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(i₂, 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

 

和max sum的算法类似，加入一个维度，用来表示分成几组

第m组的第n个数计算方式为max（m组n-1，m-1组n前的最大值）+n的值

因为只和前一组有关，所以使用滚动数组

#define _CRT_SECURE_NO_WARNINGS


#include<cstdlib>
#include<iostream>
using namespace std;


#define x 1000000

int max(int a, int b) {
    return a > b ? a : b;
}

int main()
{
    int m, n;
    while (~scanf("%d%d", &m, &n)) {
        int a[x + 1];
        int dp[x + 1] = { 0 };
        int mx[x + 1] = { 0 };
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        int t = -99999;
        for (int i = 1; i <= m; i++) {
            t = -99999999;
            for (int j = i; j <= n; j++) {
                dp[j] = max(dp[j - 1], mx[j - 1]) + a[j];
                mx[j - 1] = t;
                t = max(t, dp[j]);
            }
        }

        printf("%d\n", t);

    }



    return 0;
}