CF#571 B. Minimization （DP）

题目点我点我点我

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.

Examples

input

3 21 2 4

output

1

input

5 23 -5 3 -5 3

output

0

input

6 34 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.

题目大意：给出一个序列，可以任意调整序列的顺序，使得给出的式子的值最小 

∑k=1n−k|ai−ai+k|

解题思路：将序列升序排序后，dp[i][j]:选择i个数量为n/k+1的集合，选择j个数量为n/k的集合的最小值。n/k+1的集合有n%k组，n/k的集合有k-n%k组。

如：序列为1，2，3，4，5，6，7，8，9，10.

当k=2时，不难看出来，1，6，2，7，3，8，4，9，5，10才是最优解，

即有2个n/k的集合，分别为｛1，2，3，4，5｝，｛6，7，8，9，10｝。

这里不用考虑数量为n/k+1的集合是n%k=0，若n%k！=0的话则要考虑。

/* ***********************************************┆  ┏┓　　　┏┓ ┆┆┏┛┻━━━┛┻┓ ┆┆┃　　　　　　　┃ ┆┆┃　　　━　　　┃ ┆┆┃　┳┛　┗┳　┃ ┆┆┃　　　　　　　┃ ┆┆┃　　　┻　　　┃ ┆┆┗━┓　马　┏━┛ ┆┆　　┃　勒　┃　　┆　　　　　　┆　　┃　戈　┗━━━┓ ┆┆　　┃　壁　　　　　┣┓┆┆　　┃　的草泥马　　┏┛┆┆　　┗┓┓┏━┳┓┏┛ ┆┆　　　┃┫┫　┃┫┫ ┆┆　　　┗┻┛　┗┻┛ ┆************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <bitset>using namespace std;#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)#define pb push_back#define mp make_pairconst int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define LL long long#define ULL unsigned long long#define MS0(X) memset((X), 0, sizeof((X)))#define SelfType intSelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}#define Sd(X) int (X); scanf("%d", &X)#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())#define all(a) a.begin(), a.end()typedef pair<int, int> pii;typedef pair<long long, long long> pll;typedef vector<int> vi;typedef vector<long long> vll;inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")LL dp[5005][5005];int a[300005];int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);ios::sync_with_stdio(0);cin.tie(0);Sdd(n,k);rep(i,1,n)a[i] = read();sort(a+1,a+1+n);int big = n%k, small = k - n%k;int bsz = n/k+1, ssz = n/k;memset(dp,0x3f3f3f3f3f3f3f3f,sizeof dp);dp[0][0] = 0;rep(i,0,big){    rep(j,0,small)    {        int id = i * bsz + j * ssz;        if(i) dp[i][j] = min(dp[i][j],dp[i-1][j] + a[id] - a[id-bsz+1]);        if(j) dp[i][j] = min(dp[i][j],dp[i][j-1] + a[id] - a[id-ssz+1]);    }}printf("%I64d\n",dp[big][small]);return 0;}