一般贪心
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
7
4 3
2 3 5 9
9
3 2
3 5 7
8
Sample Output
Hint
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}.
题意大概是N个东西放到K个箱子,可以两个放到一起,求最小的箱子体积(没个箱子都一样大)
此为一道贪心,然而直接思考可能太抽象,不妨把想象过程,在思想中模拟(神探夏洛克的的思维宫殿),这时就会发现只是一道水题#.#;
我们先从大的开始放到箱子里,直到放满,剩下的小的再往里放,最小的放到能相对来说最大的,
贪心就是找到一种放的方式先开始,然后开始规划改变称最优的;
这里给一下代码
#include<stdio.h> #include<algorithm> using namespace std; long long n,k,s[100005],maxs; int main() { scanf("%lld%lld",&n,&k); for(long long i=0; i<n; i++) scanf("%lld",&s[i]); maxs=s[n-1]; for(long long i=0; i<n-k; i++) maxs=max(maxs,s[i]+s[(n-k)*2-i-1]); printf("%lld\n",maxs); return 0; }