E - Levko and Array DP + e二分

E - Levko and Array

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description:  System Crawler  (2013-11-12)

Description

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 Input

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

 dp[i]代表前i个数里面至少要修改多少个数才会符合要求

 

const int maxn = 30000;
LL a[maxn];
LL c[maxn];
LL dp[maxn];
int n,t;
bool ok(LL x)
{
    repf(i,1,n)
    {
        dp[i] = i - 1;
        repf(j,1,i-1)
            if(abs(a[i] - a[j]) <= x*(i - j))
                dp[i] = min(dp[i],dp[j] + i - j - 1);
        if(dp[i]+n-i<=t)return true;
    } 
    return dp[n] <= t;
}
    
int main() 
{
    while(cin>>n>>t)
    {
        repf(i,1,n) scanf("%I64d",&a[i]);
        LL Max = 0;
        repf(i,2,n) c[i] = a[i] - a[i - 1],Max = max(Max,abs(c[i]));
        LL L = 0;
        LL R = Max;
        LL ans = Max;
        while(L <= R)
        {
            LL mid = (L + R)/2;
            if(ok(mid))
            {
                ans = mid;
                R = mid - 1;
            }else L = mid + 1;
        }
        cout<<ans<<endl;
    }
    return 0;
}