B. Levko and Array----二分答案

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本文探讨了如何通过更改数组中最多k个元素来最小化最大相邻元素差的方法。使用二分查找结合动态规划思想，有效地解决了问题。

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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:

$\text{[math]}$ 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
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.

题目链接：http://codeforces.com/contest/360/problem/B

题目的意思是说最多改变数组中k个数，使数组满足c=max(a[i+1]-a[i]) 1<=i<n&&n>1 n<=1时 c=0;也就是说满足两个相邻的数的绝对值差的最大值最小。

这个题一开始看出是二分答案了，但是并没有想出来，去看了网上的博客，但是博客都是二分+dp，我又交了几发，终于让我搞出了二分答案的做法。

f(i)表示前i个的最小修改次数，且i不修改，枚举上一个不修改的位置(感觉我的check里还是融入了dp的思想)。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#define inf 1000000000
#define ll long long
using namespace std;
int n,k;
int a[2005],f[2005];
bool check(ll x){
    for(int i=1;i<=n;i++){
        f[i]=i-1;
        for(int j=1;j<i;j++){
            if(abs(a[i]-a[j])<=(i-j)*x)
                f[i]=min(f[i],f[j]+(i-j-1));
        }
        if(f[i]+n-i<=k)
            return 1;
    }
    return 0;
}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    ll l=0,r=2*inf;
    while(l<=r){
        int mid=(l+r)>>1;
        if(check(mid)){
            r=mid-1;
        }
        else{
            l=mid+1;
        }
    }
    printf("%I64d\n",l);
    return 0;
}