Codeforce Div-2 985 C. Liebig's Barrels

Posted 2020-11-07 断腿三郎

http://codeforces.com/contest/985/problem/C

 

C. Liebig‘s Barrels

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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

Copy

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.

 

思路：

　　先找把每k个中最小的加起来，这是必须的。然后再把符合要求的从上至下，加在一起。

　　贪心思想。

　　不懂请留言，没人看不想写得太详细，况且这题本身不难。

#include<iostream>
#include<algorithm>
using namespace std;
long long a[2000086];
bool book[2000086];
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 flag=0;
    int maxx,minn,t=n*k;
    long long ans=0;
    if(a[n]-a[1]>l){cout<<0<<endl;}
    else{
        for(int i=n;i<=n*k;i++){
            if(a[i]-a[1]>l){t=i-1;break;}
        }
        int flag=0;
        for(int j=1;j<=t;j+=k){
            ans+=a[j];flag++;
            book[j]=true;
            if(flag==n){break;}
        }
        for(int j=t;j>=1;j--){
            if(flag==n){break;}
            if(!book[j]){ans+=a[j];flag++;}
            if(flag==n){break;}
        }
        cout<<ans<<endl;
    }
}