1 //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 2 //dp[i][j]:把第i个数转成第j小的数，最小花费 3 //此题与poj 3666相似 a[i]转换成a[i]-i 4  5 #include <iostream> 6 #include <cstdio> 7 #include <cstdlib> 8 #include <algorithm> 9 #include <vector>10 #include <math.h>11 #include <memory.h>12 using namespace std;13 #define LL long long14 typedef pair<int,int> pii;15 const int inf = 0x3f3f3f3f;16 const LL MOD =100000000LL;17 const int N = 3000+10;18 const double eps = 1e-8;19 void fre() {freopen("in.txt","r",stdin);}20 void freout() {freopen("out.txt","w",stdout);}21 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;}22 23 int a[N],b[N];24 LL dp[N][N];25 int main(){26     int n;27     scanf("%d",&n);28     for(int i=1;i<=n;i++){29         scanf("%d",&a[i]);30         a[i]-=i;31         b[i]=a[i];32     }33     sort(b+1,b+1+n);34     for(int i=1;i<=n;i++){35         dp[1][i]=abs(a[1]-b[i]);36     }37     for(int i=2;i<=n;i++){38         LL minn=1e18;39         for(int j=1;j<=n;j++){40            minn=min(minn,dp[i-1][j]);41            dp[i][j]=minn+abs(a[i]-b[j]);42         }43     }44     LL ans=1e18;45     for(int i=1;i<=n;i++){46         ans=min(ans,dp[n][i]);47     }48     cout<<ans<<endl;49     return 0;50 }