Problem:
Score : 300 points

Problem Statement

We will play a one-player game using a number line and N pieces.

First, we place each of these pieces at some integer coordinate.

Here, multiple pieces can be placed at the same coordinate.

Our objective is to visit all of the M coordinates X_1, X_2, ..., X_M with these pieces, by repeating the following move:

Move: Choose a piece and let x be its coordinate. Put that piece at coordinate x+1 or x-1.

Note that the coordinates where we initially place the pieces are already regarded as visited.

Find the minimum number of moves required to achieve the objective.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

-10^5 \leq X_i \leq 10^5

X_1, X_2, ..., X_M are all different.

Input

Input is given from Standard Input in the following format:

N M
X_1 X_2 ... X_M

Output

Find the minimum number of moves required to achieve the objective.

Sample Input 1

2 5
10 12 1 2 14

Sample Output 1

5

The objective can be achieved in five moves as follows, and this is the minimum number of moves required.

Initially, put the two pieces at coordinates 1 and 10.

Move the piece at coordinate 1 to 2.

Move the piece at coordinate 10 to 11.

Move the piece at coordinate 11 to 12.

Move the piece at coordinate 12 to 13.

Move the piece at coordinate 13 to 14.

Sample Input 2

3 7
-10 -3 0 9 -100 2 17

Sample Output 2

19

Sample Input 3

100 1
-100000

Sample Output 3

0

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

typedef long value_type; /* ソートするキーの型 */

value_type med3(value_type x, value_type y, value_type z)
/* x, y, z の中間値を返す */
{
    if (x < y)
        if (y < z) return y; else if (z < x) return x; else return z; else
        if (z < y) return y; else if (x < z) return x; else return z;
}

void quicksort(value_type a[], long left, long right)
/* クイックソート
 * a     : ソートする配列
 * left  : ソートするデータの開始位置
 * right : ソートするデータの終了位置
 */
{
    if (left < right) {
        long i = left, j = right;
        value_type tmp, pivot = med3(a[i], a[i + (j - i) / 2], a[j]); /* (i+j)/2ではオーバーフローしてしまう */
        while (1) { /* a[] を pivot 以上と以下の集まりに分割する */
            while (a[i] < pivot) i++; /* a[i] >= pivot となる位置を検索 */
            while (pivot < a[j]) j--; /* a[j] <= pivot となる位置を検索 */
            if (i >= j) break;
            tmp = a[i]; a[i] = a[j]; a[j] = tmp; /* a[i],a[j] を交換 */
            i++; j--;
        }
        quicksort(a, left, i - 1);  /* 分割した左を再帰的にソート */
        quicksort(a, j + 1, right); /* 分割した右を再帰的にソート */
    }
}

int main(){
  long n,m;
  long x[100000];
  long sa[100000];
  long sum=0;
  scanf("%d %d\n",&n,&m);
  for(long i=0;i<m;i++) {
    scanf("%ld",&x[i]);
  }
  for(long i=0;i<n-1;i++) {
    sa[i]=0;
  }
  quicksort(x,0,m-1);
  if(n == 1){
    sum = x[m-1] - x[0];
  } else if(n >= m) {
    sum = 0;
  } else {
    for(long i=1;i<m;i++) {
      sum += x[i]-x[i-1];
      for(long j=0;j<n-1;j++) {
        if(x[i]-x[i-1] > sa[j]) {
          sa[j] = x[i]-x[i-1];
          break;
        }
      }
    }
    for(long j=0;j<n-1;j++) {
      sum -= sa[j];
    }
  }
  printf("%ld",sum);
  return 0;
}