Max Sum Plus Plus HDU - 1024

题目：

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.

/*******************************************
 *    dp挑战一     相信自己！！ 加油
 *  2018/8/15    15:45 
 *******************************************/
 
/*******************************************
 * https://www.cnblogs.com/dongsheng/archive/2013/05/28/3104629.html
 * 思路出处，发现这并不是一道简单题呀 注意以num[j]结尾 
 *******************************************/
 
#include<cstdio>
#include<string.h>
#include<algorithm>
int num[1000005];
int pre_max[1000005];
using namespace std;
int main(){
    int n,m;
    while(~scanf("%d%d",&m,&n)){
        for(int i=1;i<=n;i++){
            scanf("%d",&num[i]);
            pre_max[i]=0;
        }
        
        //dp
        for(int i=1;i<=m;i++){
            int temp=0;
            for(int k=1;k<=i;k++){
                temp+=num[k];
            }
            pre_max[n]=temp;
            for(int j=i+1;j<=n;j++){
                temp=max(pre_max[j-1],temp)+num[j];
                pre_max[j-1]=pre_max[n];
                pre_max[n]=max(pre_max[n],temp);
            }
        }
        printf("%d\n",pre_max[n]);
    } 
}