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.
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.
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.
4 2 1 2 2 1 2 3 2 2 3
7
2 1 0 10 10
20
1 2 1 5 2
2
3 2 1 1 2 3 4 5 6
0
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n，k，L
给你n×k个木板要你组成n个木桶，然后每一个木桶的容量差值一定要在L的范围内。
题解：
贪心做法：for（1～n）里面一定是最小的，但是怎么取最优呢，以下是SQY学长的做法：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
ll a[maxn],n,m,k,len,upper,x,y,sum,cur,i;
int main()
{
cin>>n>>k>>len;
m=n*k;
for(int i=1;i<=m;i++){
cin>>a[i];
}
sort(a+1,a+1+m);
if(a[n]-a[1]>len){
return 0*puts("0");
}
if(k==1){
for(i=1;i<=n;i++)
sum+=a[i];
}else{
while (a[upper] - a[1] <= len && upper <= m) upper++;
upper--;
x=(upper-n)/(k-1);
for( i=1,cur=1 ; i<=x ; i++,cur+=k)sum+=a[cur];
for( i=1;i<n-x;i++){
sum+=a[upper--];
}
sum+= a[x * k+1];
}
cout<<sum<<endl;
return 0;
}
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll m,n,k,len;
int main()
{
cin>>n>>k>>len;
m=n*k;
ll a[m+10];
for(int i=1;i<=m;i++){
cin>>a[i];
}
sort(a+1,a+1+m);
if(a[n]-a[1]>len){
cout<<0<<endl;
}else{
ll cnt=0,pos,sum=0;
int vis[m+10]={0};
for(int i=n;i<=m;i++){
if(a[i]-a[1]>len){
pos=i-1;break;
}
}
for(int i=1;i<=pos&&cnt<n;i+=k){
sum+=a[i];
cnt++;
vis[i]=1;
}
for(int i=pos;pos>=n&&cnt<n;i--){
if(!vis[i]){
sum+=a[i];
cnt++;
vis[i]=1;
}
}
cout<<sum<<endl;
}
return 0;
return 0;
}