HDU 1024

原题：http://acm.hdu.edu.cn/showproblem.php?pid=1024
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

Hint

Huge input, scanf and dynamic programming is recommended.
AC代码：

#include <iostream>
#include<cstdio>
#include<memory.h>
#define maxn 1000005
#define maxm 100
#define inf 0x7ffffff
using namespace std;
int num[maxn];//存放输入的数据 
int d[maxn];//定义一个记录动态的数组 
int pre[maxn];
int main()
{

    int m,n,tmp;
    while(cin>>m>>n){
        int tmp;
        for(int i=1;i<=n;++i){
            cin>>num[i];
        }
        memset(d,0,sizeof(d));
        memset(pre,0,sizeof(pre));
        for(int i=1;i<=m;++i){
            tmp=-inf;
            for(int j=i;j<=n;++j){
                d[j]=max(d[j-1],pre[j-1])+num[j];//记录状态
                pre[j-1]=tmp;//记录j-1之前的最大和 
                tmp=max(tmp,d[j]);//每次选出最大的数 更新 
            }//当前状态只与前一状态有关
        }
        cout<<tmp<<endl;
    }
 
    return 0;
}