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, S ₂, 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


        
  

Hint

Huge input, scanf and dynamic programming is recommended.

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=1e6+10;
int f[N],pre[N],a[N];
//f 记录用j个
//pre 记录用 j-1个
 int main(){
	int n,m;
	//freopen("in.txt","r",stdin);
	while(~scanf("%d%d",&m,&n)){
        memset(f,0,sizeof(f));
        memset(pre,0,sizeof(pre));
        for(int i=1;i<=n;i++)
            cin>>a[i];
        int t;int i;
        for(int j=1;j<=m;j++)
        for(t=-0x7fffffff,i=j;i<=n;i++){
            f[i]=max(f[i-1],pre[i-1])+a[i];
            pre[i-1]=t;
            t=max(t,f[i]);
        }
        printf("%d\n",t);
	}
	//fclose(stdin);
}