把一个序列转换成严格递增序列的最小花费 CF E

//把一个序列转换成严格递增序列的最小花费 CF E - Sonya and Problem Wihtout a Legend
//dp[i][j]:把第i个数转成第j小的数，最小花费
//此题与poj 3666相似 a[i]转换成a[i]-i

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <math.h>
#include <memory.h>
using namespace std;
#define LL long long
typedef pair<int,int> pii;
const int inf = 0x3f3f3f3f;
const LL MOD =100000000LL;
const int N = 3000+10;
const double eps = 1e-8;
void fre() {freopen("in.txt","r",stdin);}
void freout() {freopen("out.txt","w",stdout);}
inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}

int a[N],b[N];
LL dp[N][N];
int main(){
    int n;
    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+1+n);
    for(int i=1;i<=n;i++){
        dp[1][i]=abs(a[1]-b[i]);
    }
    for(int i=2;i<=n;i++){
        LL minn=1e18;
        for(int j=1;j<=n;j++){
           minn=min(minn,dp[i-1][j]);
           dp[i][j]=minn+abs(a[i]-b[j]);
        }
    }
    LL ans=1e18;
    for(int i=1;i<=n;i++){
        ans=min(ans,dp[n][i]);
    }
    cout<<ans<<endl;
    return 0;
}