Minimization - Codeforces 571 B

Minimization

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You've got array A, consisting of n integers and a positive integer k. Array A is indexed by integers from 1 to n.

You need to permute the array elements so that value

became minimal possible. In particular, it is allowed not to change order of elements at all.

Input

The first line contains two integers n, k (2 ≤ n ≤ 3·105, 1 ≤ k ≤ min(5000, n - 1)).

The second line contains n integers A[1], A[2], ..., A[n] ( - 109 ≤ A[i] ≤ 109), separate by spaces — elements of the array A.

Output

Print the minimum possible value of the sum described in the statement.

Sample test(s)

input

3 2
1 2 4

output

1

input

5 2
3 -5 3 -5 3

output

0

input

6 3
4 3 4 3 2 5

output

3

Note

In the first test one of the optimal permutations is 1 4 2.

In the second test the initial order is optimal.

In the third test one of the optimal permutations is 2 3 4 4 3 5.

思路：可以将最后形成的这个序列拆成k组，其中ai和ai+k为一组。将原序列排序后，将其拆成连续的k组，然后即可拼成这个最后的序列。这k组中，有a=n%k组的长度为len1=n/k+1,有b=k-n组的长度的len=n/k。然后用dp[i][j]去求最小解。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int T,t,n,m,a,b,len1,len2,ans;
int num[300010];
int dp[5010][5010],INF=1e9;
int main()
{
    int i,j,k,u;
    INF=INF*2+10;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
       scanf("%d",&num[i]);
    len2=n/m;
    len1=len2+1;
    a=n%m;
    b=m-a;
    sort(num+1,num+1+n);
    for(i=0;i<=a;i++)
       for(j=0;j<=b;j++)
       {
           dp[i][j]=INF;
           k=i*len1+j*len2;
           if(i>0)
             dp[i][j]=min(dp[i][j],dp[i-1][j]+num[k]-num[k-len1+1]);
           if(j>0)
             dp[i][j]=min(dp[i][j],dp[i][j-1]+num[k]-num[k-len2+1]);
           if(i+j==0)
             dp[i][j]=0;
       }
    printf("%d\n",dp[a][b]);
}