codeforces 604B (贪心)

B. More Cowbell

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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}.

写的时候傻逼的把数据范围看错，敲了一发二分的，然后又把题目理解错，以为是从头到尾中间截

取的．后来仔细看了看题目数据范围原来每个背包最多放两个，有k*2-n个背包只需要放一个，先

贪心的把最大的先放了，然后剩下的背包每个放两个物品，每次选择一个当前最大的和最小的．

#include <bits/stdc++.h>using namespace std;#define maxn 111111int n, k;int a[maxn];int main () {    while (cin >> n >> k) {        k = min (k, n);        for (int i = 1; i <= n; i++)            cin >> a[i];        int ans = 0;        for (int i = 1; i <= k*2-n; i++)            ans = max (a[n-i+1], ans);        for (int i = 1; i <= n*2-k*2; i++) {            ans = max (a[i]+a[2*n-2*k-i+1], ans);        }        cout << ans << endl;    }    return 0;}