HDU - 1024 Max Sum Plus Plus（DP）

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 … Sx, … Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + … + Sj (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(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead.

Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 … Sn.
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

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1000010;
int a[N],f[N];
int main()
{
    
	int n,m;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
    
		memset(f,0,sizeof f);
		for(int i=1;i<=n;i++) scanf("%d",&a[i]);
		for(int i=1;i<=m;i++)
		{
    
			int sum=0;
			for(int j=1;j<=i;j++)
				sum+=a[j];
			f[n]=sum;
			for(int j=i+1;j<=n;j++)
			{
    
				sum=max(sum,f[j-1])+a[j];
				f[j-1]=f[n];
				f[n]=max(f[j-1],sum);
			}
		}
		printf("%d\n",f[n]);
	}
	return 0;
}