C. Liebig's Barrels_print single integer — maximal total sum of the vo-CSDN博客

本文链接：https://blog.csdn.net/wxl7777/article/details/80420329

You have m = n·k wooden staves. The i-th stave has length ai. You have to assemble n barrels consisting of k staves each, you can use any k staves to construct a barrel. Each stave must belong to exactly one barrel.

Let volume vj of barrel j be equal to the length of the minimal stave in it.

$\text{[math]}$

You want to assemble exactly n barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed l, i.e. |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

Print maximal total sum of volumes of equal enough barrels or 0 if it's impossible to satisfy the condition above.

Input

The first line contains three space-separated integers n, k and l (1 ≤ n, k ≤ 105, 1 ≤ n·k ≤ 105, 0 ≤ l ≤ 109).

The second line contains m = n·k space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — lengths of staves.

Output

Print single integer — maximal total sum of the volumes of barrels or 0 if it's impossible to construct exactly n barrels satisfying the condition |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

   Examples 
 
     input 
    
      Copy 
    
4 2 1
2 2 1 2 3 2 2 3

     output 
    
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7

     input 
    
      Copy 
    
2 1 0
10 10

     output 
    
      Copy 
    
20

     input 
    
      Copy 
    
1 2 1
5 2

     output 
    
      Copy 
    
2

     input 
    
      Copy 
    
3 2 1
1 2 3 4 5 6

     output 
    
      Copy 
    
0

Note

In the first example you can form the following barrels: [1, 2], [2, 2], [2, 3], [2, 3].

In the second example you can form the following barrels: [10], [10].

In the third example you can form the following barrels: [2, 5].

In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough.

题意：给你K*n木板每次拿出k快组成n个木桶每个木桶的体积等于那个最短的木板长度求最大体积（所有桶的）。

思路:排序加贪心（每次尽可能让小的在一起）。

#include<bits/stdc++.h>

using namespace std;
const int maxn=1e5+50;
int a[maxn];
int vis[maxn];
int main(){
    int n,k,l;
    cin>>n>>k>>l;
    for(int i=1;i<=n*k;i++)
        cin>>a[i];
    sort(a+1,a+n*k+1);
    int pos;
    if(a[n]-a[1]>l){
        cout<<'0'<<endl;
         return 0;
    }else{
        for(int i=n;i<=n*k;i++){
            if(a[i]-a[1]>l){
                pos=i-1;
                break;
            }
        }
        long long ans=0;
        int cnt=0;
        for(int i=1;i<=pos;i+=k){
            ans+=a[i];
            cnt++;
            vis[i]=1;
            if(cnt==n)
                break;
        }
        if(cnt==n)
            cout<<ans<<endl;
        else{
            for(int i=pos;i>=1;i--){
                if(!vis[i]){
                    ans+=a[i];
                    cnt++;
                }
                if(cnt==n)
                    break;
            }
            cout<<ans<<endl;
        }

    }
    return 0;
}