Maximum Sum(前缀和)

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Maximum Sum
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given an array a1,a2,…,an1,2,…,, where all elements are different.

You have to perform exactly k operations with it. During each operation, you do exactly one of the following two actions (you choose which to do yourself):

  • find two minimum elements in the array, and delete them;
  • find the maximum element in the array, and delete it.

You have to calculate the maximum possible sum of elements in the resulting array.

Input

The first line contains one integer t (1≤t≤1041≤≤104) — the number of test cases.

Each test case consists of two lines:

  • the first line contains two integers n and k (3≤n≤2⋅1053≤≤2⋅105; 1≤k≤999991≤≤99999; 2k<n2<) — the number of elements and operations, respectively.
  • the second line contains n integers a1,a2,…,an1,2,…, (1≤ai≤1091≤≤109; all ai are different) — the elements of the array.

Additional constraint on the input: the sum of n does not exceed 2⋅1052⋅105.

Output

For each test case, print one integer — the maximum possible sum of elements in the resulting array.

Example
input
Copy
6
5 1
2 5 1 10 6
5 2
2 5 1 10 6
3 1
1 2 3
6 1
15 22 12 10 13 11
6 2
15 22 12 10 13 11
5 1
999999996 999999999 999999997 999999998 999999995
output
Copy
21
11
3
62
46
3999999986
Note

In the first testcase, applying the first operation produces the following outcome:

  • two minimums are 11 and 22; removing them leaves the array as [5,10,6][5,10,6], with sum 2121;
  • a maximum is 1010; removing it leaves the array as [2,5,1,6][2,5,1,6], with sum 1414.

2121 is the best answer.

In the second testcase, it's optimal to first erase two minimums, then a maximum.

复制代码
//Maximum Sum
//利用前缀和维护数组的总和
//枚举每一种情况
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=3e5+10,mod=1e9+7;
int n,t,a[N],f[N],res,num,ans,m,s[N];
bool vis[N];
signed main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        res=0;
        cin>>n>>num;
        for(int i=1;i<=n;i++) cin>>a[i];
        sort(a+1,a+1+n);
        for(int i=1;i<=n;i++) s[i]=s[i-1]+a[i];
        for(int i=0;i<=num;i++){
            res=max(res,s[n-num+i]-s[2*i]);
        }
        cout<<res<<endl;
    }
    return 0;
}
复制代码

 


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本文作者：Sakurajimamai
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