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

 //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 = +;
 const double eps = 1e-;
 void fre() {freopen("in.txt","r",stdin);}
 void freout() {freopen("out.txt","w",stdout);}
 inline int read() {int x=,f=;char ch=getchar();while(ch>''||ch<'') {if(ch=='-') f=-; ch=getchar();}while(ch>=''&&ch<='') {x=x*+ch-'';ch=getchar();}return x*f;}

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