C. Liebig's Barrels

最新推荐文章于 2020-07-14 15:38:12 发布

转载最新推荐文章于 2020-07-14 15:38:12 发布 · 99 阅读

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原文链接：http://www.cnblogs.com/qldabiaoge/p/9071432.html

本文介绍了一种关于桶体积最大化的算法实现，该算法通过排序和贪心策略，在限定条件下计算能够组成的桶的最大总体积。文章包含具体实现代码，详细解释了如何确保桶之间的体积差异不超过指定限制。

You have m = n·k wooden staves. The i-th stave has length ai. You have to assemble nbarrels 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 and1 ≤ y ≤ n.

Examples

input

Copy

4 2 1
2 2 1 2 3 2 2 3

output

Copy

input

Copy

2 1 0
10 10

output

Copy

input

Copy

1 2 1
5 2

output

Copy

input

Copy

3 2 1
1 2 3 4 5 6

output

Copy

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.

诸事不顺，操

一个贪心，其实就是分为n堆数，每堆数的最小值相差不能大于limit ，

求出n堆数最小值的和

upper_bound 返回的是第一个大于的数，减去1就是小于等于的数了

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 long long a[maxn];
 5 int n, k, limit;
 6 int main() {
 7     scanf("%d%d%d", &n, &k, &limit);
 8     for (int i = 0 ; i < n * k ; i++)
 9         scanf("%lld", &a[i]);
10     sort(a, a + n * k );
11     int temp = upper_bound(a, a + n * k, a[0] + limit) - a;
12     long long ans = 0;
13     int sum = n * k;
14     if (temp >= n) {
15         int temp1=temp;
16         while(sum > temp && sum - temp >= k - 1) {
17             sum -= k - 1;
18             ans += a[--temp1];
19         }
20         for (int i = 0 ; i * k < temp1 ; i++)
21             ans += a[i * k];
22     }
23     printf("%lld\n", ans);
24     return 0;
25 }

转载于:https://www.cnblogs.com/qldabiaoge/p/9071432.html