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：n个水桶，每个桶k快板，且装水具有短板效应，所有桶之间能装的水不能超过l，求出n个桶最大总装水量的拼装桶的方式，共给你k*n快板
题解：贪心，排一下序，upper_bound找出排序后a[0]到a[0]+l这部分有多少块板pos，每个桶的最短板必须从这里面取，当pos<n的时候就不满足n个桶之间装水量<=l了。当pos==n时，这部分班对应n个桶的最短板，当pos>n时，要使短板经量的大，所以小的板经量组装在一个桶，既依次取最短板a[0],a[0+k},a[0+2k].....,又最短板只能从0到pos内取，所以再从pos-1，一直往下取，直到取了n快最短板
注意：ans 会爆int，所以要用long long
代码：
#include <bits/stdc++.h>
using namespace std;
int n, k ,l;
const int M = 2e5+10;
long long a[M];
int vis[M];
long long ans;
int main()
{
cin>>n>>k>>l;
memset(vis,0,sizeof(vis));
ans =0;
for(int i=0;i<n*k;i++)
{
cin>>a[i];
}
sort(a,a+n*k);
int pos = upper_bound(a,a+n*k,a[0]+l)-a;
//cout<<pos<<" "<<a[pos]<<endl;
int p=0;
if(pos<n)
cout<<0<<endl;
else if(pos==n)
{
for(int i=0;i<pos;i++)
{
ans+=a[i];
}
cout<<ans<<endl;
}
else if(pos>n)
{
for(int i=0;i<pos;i+=k)
{
ans+=a[i];
p++;
vis[i] = 1;
if(p==n)
break;
}
if(p<n)
{
for(int i=pos-1;i>=0;i--)
{
if(p==n)
break;
if(vis[i])
continue;
ans+=a[i];
vis[i] = 1;
p++;
if(p==n)
break;
}
}
cout<<ans<<endl;
}
}