Max Sum Plus Plus HDU - 1024

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, 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


        
  

Hint

Huge input, scanf and dynamic programming is recommended.

 

好久没有写过博客了，最近写的题比较少太水了，还要打蓝桥杯（贪心杯）和天梯赛 

天梯赛擦线省三，险之又险，说到底还是自己太菜，蓝桥就拿了个国赛的名额。

好吧，来说说这道题，相较于之前maxsum 的线性dp这个骚了点 划分了区间。类似与背包问题

dp方程的第一个元素的意思是(j-1个元素)分划为i组，最后一组加上num[j]的值             第二个元素的意思是（j-1个元素）分划为i-1组，最后一个元素独立成为i组

 

max数组是用来储存上一次分组最优的情况

#include
#include
#include 
using namespace std;
const int inf=0x7fffffff;
const int maxn=1e6+100;

int num[maxn];
int Max[maxn];
int dp[maxn];

int main()
{
	int m,n;
	while(~scanf("%d%d",&m,&n))
	{
		for(int i=1;i<=n;i++)
		scanf("%d",&num[i]);
		memset(Max,0,sizeof(Max));
		memset(dp,0,sizeof(dp));
		int maxx;
		for(int i=1;i<=m;i++)
		{
			maxx=-inf;
			for(int j=i;j<=n;j++)
			{
				dp[j]=max(dp[j-1]+num[j],Max[j-1]+num[j]);
				Max[j-1]= maxx;
				maxx=max(maxx,dp[j]);
			}
		}
		printf("%d\n",maxx);
	}
}