Max Sum Plus Plus

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, S₃, 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 _xis 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.

Examples

Input

1 3 1 2 3
2 6 -1 4 -2 3 -2 3

Output

6
8

正确解法：

dp太难了，妈妈救我。

题目的意思是，在n个数字中找到m个不相交的序列，求这序列和最大。

设前i个序列中最后一位数字的下标为j的序列和为 f[i][j]

f[i][j]=max(f[i][j-1],f[i-1][k])+a[j]; （i-1<=k< j）

在找 f[i][j] 时，最后一位数字下标为j时，要么这个数字跟着前一个序列 f[i][j-1]+a[j]，要么这个数字新开一个序列 找一个比 i-1大的 比 j 小的序列的最大数加上a[j]。

原文地址：https://www.cnblogs.com/Kaike/p/10008784.html