codeforces 713C C. Sonya and Problem Wihtout a Legend(dp)（将一个数组变成严格单增数组的最少步骤）

E. Sonya and Problem Wihtout a Legend

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Sonya was unable to think of a story for this problem, so here comes the formal description.

You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.

Next line contains n integer ai (1 ≤ ai ≤ 109).

Output

Print the minimum number of operation required to make the array strictly increasing.

Examples

input

output

input

output

Note

In the first sample, the array is going to look as follows:

2 3 5 6 7 9 11

|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9

And for the second sample:

1 2 3 4 5

|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12

题意：

给一个序列，可以给每一个数加减一个数，代价为他们改变的数的绝对值，那么要求用最小代价把序列变成严格递增的

思路：

有一个非严格递增的版本：POJ3666

 而本题的严格如何转化非严格，直接加一行代码，a[i]-=i；

原理

推导过程大致如下：a[i] < a[i+1] —> a[i] <= a[i+1]-1 —> a[i]-i <= a[i+1]-(i+1)—> b[i]<=b[i+1]

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 3e3+5;
const int M = 24005;
int n,a[N],b[N];
ll dp[N][N];

int  main(){
  scanf("%d",&n);
  for(int i=1; i<=n; i++){
    scanf("%d",&a[i]);
    a[i]-=i;
    b[i]=a[i];
  }
  sort(b+1, b+n+1);
  for(int i=1; i<=n; i++){
    ll minn=dp[i-1][1];
    for(int j=1; j<=n; j++){
      minn=min(minn, dp[i-1][j]);
      dp[i][j]=abs(a[i]-b[j])+minn;
    }
  }
  ll ans=dp[n][1];
  for(int i=2; i<=n; i++){
    ans=min(ans, dp[n][i]);
  }
  printf("%lld
",ans);
  return 0;
}