CF985C.Liebig's Barrels
普及/提高-
通过率:0%
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题目描述
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.
输入格式
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.
输出格式
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 .
输入输出样例
输入#1
4 2 1 2 2 1 2 3 2 2 3
输出#1
7
输入#2
2 1 0 10 10
输出#2
20
输入#3
1 2 1 5 2
输出#3
2
输入#4
3 2 1 1 2 3 4 5 6
输出#4
0
说明/提示
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.