Problem:
Score : 600 points

Problem Statement

You are given an integer sequence A of length N and an integer K.
You will perform the following operation on this sequence Q times:

Choose a contiguous subsequence of length K, then remove the smallest element among the K elements contained in the chosen subsequence (if there are multiple such elements, choose one of them as you like).

Let X and Y be the values of the largest and smallest element removed in the Q operations. You would like X-Y to be as small as possible.
Find the smallest possible value of X-Y when the Q operations are performed optimally.

Constraints

1 \leq N \leq 2000

1 \leq K \leq N

1 \leq Q \leq N-K+1

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K Q
A_1 A_2 ... A_N

Output

Print the smallest possible value of X-Y.

Sample Input 1

5 3 2
4 3 1 5 2

Sample Output 1

1

In the first operation, whichever contiguous subsequence of length 3 we choose, the minimum element in it is 1.
Thus, the first operation removes A_3=1 and now we have A=(4,3,5,2).
In the second operation, it is optimal to choose (A_2,A_3,A_4)=(3,5,2) as the contiguous subsequence of length 3 and remove A_4=2.
In this case, the largest element removed is 2, and the smallest is 1, so their difference is 2-1=1.

Sample Input 2

10 1 6
1 1 2 3 5 8 13 21 34 55

Sample Output 2

7

Sample Input 3

11 7 5
24979445 861648772 623690081 433933447 476190629 262703497 211047202 971407775 628894325 731963982 822804784

Sample Output 3

451211184

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
//#include<string>
using namespace std;
int n,k,q,y,a[2001],b[2001],t,ll,s[2001],ss,ans=2e9;
bool cmp(int a,int b){
  return a<b;
}
int main(){
  scanf("%d %d %d",&n,&k,&q);
  for(register int i=1;i<=n;i++)
    scanf("%d",&a[i]);
  for(register int j=1;j<=n;j++){
  	y=a[j];//枚举能被删除的最小元素y
  	memset(s,0,sizeof(s));
  	ss=0;
	t=0;
	for(register int i=1;i<=n+1;i++){
  	   if(i==n+1||a[i]<y){//如果a[i]为一段的边界
		 sort(b+1,b+t+1,cmp);
		 ll=t-k+1;//当前这段能删的数的个数 
		 for(register int l=1;l<=ll;l++)
		    s[++ss]=b[l];//把能删的数存在s数组里
		 memset(b,0,sizeof(b));
		 t=0;//清空b数组 
	   }
	   else{
	   	 b[++t]=a[i];//不是边界就加入到当前这段 
	   }
	}
	if(ss<q){//如果能被删的数不足q个，就直接枚举下个y 
	  continue;
	}
	  sort(s+1,s+ss+1,cmp);
	  //printf("%d %d\n",s[q],y);
	  ans=min(ans,s[q]-y);//s[q]就是x,这里是更新ans
  }
  printf("%d\n",ans);
  return 0;
}