)return 0;
    int ans = 0;
    int st = 1,ed = n;
    while(1)
    {
        if(st>ed)break;
        if(a[st]+a[ed]<=x)
        {
            st++,ed--;
            ans++;
        }
        else
        {
            ed--;
            ans++;
        }
    }
    if(ans<=k)return 1;
    return 0;
}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    int l = 1,r = 2* 1000000;
    while(l<=r)
    {
        int mid = (l+r)/2;
        if(check(mid))r = mid-1;
        else l = mid+1;
    }
    printf("%d\n",l);
}