hdu 1024Max Sum Plus Plus

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 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

题目是看懂了，但代码还没理解！！

#include
using namespace std;
#define max(a,b) a>b?a:b
#define MAXN 1000000
int num[MAXN+50],now[MAXN+50],pre[MAXN+50];
int main()
{
 int m,n,i,j,k,ans,max_pre;
    while(cin>>m>>n)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&num[i]);                    //数据太多所以用scanf 
        memset(now,0,sizeof(now));
        memset(pre,0,sizeof(pre));
        for(i=1;i<=m;i++)
        {
   max_pre=-99999999;
            for(j=i;j<=n;j++)
            {
    
                now[j]=max(now[j-1]+num[j],pre[j-1]+num[j]);
                pre[j-1]=max_pre;
                if(now[j]>max_pre)
                    max_pre=now[j];
             //cout<   }
   cout<  }
        cout<    }
 
    return 0;
}