CF 433C Ryouko's Memory Note

CF 433C  Ryouko's Memory Note

C. Ryouko's Memory Note

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.

Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.

Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from pagex to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is .

Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.

Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.

Input

The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).

The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).

Output

Print a single integer — the minimum number of pages Ryouko needs to turn.

Sample test(s)

input

4 61 2 3 4 3 2

output

3

input

10 59 4 3 8 8

output

6

Note

In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.

In the second sample, optimal solution is achieved by merging page 9 to 4.

执行一次操作 将相邻的纸上的信息完全复制到另一张纸上。。来减少翻得书的页数 

用vector来存储  相邻的数  排除相邻的数相等的情况  记录下当前所需翻的页数

将相邻的页数排序 取出中位数 替代当前的数  与之前所需要翻得页数进行比较 取最小值

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#define eps 1e-8#define op operator#define MOD  10009#define MAXN  100100#define FOR(i,a,b)  for(int i=a;i<=b;i++)#define FOV(i,a,b)  for(int i=a;i>=b;i--)#define REP(i,a,b)  for(int i=a;i<b;i++)#define REV(i,a,b)  for(int i=a-1;i>=b;i--)#define MEM(a,x)    memset(a,x,sizeof a)#define ll __int64using namespace std;int a[MAXN];vector<int> v[MAXN];int main(){//freopen("ceshi.txt","r",stdin);    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        int mx=0;        for(int i=0;i<m;i++)        {            scanf("%d",&a[i]);            mx=max(mx,a[i]);        }        ll sum=0;        for(int i=1;i<m;i++)        {            if(a[i]==a[i-1])  continue;            v[a[i]].push_back(a[i-1]);            v[a[i-1]].push_back(a[i]);            sum+=(abs(a[i]-a[i-1]));        }        ll mi=sum;//    cout<<sum<<endl;        vector<int> x;        for(int i=1;i<=mx;i++)        {            ll temp=sum;            if(v[i].size()==0)  continue;            x=v[i];            sort(x.begin(),x.end());            int y=x[x.size()/2];//被替换之后的数            for(int j=0;j<x.size();j++)            {                temp+=(abs(x[j]-y)-abs(i-x[j]));            }            mi=min(mi,temp);        }        printf("%I64d\n",mi);    }    return 0;}