Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18190 Accepted Submission(s): 5955

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.

Author
JGShining（极光炫影）

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Statistic | Submit | Discuss | Note

数据好像很弱 O（n^2）能过

dp[i][j][0] 表示前i个数组成j段，且不选入第i个数
dp[i][j][1]表示前i个数组成j段，且选入第i个数

dp[i][j][0] = max (dp[i - 1][j][1], dp[i - 1][j][0])
dp[i][j][1] = max (dp[i - 1][j][1], dp[i - 1][j - 1][0], dp[i - 1][j - 1][1]) + a[i];

对于j=1的情况，就是最大子段和
数据有点大，用下滚动数组

/*************************************************************************
    > File Name: dp14.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月13日 星期五 20时39分01秒
 ************************************************************************/

#include