CodeForces 13C Sequence

C - Sequence

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

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Description

Little Petya likes to play very much. And most of all he likes to play the following game:

He is given a sequence of N integer numbers. At each step it is allowed to increase the value of any number by 1 or to decrease it by 1. The goal of the game is to make the sequence non-decreasing with the smallest number of steps. Petya is not good at math, so he asks for your help.

The sequence a is called non-decreasing if a₁ ≤ a₂ ≤ ... ≤ a_N holds, where N is the length of the sequence.

Input

The first line of the input contains single integer N (1 ≤ N ≤ 5000) — the length of the initial sequence. The following N lines contain one integer each — elements of the sequence. These numbers do not exceed 10⁹ by absolute value.

Output

Output one integer — minimum number of steps required to achieve the goal.

Sample Input

Input

5
3 2 -1 2 11

Output

4

Input

5
2 1 1 1 1

Output

1

 

///状态转移方程  f[i][j]=min(f[i-1][j]+abs(a[i]-b[j]),f[i][j-1])
///f[i][j]表示把前i个数变成递增序列，第i个数变成b[j]的最少步数。
///为什么把序列中的数变成原序列中的数能得到最优解，这个猜想还不太明白的说。

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
long long  f[5005];
int a[5005],b[5005];
int main()
{
   int n;
   while(scanf("%d",&n)!=EOF)
   {
       for(int i=0;i<=n;i++)
       f[i]=0;
       for(int i=1;i<=n;i++)
       {
           scanf("%d",&a[i]);
           b[i]=a[i];
       }
       sort(b+1,b+1+n);
       for(int i=1;i<=n;i++)
       for(int j=1;j<=n;j++)
       {
           f[j]+=abs(a[i]-b[j]);
           if(j>1)
           f[j]=min(f[j-1],f[j]);
       }
       cout<<f[n]<<endl;
   }
}