The fastfood chain McBurger owns several restaurants along a highway. Recently, they have decided to build several depots along the highway, each one located at a restaurent and supplying several of the restaurants with the needed ingredients. Naturally, these depots should be placed so that the average distance between a restaurant and its assigned depot is minimized. You are to write a program that computes the optimal positions and assignments of the depots.
To make this more precise, the management of McBurger has issued the following specification: You will be given the positions of nrestaurants along the highway as n integers $d_1 < d_2 < \dots < d_n$ (these are the distances measured from the company’s headquarter, which happens to be at the same highway). Furthermore, a number $k (k \leŸ n)$ will be given, the number of depots to be built.

The k depots will be built at the locations of k different restaurants. Each restaurant will be assigned to the closest depot, from which it will then receive its supplies. To minimize shipping costs, the total distance sum, defined as

\begin{displaymath}\sum_{i=1}^n \mid d_i - (\mbox{position of depot serving restaurant }i) \mid
\end{displaymath}

must be as small as possible.

Write a program that computes the positions of the k depots, such that the total distance sum is minimized.

Input 

The input file contains several descriptions of fastfood chains. Each description starts with a line containing the two integers n and k. n and kwill satisfy $1 \leŸ n
\leŸ 200$, $1 \leŸ k Ÿ\le 30$, $k \le n$. Following this will n lines containing one integer each, giving the positions di of the restaurants, ordered increasingly.

The input file will end with a case starting with n = k = 0. This case should not be processed.

Output 

For each chain, first output the number of the chain. Then output an optimal placement of the depots as follows: for each depot output a line containing its position and the range of restaurants it serves. If there is more than one optimal solution, output any of them. After the depot descriptions output a line containing the total distance sum, as defined in the problem text.
Output a blank line after each test case.

Sample Input 

6 3
5
6
12
19
20
27
0 0

Sample Output 

Chain 1
Depot 1 at restaurant 2 serves restaurants 1 to 3
Depot 2 at restaurant 4 serves restaurants 4 to 5
Depot 3 at restaurant 6 serves restaurant 6
Total distance sum = 8
#include<cstdio>
#include<cstring>
#include<algorithm>
#define mem(name,value) memset(name,value,sizeof(name))
#define FOR(i,n) for(int i=1;i<=n;i++)
using namespace std;
const int maxn = 200+10;
const int maxk = 30+10;
const int inf = 0x3f3f3f3f;
int d[maxk][maxn],des[maxn],a1[maxk];
typedef pair<int,int>pii;
pii p[maxk][maxn],p1[maxk];
int dp(int cnt,int n){
    if(d[cnt][n]!=-1) return d[cnt][n];
    int &ans = d[cnt][n];
    if(cnt==0) return ans = (n ? inf : 0);
    ans = inf;
    for(int i=cnt;i<=n;i++){
        int len = (n-i)/2, v = i+len;
        int s = 0;
        for(int j=v+1;j<=n;j++) s += des[j] - des[v];
        for(int j=v-1;j>=i;j--) s += des[v] - des[j];
        int tmp = dp(cnt-1,i-1) + s;
        if(tmp < ans){
            ans = tmp;
            p[cnt][n] = make_pair(i,v);
        }
    }
    return ans;
}
void print_ans(int cnt,int n){
    if(cnt==0) return ;
    a1[cnt] = p[cnt][n].second;
    p1[cnt] = make_pair(p[cnt][n].first,n);
    print_ans(cnt-1,p1[cnt].first-1);
}
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out2.txt","w",stdout);
    int n,k,kase=0;
    while(~scanf("%d%d",&n,&k) && n){
        printf("Chain %d\n",++kase);
        mem(d,-1);
        des[0] = 0;
        for(int i=1;i<=n;i++){
            scanf("%d",&des[i]);
        }
        int solve = dp(k,n);
        print_ans(k,n);
        for(int i=1;i<=k;i++){
            printf("Depot %d at restaurant %d serves ",i,a1[i]);
            if(p1[i].first==p1[i].second) printf("restaurant %d\n",p1[i].first);
            else printf("restaurants %d to %d\n",p1[i].first,p1[i].second);
        }
        printf("Total distance sum = %d\n\n",solve);
    }
    return 0;
}
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