B. Levko and Array

B. Levko and Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Levko has an array that consists of integers: a₁, a₂, ... , a_n. But he doesn’t like this array at all.

Levko thinks that the beauty of the array a directly depends on value c(a), which can be calculated by the formula:

The less value c(a) is, the more beautiful the array is.

 

It’s time to change the world and Levko is going to change his array for the better. To be exact, Levko wants to change the values of at most k array elements (it is allowed to replace the values by any integers). Of course, the changes should make the array as beautiful as possible.

Help Levko and calculate what minimum number c(a) he can reach.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2000). The second line contains space-separated integers a₁, a₂, ... , a_n ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

A single number — the minimum value of c(a) Levko can get.

Sample test(s)

input

5 2
4 7 4 7 4

output

0

input

3 1
-100 0 100

output

100

input

6 3
1 2 3 7 8 9

output

1

Note

In the first sample Levko can change the second and fourth elements and get array: 4, 4, 4, 4, 4.

In the third sample he can get array: 1, 2, 3, 4, 5, 6.

//如果mid 可以满足 那么mid+1也是可以满足，这样就可以二分答案、
//这样就可以二分答案、注意二分的最大值要超过 2*10^9
// 然后就可以dp判断这个mid 是否满足
// dp[i] , 表示 计算到 a[i] ， a[i] 不变时需要最少的修改次数才可以满足 相邻差 mid
// 转移方程 dp[i] = min(dp[i],dp[j]+(i-j-1)) ; j < i && a[i]-a[j] <= (i-j)*mid ；
// 意思：如果a[i]-a[j] 满足 (i-j)*mid 那么 通过改变所有 j+1~i-1的数是可以满足条件的

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <algorithm>
#define maxn 2002
#define INF 2000000001
#define LL long long
using namespace std;

int n , k ;
LL a[maxn] ;
int dp[maxn] ;
int abs1( int a ){ return a >= 0 ? a : -a ;}
int min1( int a , int b ){ return a > b ? b : a ;}
bool find( LL mid ) 
{
    int i , j , u ;
    for( i = 1 ; i <= n+1 ; i++)
        dp[i] = i-1 ;
        dp[1] = 0 ;
    for( i = 2 ; i <= n ;i++ )
        for( j = 1 ; j < i ;j++ )if(abs1(a[i]-a[j]) <= (i-j)*mid)
             dp[i] = min1(dp[i],dp[j]+(i-j-1)) ;
    for( i = 1 ; i <= n ;i++)
        dp[n+1] = min1(dp[n+1],n-i+dp[i]) ;
    if(dp[n+1] <= k)return true ;
    return false ;
}
int main()
{
    int i ;
    LL mid , L , R ;
    //freopen("in.txt","r",stdin) ;
    while( scanf("%d%d" ,&n , &k ) != EOF )
    {
        for( i = 1 ; i <= n ;i++)
            scanf("%I64d" , &a[i]) ;
        L = 0 ; R = INF ; // 二分答案
        while( L <= R )
        {
            mid = (L+R)>>1 ;
            if(find(mid))R = mid-1 ;
            else L = mid+1 ;
        }
        printf("%d
",R+1) ;
    }
    return 0 ;
}