C. Liebig's Barrels （贪心）

C. Liebig's Barrels （贪心）

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 12 2 1 2 3 2 2 3

output Copy

7

input Copy

2 1 010 10

output Copy

20

input Copy

1 2 15 2

output Copy

2

input Copy

3 2 11 2 3 4 5 6

output Copy

0

思路：

一道很贪心的贪心题。

比赛时，第一次思路想简单了，我也很感慨我当时能回归正途。

先找出符合 l  的最长的木板，然后木桶装水最多为l，就用比l长的（k-1）块木板，从 l 到 最小 开始组合。如果还有剩余的，然后再从大到小选出k个一组，组合。

代码：

#include using namespace std;const int maxn=100010;int a[maxn];int main(){ int n,m,l; scanf("%d%d%d",&n,&m,&l); long long sum5=0; for(int i=1;i<=n*m;i++) { scanf("%d",&a[i]); sum5+=a[i]; } sort(a+1,a+1+n*m); if(m==1) { if(a[n*m]-a[1]<=l) { printf("%I64d\n",sum5); } else printf("0\n"); } else { int st=0; priority_queue G; for(int i=n*m;i>=n;i--) { if(a[i]-a[1]<=l) { st=i; break; } G.push(a[i]); } if(st==0) { printf("0\n"); } else { long long ans=0; int su=G.size(); int ans1=su/(m-1); for(int i=st;i>=st-ans1+1;i--) { int sum=a[i]; for(int j=1;j<=m-1;j++) { int sum1=G-();G.pop(); sum=min(sum,sum1); } ans+=sum; } for(int i=1;i

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