Codeforces Round #334（div.2）（新增不用二分代码） B

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 1
2 5

output

7

input

4 3
2 3 5 9

output

9

input

3 2
3 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}.

贪心+二分（不用二分也可以）

还有，每个容器只能放两个，要符合条件，当然得最大的加最小的

当然，输入的数字已经是非递减的了，更多解释可以看代码注释

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN = 100005;
int n, k;
LL a[MAXN];
int i,j;
bool BS(LL key)
{
    int sum=0;
    int l=1;
    int r=n;
    while(l<=r)
    {
        if(l!=r&&a[l]+a[r]<=key)
        {
            l++;
            r--;
        }
        else
        {
            r--;//哎呀，大了，把右界限减少一点
        }
        sum++;
        if(sum>k) return 0;//超过限制
    }
    return 1;
}
int main()
{
    cin>>n>>k;
    for(i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    LL l = a[n], r = 1e12, ans;
    while (l <= r)
    {
        LL mid = (l + r) >> 1;
        if (BS(mid))
        {
            ans = mid;
            r = mid - 1;//符合条件，减小右边界，我们要的是最优的
        }
        else l = mid + 1;//不符合，左边界推，增大数值
    }
    cout<<ans<<endl;
    return 0;
}

　好啦，窝萌不用二分怎么办呢，这样想，反正容器最多放二个，那么我就放K个进去，不过是从后往前放的，比如是 1 2 3 4 5   K=3  那么我先把 3 4 5放进去，把它写成公式就是还剩下n-k个，接下来还是最大的和最小的相加（相对），如 2+3  1+4  最后一个是最大的数，窝萌把它和之前的和比较 ，得出 5 。（1 4）（2 3）（5）这样符合条件

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int main()
{
    int n,k;
    int a[100010];
    int i,j;
    int ans1,ans2;
    int sum1=0,sum2=0;
    cin>>n>>k;
    for(i=0;i<n;i++)
    {
        cin>>a[i];
    }
    sum1=a[n-1];
    ans1=n-k-1;ans2=n-k;
    for(i=ans1;i>=0;i--)
    {
       sum1=max(sum1,a[ans1--]+a[ans2++]);
    }
    cout<<sum1<<endl;
    return 0;
}