Problem Description

Now
I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a
brave ACMer, we always challenge ourselves to more difficult problems.
Now you are faced with a more difficult problem.

Given a consecutive number sequence S₁, S₂, S₃, S₄ ... S_x, ... S_n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S_x ≤ 32767). We define a function sum(i, j) = S_i + ... + S_j (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + sum(i_m, j_m) maximal (i_x ≤ i_y ≤ j_x or i_x ≤ j_y ≤ j_x is not allowed).

But
I`m lazy, I don't want to write a special-judge module, so you don't
have to output m pairs of i and j, just output the maximal summation of
sum(i_x, j_x)(1 ≤ x ≤ m) instead. ^_^

 

Input

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

 

Output

Output the maximal summation described above in one line.

 

Sample Input

 

Sample Output

 

没做出来。。只能想到二维的，滚动数组果然厉害.

/**题意：n个数字分成m段的最大和*/

///分析：dp[i][j] 代表以a[j]结尾的前j个数字被分成 i 段得到的最大和

///可以得到: 1.如果a[j]单独成段 dp[i][j] = dp[i-1][k] + a[j] 其中 1<=k<j 意思是前1- k 个数字组成了i-1段.

///          2.如果a[j]并入第i段 那么dp[i][j]=dp[i][j-1]+a[j] 分析可得 dp[i][j] = max(1,2)

///但是这题的条件是不允许这样做的 ，首先枚举 i , j ,k 的时间复杂度是 O(n^3) dp数组的空间要 O(n^2)，而数据量已经

///到达了 1000000 显然不允许.于是，这里就要用一个新的思想了:滚动数组.

///我们可以看到dp[i][j]只和是否包含a[j]相关，所以这里我们可以用两个一维数组存当前状态与前一个状态.

///新的状态: dp[j]表示以a[j]结尾的前i段的最大和,pre[j]表示前j个数组成前i段的最大和,不一定包括a[j]

///dp[j] = max(dp[j-1]+a[j],pre[j-1]+a[j])

#include<stdio.h>

#include<iostream>

#include<string.h>

#include<math.h>

#include<algorithm>

using namespace std;

const int N = ;

int a[N];

int dp[N];

int pre[N];

int main()

{

    int m,n;

    while(scanf("%d%d",&m,&n)!=EOF){

        for(int i=;i<=n;i++){

            scanf("%d",&a[i]);

            dp[i]=pre[i]=;

        }

        dp[]=pre[]=;

        int mam;

        for(int i=;i<=m;i++) {///枚举每一段

            mam = -0x7fffffff;

            for(int j=i;j<=n;j++){

                dp[j] = max(dp[j-]+a[j],pre[j-]+a[j]);

                pre[j-] = mam;  ///表示前 j-1 个数组成i段能够表示的最大和

                mam=max(dp[j],mam);

            }

        }

        printf("%d\n",mam);

    }

    return ;

}