Max Sum Plus Plus

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 S ₁, S ₂, S ₃, S ₄ ... 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(i ₁, j ₁) + sum(i ₂, j ₂) + sum(i ₃, j ₃) + ... + 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 S ₁, S₂, S ₃ ... 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

 题目大意：从n个数中取m段和最大

dp[i][j]表示从前i个数中取j段（第i个数一定取), 所以i可以单独一段，也可以和前面构成连续的一段。推出状态转移方程dp[i][j]=max(dp[i-1][j]+a[i],dp[k][j-1]+a[i]) ,k的范围为j-1到i-1

但是这道题n和m比较大，二维的状态显然不行，所以需要降维。

#include <bits/stdc++.h>
using namespace std;
int a[100005],pre[100005],dp[100005];  
int main()
{
    int n,m,k;
    ios::sync_with_stdio(false);
    while(cin>>m>>n)
    {
        for(int i=1;i<=n;i++) cin>>a[i];
        memset(pre,0,sizeof(pre));
        memset(dp,0,sizeof(dp));
        int maxx;
        for(int i=1;i<=m;i++)   //分成i段 
        {
            maxx=-99999999;
            for(int j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1],pre[j-1])+a[j];    //取到第j个元素时,可以分成i段的最大值 
                pre[j-1]=maxx;     //用前j-1个元素分成i段的最大值 
                maxx=max(maxx,dp[j]);
            }
        }
        cout<<maxx<<endl;
    }
}