Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

1 3 1 2 3
2 6 -1 4 -2 3 -2 3

Sample Output

6
8

Hint

Huge input, scanf and dynamic programming is recommended.

给定一个数组，求M段连续的子段和最大。dp[i][j]表示前i段选择第j个元素的最优解。

dp[i][j]=max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j

由于k<j，所以我们能够用两个一维数组，一个dp[i]记录当前行状态。一个d[i]记录下一行可选的最大值。

用64位会超时。但int能水过。。

</pre><pre name="code" class="cpp">/*
dp[i][j]=max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j
第j个元素选i段区间，对于当前元素a[j],能够把它接到第i段的第j-1位置构成一段。
或者单独成一段。

观察这个方程。对于dp[i][j-1]这一项，它和dp[i][j]同i,就是在一段。
用一维数组就能记录；
dp[i-1][k]是i-1段j位置前能取的最大值。我们能够在计算当前第i段时，
把dp[i-1][k]记录下来，取最大值就好了。
*/

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define ll __int64
const int inf=0x7fffffff;
#define N 1000010
int a[N];
int pre[N];        //即dp[i-1][k]。记录j位置前可选值
int dp[N];
int main()
{
    int i,j,n,m;
    int tmp,ans;
    while(~scanf("%d%d",&m,&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            dp[i]=0;
            pre[i]=0;
        }
        pre[0]=dp[0]=0;
        ans=-inf;
        for(i=1;i<=m;i++)
        {
            tmp=-inf;
            for(j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1]+a[j],pre[j-1]+a[j]);
                pre[j-1]=tmp;      //tmp记录的是当前段j位置的最大值，
                tmp=max(tmp,dp[j]);//而数组pre[]是记录j-1位置的，所以要先取tmp,再更新tmp
            }
        }
        for(i=m;i<=n;i++)
            ans=max(ans,dp[i]);
        printf("%d\n",ans);
    }
    return 0;
}