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

题意：n个东西，k个箱子，一个箱子能放两个东西，东西有重量，一个箱子里面东西的重量就是他的容量，求最小容量方法中的最大容量

思路：分情况：如果n<=k，那么ans为a
，如果n>k，那么就要贪心了，选取前n-k个东西，然后将其倒序插入剩余的东西中，过程中取最大值即可
总结：这道题改了不少，一开始没看到一个箱子只能装两个东西，然后又理解有误，WA了好多，最后把自己绕晕了，好久才改对。衰= =

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 100100
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int a[MAXN];
int main()
{
int n,k,i,j;
int ans;
while(scanf("%d%d",&n,&k)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
if(n<=k)
{
printf("%d\n",a[n-1]);
continue;
}
ans=a[n-1];
int cnt=n-k;
int j=cnt;
for(i=cnt-1;i>=0;i--)
ans=max(ans,a[i]+a[j++]);
printf("%d\n",ans);
}
return 0;
}

几组测试数据：

2 1
2 54 3
2 3 5 92 1
2 53 2
3 5 76 3
1 2 3 4 5 6
5 3
1 1 2 2 84 2
1 2 3 5