CodeForces - 361D Levko and Array二分加dp

D. 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, ... , an. 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, ... , an( - 109 ≤ ai ≤ 109).

Output

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

Examples input

5 2
4 7 4 7 4
      

output

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.

二分能得到的最大c（a），

check函數裡，對于每個c（a）,dp[i]表示前i個最少的更換次數

#include<stdio.h>
#include<algorithm>
#define rep(i,a,n) for(i=a;i<n;i++)
#define ll long long 
using namespace std;
int cmp(ll a,ll b){
	return a<b;
}
ll n,k,a[2005],dp[2005];
int check(ll x)
{
	int i,j;
	dp[0]=0;
	rep(i,1,n)
	{
		dp[i]=i;
		for(j=i-1;j>=0;j--)
		{
			if(abs(a[i]-a[j])<=x*(i-j))
			{
				dp[i]=min(dp[i],dp[j]+i-j-1);
			}
		}
		if(dp[i]+n-i-1<=k)
		return 1;
	}
	if(dp[n-1]<=k)
	return 1;
	return 0;
}
int main()
{
	ll le,ri,mid,i,ans;
	scanf("%lld %lld",&n,&k);
	rep(i,0,n)
	scanf("%lld",&a[i]); 
	le=0;ri=2000000000;
	while(le<=ri)
	{
		mid=(le+ri)>>1;
		if(check(mid))
		{
			ri=mid-1;
			ans=mid;
		}
		else
		le=mid+1;
	}
	printf("%lld\n",ans);
	return 0;
}