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

题目链接：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 intok boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more thantwo 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 integers_i. In fact, he keeps his cowbells sorted by size, sos_i - 1 ≤ s_i for anyi > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of sizes if and only if the sum of their sizes does not exceeds. Given this information, help Kevin determine the smallests for which it is possible to put all of his cowbells intok 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 integerss₁, 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 intok boxes of size s.

Sample test(s)

Input

2 12 5

Output

7

Input

4 32 3 5 9

Output

9

Input

3 23 5 7

Output

8

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}.

二分+贪心

题意：n个物品放入k个盒子中，每个盒子最多放入两个物品，要求物品权值小于盒子的权值才能放入，求将物品全部放入盒子时盒子的最小权值是多少。

数据：(1 ≤ n ≤ 2·k ≤ 100 000）

解题方法：二分+贪心

#include<stdio.h>#include<algorithm>using namespace std;long long int s[100100],n,k;int check(int x){    long long int xx=1,yy=n,ans=0,i;    for(i=1;i<=n;i++)    if(x<s[i]) return 0;//单个物品超过x显然是不合理的    while(1)    {        if(xx>yy) break;        if(s[xx]+s[yy]>x)        {            yy--;ans++;        }//如果两个物品相加大于x，那么这个大物品单独放在一个盒子里        else        {            ans++; xx++;yy--;        }    }    if(ans<=k) return 1;    else return 0;}int main(){    long long int i,j;    while(scanf("%lld%lld",&n,&k)!=EOF)    {      for(i=1;i<=n;i++)      scanf("%lld",&s[i]);      long long int l=1,r=2*1000000,mid;      while(l<=r)      {          mid=(r+l)/2;          if(check(mid)) r=mid-1;          else l=mid+1;      }      printf("%lld\n",l);//注意输出的是l    }}